2. Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line 3 x + y + 2 = 0.

 2. Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line [sqrt(3)] x + y + 2 = 0.

[sqrt(3)] x + y + 2 = 0

[sqrt(3)] x + y = (-2)  --------------(1)

we require the RHS to be no negative.

so multiply by (-1)

-[sqrt(3)]x +(-1)y = 2

a = { -[sqrt(3)] } , b = (-1)

sqrt [(a^2) +(b^2)] = sqrt[3+1]  =sqrt[4] = 2

so divide each term in (1) with 2

 [ - {sqrt(3)} /2 ]x +(-1/2)y =1

use x cos θ + y sin θ = p 

to get 

cos θ =[ - {sqrt(3)} /2 ]

 sin θ = (-1/2)

so  θ is in the third quadrant

θ = pi + ( pi /6 ) =[7 pi/6]

using RHS

p =1

 Find the values of k for which the line 

(k–3) x – (4 –( k^ 2) ) y + (k^2) –7k + 6 = 0 is

(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.

slope of the given line is 

m = [k-3] / [ 4 –( k^ 2)  ]

(a)

When parallel to x axis

slope = 0

hence [k-3] / [ 4 –( k^ 2)  ] = 0

k-3 = 0

k = 3

(b) Parallel to the y-axis,

 

When parallel to y axis

slope = infinity

hence [k-3] / [ 4 –( k^ 2)  ] = infinity

hence  

[ 4 –( k^ 2)  ] = 0

( k^ 2) = 4

k = 2 or (-2)

 

(c) Passing through the origin.

when line passes through the origin (0,0) 

the (k–3) x – (4 –( k^ 2) ) y + (k^2) –7k + 6 = 0 is satisfied by (0,0)

 0 - 0 + (k^2) –7k + 6 = 0

(k^2) –7k + 6 = 0

[k-6][k-1]  = 0

k =6 , k = 1

=================================================

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

 

solution 

 21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

 

19. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

solution

18.Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror.

solution 

 

17. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle 

 solution

 14. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the 

line x + y = 4 ?

solution

12.Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes

solution

 

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

solution

 

8. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 

and x – k = 0

solution

9. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and
2x – y – 3 = 0 may intersect at one point.

 solution

6. Find the equation of the line parallel to y-axis and drawn through the point of
intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

solution

 

4. What are the points on the y-axis whose distance from the line
[x/3] + [y/4]=1 is 4 units.

solution

 

 3. Find the equations of the lines, which cut-off intercepts on the axes whose sum
and product are 1 and – 6, respectively.

solution

 2. Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line [sqrt(3)] x + y + 2 = 0.

solution 

 

 Find the values of k for which the line 

(k–3) x – (4 –( k^ 2) ) y + (k^2) –7k + 6 = 0 is

(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.

 solution 

ncert cbse chapter 9 sequences and series miscellaneous exercise

 

 32.

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on the second day. 4 more workers dropped out on the third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was finished

solution

 

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Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

 

 ncert cbse chapter 10 straight lines miscellaneous exercise

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3. 

let m be the slope of the required line

using formula slope = [(- coefficient of x) / (coefficient of y)]

Slope of  the given line [ x – 2y = 3 ] 

is m2 = [(-1)/(-2)] = [1/2]

using formula for angle between two lines

tanu = | [m1 - m2] / [ 1 + m1*m2]| 

given angle is 45 degrees

tan(45 degrees) = | [m - (1/2)] / [ 1 + m*(1/2)]| 

1  = | [2m-1] / [2+m] |

because of the absolute value sign

(+1) = [2m-1] / [2+m]  OR  (-1) = [2m-1] / [2+m] 

solving

2+m = 2m -1  or  -2-m = 2m-1

m=3  or m=(-1/3)

Required line passes through (3, 2) 

using point slope form

equation is

[y-2] =[3][x-3]  or [y-2] =[(-1/3)][x-3]

y-2 =3x-9   or 3y-6 =(- x)+3

3x-y-7=0  or x+3y-9=0

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

 

solution 

 21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

 

Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the
line to be a plane mirror.

solution 

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

solution

ncert cbse chapter 9 sequences and series miscellaneous exercise

 

 32.

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on the second day. 4 more workers dropped out on the third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was finished

solution

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.

 

A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

 

 ncert cbse chapter 10 straight lines miscellaneous exercise

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

Since A lies on the x axis, the y coordinate of A is 0

Let A =(k,0) 

let P=(1,2)

Q =(5,3)

both PA and AQ make equal angles with the normal line (perpendicular to the x axis) at A due to  the law of reflection.

If AQ makes an angle u with the positive direction of the x axis, then PA makes an angle of (180-u) with the positive direction of the x axis.

using the definition of slope

 

for PA, P=(1,2), A =(k,0)

tan(u) =[0-2] / [k-1] --------------------(1)

for QA , Q =(5,3) , A =(k,0)

tan[180-u]  = [0-3] / [k-5]

but tan[180-u]  = {-tan (u)}

we get 

{-tan (u)} = [0-3] / [k-5] ------------(2)

using eqn (1)  in  (2)

-{[0-2] / [k-1] } = [0-3] / [k-5]

2(k-5) = (-3)(k-1)

2k-10 = -3k+3

5k =13

k =[13/5]

therefore

A=( (13/5)  , 0 )

-----------------------------------------------------------------------------------------------------------------

21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

 

first line is

[ 9x + 6y – 7 = 0 ] dividing  by 3 to make the coefficients in the two given equations the same

3x +2y -(7/3) =0 -------------(1)

second line is  

 3x + 2y + 6 =0 --------------(2)

 

A line parallet to these lines will be of the form

3x+2y +k =0 ----------------(3) 

distance between (1) and (3)

d1 = | k +(7/3) | / sqrt[(3^2) +(2^2)]

distance between (2) and (3)

d2 = | k -6 | / sqrt[(3^2) +(2^2)]

 

for equidistant line d1 = d2

 equating d1 and d2 and cancelling off the denominator

| k +(7/3) | = | k -6 | 

k +(7/3) =  k -6   OR  k +(7/3) = -(k -6 )

(7/3) =(-6) which is not possible

OR 

k +(7/3) = -(k -6 )

2k =6 -(7/3) 

2k = (11/3)

k = (11/6)

substitute in (3)

3x+2y +(11/6) =0

 

18 x  + 12y +11 =0

 

 

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

 

solution 

 21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

ncert cbse chapter 9 sequences and series miscellaneous exercise

 

 32.

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on the second day. 4 more workers dropped out on the third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was finished

solution

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.

A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

We have to find the point of intersection of the lines

2x – 3y + 4 = 0

3x + 4y – 5 = 0  

[ 2x - 3y = (-4) ] x 3

[ 3x + 4y = 5 ] x 2

6x -9y = (-12)

6x +8y =10

subtracting

-17y =-22

 

y=(22/17)

substitute in 

3x + 4y = 5

3x = 5 -{88/17}=[-3/17]

x =(-1/17)

 

The person is standing at [ (-1/17) , (22/17) ] 

The shortest path is perpendicular to 6x – 7y + 8 = 0

Perpendicular line is 7x+6y+k=0

This passes through [ (-1/17) , (22/17) ] if

7 (-1/17) +6(22/17) +k =0

k = (-125/17)

Perpendicular line is 7x+6y+k=0

changes to  7x+6y+(-125/17)=0 

or 119x+102y-125=0 

119x+102y=125 is the required path.

 

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

 

ncert cbse chapter 9 sequences and series miscellaneous exercise

 

 32.

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on the second day. 4 more workers dropped out on the third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was finished

solution

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.

find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

 ncert cbse chapter 8 binomial theorem miscellaneous exercise 

 3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

using binomial expansion

 (1+2x)^6 = 1 +C(6,1)(2x)+C(6,2)(2x)^2+C(6,3)(2x)^3 +C(6,4)(2x)^4 

+C(6,5)(2x)^5+C(6,6)(2x)^6

 C(6,1) = 6 =C(6,5)

C(6,2)=15=C(6,4)

C(6,3)=20

C(6,6)=1

(1+2x)^6 =

 1 +6(2x) +15(4(x^2))+20(8(x^3))+15(16(x^4))+6(32(x^5))+1(32(x^6))

=1+12x+60(x^2)+160(x^3)+240(x^4)+192(x^5)+32(x^6)

 

(1-x)^7=

1-C(7,1)(x)+C(7,2)(x^2)-C(7,3)(x^3)+C(7,4)(x^4)

-C(7,5)(x^5)+C(7,6)(x^6)-C(7,7)(x^7) 

C(7,1)=7=C(7,6) 

C(7,2)=21=C(7,5) 

C(7,3)=35=C(7,4)  

C(7,7)=1

 (1-x)^7=1 -7x+21(x^2)-35(x^3)+35(x^4)-21(x^5)+7(x^6)-1(x^7)

 

now multiplying the two expansions and concentrate only on the terms containing

(x^5)

1* {-21(x^5)}+{+12x}*{+35(x^4}+{+60(x^2)}*{-35(x^3)}+{+160(x^3)}{+21(x^2)}+{+240(x^4)}{-7x}+{+192(x^5)}{1}

 

coefficient of  (x^5) is

1*(-21)+(12)*(35)+(60)(-35)+(160)(21)+(240)(-7) +192*1

=(-21)+420-2100+3360-1680+192

=171

 

chapter 8 binomial theorem miscellaneous exercise

 1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

 

2.  Find a if the coefficients of (x^2)  & (x^3) in the expansion of {(3+ax)^9} are equal 

solution

  3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

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The english alphabet has 5 vowels and 21 consonants. How many words with 2 different vowels and 2 different consonants can be formed

ncert cbse chapter 7 permutations miscellaneous exercise

6.The english alphabet has 5 vowels and 21 consonants. How many words with 2 different vowels and 2 different consonants can be formed

Because there are 5 vowels,

the 2 vowels can be selected in C(5,2) = [(5*4)/(1*2)] = 10 ways

Because there are 21 consonants

the 2 consonants can be selected in C(21,2) = [(21*20)/(1*2)] = 210 ways

After selecting the 4 required letters now we can arrange them in

(4!) = 4*3*2*1 = 24 ways

Required number of words = 10*210*24 =50400

7. In an examination  a question paper consists of two parts containing 5 and 7 questions respectively. A  student is required to answer 8 questions in all selecting atleast 3  from each part.  In how many ways can a student select the questions?

let the selections be denoted by (x,y) where x denotes the questions from part I

and y the questions from part II

atleast 3 means 3 or more

(3,5) or (4,4) or (5,3) are the different possibilities with their respective number of

possible selections being

C(5,3) * C(7,5) = [(5*4)/(1*2)]*[(7*6)/(1*2)]=210 ways

C(5,4) * C(7,4) = [5]*[(7*6*5)/(1*2*3)]=175 ways

C(5,5) * C(7,3) = [1]*[(7*6*5)/(1*2*3)]=35 ways

Total number of ways = 210+175+35 =420 ways

Here note that C[n,r] =C[n,(n-r)]

gives

C[5,3]=C[5,2]

C[7,5]=C[7,2]

C[5.4]=C[5,1]=5

C[7,4]=C[7,3]

ncert cbse 11th mathematics chapter  7 permutations exercise 7.3

11.In how many ways can the letters of the word PERMUTATIONS be arranged if

i)words start with P and end with S

ii)vowels are all together

iii)there are always 4 letters between P and S

solution

ncert cbse 11th mathematics chapter  7 permutations miscellaneous exercise

1.How many words each of 2 vowels and 3 consonants can be formed using the letters of the word DAUGHTER

solution

2. How many words of can be formed using all the letters of the word

 EQUATION  so that the vowels and consonants occur together?

solution

4. If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

solution

6.How many 6 digit numbers can be formed using 0,1,3,5,7,9 which are divisible by 10 and no digit is repeated?

solution

6.The english alphabet has 5 vowels and 21 consonants. How many words with 2 different vowels and 2 different consonants can be formed

solution

7. In an examination  a question paper consists of two parts containing 5 and 7 questions respectively. A  student is required to answer 8 questions in all selecting atleast 3  from each part.  In how many ways can a student select the questions?

solution

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

ncert cbse chapter7permutations miscellaneous exercise

4. If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

first arrange the 11 letters in alphabetical order

AA   2 A's

E

II  2 I's

M

NN 2 N's

O

T

X

Therefore before the first word starting with E, words starting with the letter

A will be arranged in alphabetical order in the dictionary.

fix one A in the first slot

remaining 10 letters contain 2 I's and 2 N's [ one of the 2 A's is used for the first slot ]

So number of words starting with the letter A

= [(10!)/ {(2!)*(2!)}] =907200

6.How many 6 digit numbers can be formed using 0,1,3,5,7,9 which are divisible by 10 and no digit is repeated?

The one's place should contain 0 for the number to be divisible by 10

Fix the digit 0 in the one's place Therefore the one's place can be filled in only one way.

The other 5 slots can be filled in 5! = 120 ways

Answer = 1*120 = 120

ncert cbse 11th mathematics chapter  7 permutations exercise 7.3

11.In how many ways can the letters of the word PERMUTATIONS be arranged if

i)words start with P and end with S

ii)vowels are all together

iii)there are always 4 letters between P and S

solution

ncert cbse 11th mathematics chapter  7 permutations miscellaneous exercise

1.How many words each of 2 vowels and 3 consonants can be formed using the letters of the word DAUGHTER

solution

2. How many words of can be formed using all the letters of the word

 EQUATION  so that the vowels and consonants occur together?

solution

4. If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

solution

6.How many 6 digit numbers can be formed using 0,1,3,5,7,9 which are divisible by 10 and no digit is repeated?

solution

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

ncert cbse 11th maths excercise 7.3 chapter 7

ncert cbse 11th maths  excercise 7..3 chapter 7

exercise 7.3

11.In how many ways can the letters of the word PERMUTATIONS be arranged if

i)words start with P and end with S

ii)vowels are all together

iii)there are always 4 letters between P and S

There are 12 letters in the word PERMUTATIONS

i)words start with P and end with S

If the word start with P and end with S

first slot and 12th slot can be filled in 1 way each.

After removing P and S,

For the remaining 10 slots there are 10 items

with  2 T's,

So number of words starting with P and end with S

= 1 *[(10!)/(2!)]*1 = 1814400

ii)vowels are all together

vowels in the given word are E,U,A,I,O

If the vowels are together, think of the 5 vowels as a single block.

Inside the block the vowels can be arranged in (5!) ways

Now this block of vowels with the remaining 7 consonants

make up 7+1=8 pieces with 2 T's

which can be arranged in [(8!)/(2!)] ways

Number of words in which the vowels are together = (5!)*[(8!)/(2!)]

=120*20160=2419200

iii)there are always 4 letters between P and S

When there are 4 letters between P and S

one particular arrangement will be

P x x x x S # # # #

in this example first and fifth slots can be filled in 1 way each

and the remaining 10 slots can be filled with 10 remaining letters

containing 2 T's in = 1 *[(10!)/(2!)]*1 = 1814400 ways

Now possible re-arrangements  are

1        P x x x x S # # # ###

2        #P x x x x S # # # ##

3         ##P x x x x S # # ##

4        ###P x x x x S # ##

5        ####P x x x x S##

6        #####P x x x x S#

7        ######P x x x x S

Now the position of P and S can be interchanged

8       S x x x x P # # # ###

9        #S x x x x P # # # ##

10         ##S x x x x P # # ##

11        ###S x x x x P # ##

12        ####S x x x x P##

13        #####S x x x x P#

14        ######Sx x x x P

A total of 14 arrangements

number of words where

there are always 4 letters between P and S

=1814400*14

=25401600

 

ncert cbse 11th mathematics chapter  7 permutations exercise 7.3

11.In how many ways can the letters of the word PERMUTATIONS be arranged if

i)words start with P and end with S

ii)vowels are all together

iii)there are always 4 letters between P and S

solution

ncert cbse 11th mathematics chapter  7 permutations miscellaneous exercise

1.How many words each of 2 vowels and 3 consonants can be formed using the letters of the word DAUGHTER

solution

2. How many words of can be formed using all the letters of the word

 EQUATION  so that the vowels and consonants occur together?

solution

4. If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

solution

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

tan15 degrees and sin 75 degrees

tan15 degrees and sin 75 degrees

11th cbse ncert trigonometry

5.Find the values of sin 75degrees and tan 15degrees

using trigonometry formula trigonometry identities

sin(A+B) =sinAcosB+cosAsinB


choose A=45degrees B=30degrees

sin(75degrees)
sin(45+30) =sin45cos30+cos45sin30 ( angles in degree )

=[1/(sqrt(2)][sqrt(3)/2] +[1/(sqrt(2)][1/2]

= {[sqrt(3) +1} / {2*sqrt(2)}

using trigonometry formula trigonometry identities 

tan(A-B) = [tanA - tanB] / [1 + tanAtanB]

choose A=45degrees B=30degrees

tan 15degrees

tan[45-30] = [tan45 -tan30]  / [1+tan45tan30] ( angles in degree )

=[1-{1/sqrt(3)}] / [1+(1){1/sqrt(3)}]

=[ sqrt(3) -1 ] / [sqrt(3)} + 1] introduce conjugate

={ [ sqrt(3) -1 ]^2 } / { [sqrt(3) + 1] [ sqrt(3) -1 ] }

={ [ sqrt(3) -1 ]^2 } / { [sqrt(3)}^2 - 1] }
 using identities (a-b)^2 and (a+b)(a-b)
={ [sqrt(3)]^2 - 2*sqrt(3) + 1 } / {3 - 1}

={3-2*sqrt(3) +1} / [2 ]

= [ 4-2*sqrt(3) ] / 2

={ 2 * [2- sqrt(3)]} / 2

= [2 - sqrt(3)]

3.3

5.Find the values of sin 75degrees and tan 15degrees

solution

6. prove that cos[(pi/4)-x]cos[(pi/4)-y]- sin[(pi/4)-x]sin[(pi/4)-y] = sin(x+y)
 solution

10. prove that sin[(n+1)x]sin[(n+2)x] +cos[(n+1)x]cos[(n+2)x] =cosx
 solution

11. prove that cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx

solution

12.(sin6x)^2 - (sin4x)^2 = sin2x sin10x

solution

13.(cos2x)^2  - (cos6x)^2 = sin4x sin8x
 solution

14. Prove that sin2x+2sin4x+sin6x = 4[(cosx)^2]sin4x
solution

15.prove that cot4x[sin5x+sin3x]=cotx[sin5x-sin3x]
 solution

16.Prove that [cos9x -cos5x] / [sin17x - sin3x ] = -sin2x / cos10x
solution

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
solution

18. Prove that [sinx -siny] / [cosx +cosy] = tan[(x-y)/2]
solution

19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x
solution

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution


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trigonometry problem on ncert cbse 11th miscellaneous

trigonometry problem on ncert cbse 11th miscellaneous exercise

Prove that 2cos(pi/13)cos(9pi/13)+cos (3pi/13)+cos(5pi/13) = 0

using the formula 2cosxcosy = cos(x+y)+cos(x-y)
trigonometry identities

LHS= cos[(pi/13) +(9pi/13)]+cos[(pi/13) -(9pi/13)]+cos (3pi/13)+cos(5pi/13)

= cos (10pi/13) + cos(-8pi/13) +cos (3pi/13)+cos(5pi/13)

= cos (10pi/13) + cos(8pi/13) +cos (3pi/13)+cos(5pi/13) because cos(-x)=cosx

= cos[pi - (3pi/13)] +cos[pi - (5pi/13)]+cos (3pi/13)+cos(5pi/13)

= -cos (3pi/13)- cos(5pi/13)+cos (3pi/13)+cos(5pi/13) because cos(pi-x)= -cosx

= 0


1.Prove that 2cos(pi/13)cos(9pi/13)+cos (3pi/13)+cos(5pi/13) = 0
solution 

2. Prove that ( sin3x + sinx ) sinx + (cos3x - cosx) cosx = 0

solution

3. Prove that (cosx +cosy)^2 + ( sinx - siny )^2 = 4 { cos[(x+y)/2] }^2

solution

4. Prove that (cosx - cosy)^2 + ( sinx - siny )^2 = 4 { sin[(x-y)/2] }^2

solution 



5.Show that sinx +sin3x+ sin5x +sin7x = 4cosx cos2x sin 4x

solution



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positive determinant from miscellaneous probability questions ncert cbse 12th mathematics

positive determinant and electronic assembly with two subsystems question from miscellaneous probability questions ncert cbse 12th mathematics

If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive?

There are 4 entries in the second order determinant ,each of which can be filled in two ways with either 0 or 1

Such determinants can be constructed in 2⁴ ways ( (2^4) ways)

Therefore if S is the sample space n(S) = (2^4) = 16

Let E be the event that the value of the selected determinant is positive. 

Required probability = P(E) = [n(E)] / [n(S)] = 3 /16

=======================================

An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known

P(A fails) = 0.2 P(B fails alone) = 0.15 P(A and B fail) = 0.15

Evaluate the following probabilities P(A fails|B has failed) ; P(A fails alone)

P(B fails) =P(A and B fail together ) + P(B fails alone) = 0.15+0.15 = 0.3

P(A fails|B has failed) = [P(A and B fail)] / [P(B fails ) ] = [0.15 / 0.30 =(1/2) = 0.5

P(A fails alone) = P(A fails) - P(A and B fail together ) = 0.2 – 0.15 = 0.05

link to index of other miscellaneous problems on probability of cbse ncert 12th mathematics

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index of miscellaneous problems on probablity

index of miscellaneous problems on probablity for cbse xii mathematics

problem 2
A couple has two children, Find the probability that both children are males, if it is known that at least one of the children is male.Also find the probability that both children are females, if it is known that the elder child is a female.
solution to miscellaneous problem 2 on probability of cbse class 12 mathematics

problem 3
Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.
solution to miscellaneous problem 3 on probability of cbse class 12 mathematics

problem  4
Suppose that 90% of people are right handed. What is the probability that at most 6 of a random sample of 10 people are right handed?
solution to miscellaneous problem 4 on probability of cbse class 12 mathematics

problem

In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is (5/6). What is the probability that he will knock down fewer than 2 hurdles?

solution to hurdles problem of cbse class 12th probability 

problem

If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive?

solution to positive determinant question from miscellaneous probability problem in cbse ncert xii mathematics

problem

An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known

P(A fails) = 0.2 P(B fails alone) = 0.15 P(A and B fail) = 0.15

Evaluate the following probabilities P(A fails|B has failed) ; P(A fails alone)

 answer to electronic assembly question from miscellaneous probability problem in cbse ncert xii mathematics

problem

How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

finding the value of n in binomial distribution given the probability question for cbse ncert 12th mathematics

problem
Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of a certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?
solution to miscellaneous problem  on probability of cbse class 12 mathematics


problem

Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls.One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black
solution to miscellaneous problem  on probability of cbse class 12 mathematics
.

index of more problems on baye's theorem for ncert cbse xii mathematics

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miscellaneous problem 3 on bayes theorem of probability

miscellaneous problem 3 on baye's theorem on probability for class xii of cbse , ncert, mathematics

Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

Let E1 be the event that a male is selected.
Let E2 be the event that a female is selected.

Let A be the event that the selected person has grey hair,

Given that there are equal number of males and females

P(E1) = (1/2)
P(E2) = (1/2)

P(A/E1) = (5/100)  [given 5% of men have grey hair ]
P(A/E2) = (0.25/100)  [given 0.25% of women have grey hair ]

Required probability = P[ selected person is a male given that selected person has grey hair]

Required probability = p[E1/A]

                          P( E1 ) P( A / E1 )
P ( E1 / A ) = ---------------------------------------------
                P( E1 ) P( A / E1 ) +P( E2 ) P( A / E2 )


P ( E1 / A ) = [(1/2)(5/100)] / {[(1/2)(5/100)] + [(1/2)(0.25/100)]}

P ( E1 / A ) =  [5] / [5+0.25] = ( 100 / 105 ) = (20/21)

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index of more problems on baye's theorem for ncert cbse xii mathematics
In case the above is illegible use the following picture

miscellaneous problem 2 on conditional probability

miscellaneous problem 2 on conditional probability


A couple has two children, Find the probability that both children are males, if it is known that at least one of the children is male.Also find the probability that both children are females, if it is known that the elder child is a female.

Let (x,y) denote the childre, where x stands for the elder child and y stands for the younger child.

Let M stand for male child and F stand for female child.

Sample space S = { (M,M) (M,F), (F,F),(F,M) }

let A denote the event that both children are males.
let B denote the event that at least one is a male.

A = {(M,M)}
B = {(M,M) (M,F),(F,M) }

interesection of A and  B, A ∩ B = {(M,M)}

P[ A ? B ] = (1 / 4)

P[B]  = (3 /4)

P [ both children are males given that at least one of the childre is a male ] = P [ A / B ]

                    P[ A ∩ B ]
P[ A / B ] = -------------
                       P[B] 

P[ A / B ] = (1/4) / (3 /4)

P[ A / B ] = ( 1/3 )

let E denote the event that both children are females.
let F denote the event that the elder child is a  female.

E = { (F,F) }
F = {(F,F),(F,M)}

interesection of E and  F, E ? F = {(F,F)}

P[E ∩ F] = ( 1/4 )
P[F] = ( 2/4 )

P[ both children are females given that the elder child is a female ] = P[ E/F ]


                   P[ E ∩ F ]
P[ E/F ] =  -------------------
                       P[F]


P[ E/F ] = ( 1/4 ) / ( 2/4 )  = ( 1/2 )

index of more problems on baye's theorem for ncert cbse xii mathematics
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problem 10 and problem 11 of bayes theorem for ncert bayes theorem

Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Let E1 be the event that the girl threw 1 or2 or 3,or 4 with the dice .

let E2 be the event that the girl threw 5 or 6 with the die.

let A be the event that the girl obtained exactly one head .


P( E1 ) = ( 4 / 6 ) { the girl threw 1 or2 or 3,or 4 with the dice }

P( E2 ) = ( 2 / 6 ) { the girl threw 5 or 6 with the die }

P( A / E1 ) = ( 2 / 4 ) [ {HT,TH} out of {HT,TH,TT,HH}]

P( A / E2 ) = ( 3 / 8 ) [ {HTT,THT,TTH} out of {HTT,THT,TTH,HHH,TTT,THH,HHT,HTH}]


Required probability = P [ she threw 1, 2, 3 or 4 with the die given that she obtained exactly one head ]

Required probability = P [ E1 / A ]




P ( E1 / A ) = [( 4/6)( 2/4)] / { [( 4/6)( 2/4) ] + [( 2 / 6 )( 3 / 8 )] }


P ( E1 / A ) =  [ (1/3) ] / {[(1/3)]+[1/8]} =  ( 8 / 11 )
index of more problems on baye's theorem for ncert cbse mathematics

problem 11
A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

Let E1 be the event that a selected item was produced by A .

let E2 be the event that a selected item was produced by B.

let E3 be the event that a selected item was produced by C.

let A be the event that the selected item was defective .


P( E1 ) = ( 50 / 100 )

P( E2 ) = ( 30 / 100 )

P( E3 ) = ( 20 / 100 )

P( A / E1 ) = ( 1 / 100 )

P( A / E2 ) = ( 5 / 100 )

P( A / E3 ) = ( 7 / 100 )


Required probability = P [ item was produced by A given that the item is defective ]

Required probability = P [ E1 / A ]


                  
P ( E1 / A ) = [( 50 / 100 )( 1 / 100 )] / { [(50/100)(1/100)]+[(30/100)(5/100)] +[(20/100)(7/100)] }


P ( E1 / A ) =  [ 50 ] / {[50]+[150]+[140]} =  ( 5 / 34 )

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bayes theorem problems 8 and 9 for cbse ncert

problem 8

A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further,2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

Let E1 be the event that the chosen item was produced by machine A .

let E2 be the event that  chosen item was produced by machine B.

let A be the event that the  item is defective .


P( E1 ) = ( 60 / 100 )

P( E2 ) = ( 40/ 100 )

P( A / E1 ) = ( 2 / 100 )

P( A / E2 ) = ( 1 / 100 )


Required probability = P [ item was produced by machine B given that the item was defective ]

Required probability = P [ E2 / A ]






P ( E2 / A ) = [( 40 / 100 )( 1 / 100 )] / { [( 60 / 100 )( 2 / 100 )] + [( 40 / 100 )( 1 / 100 )] }


P ( E2 / A ) =  [40] / {[120]+[40]} =  ( 1 / 4 )
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problem 9

Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Let E1 be the event that the first group wins .

let E2 be the event that the second group wins.

let A be the event that the  new product was introduced .


P( E1 ) = ( 0.6 )

P( E2 ) = ( 0.4 )

P( A / E1 ) = ( 0.7 )

P( A / E2 ) = ( 0.3 )


Required probability = P [ second group had won given that the new product was introduced ]

Required probability = P [ E2 / A ]


    



P ( E2 / A ) = [( 0.4 )( 0.3)] / { [( 0.6 )( 0.7 )] + [( 0.4 )( 0.3 )] }


P ( E2 / A ) =  [0.12] / {[0.42]+[0.12]} =  ( 12 / 54 ) = ( 2 / 9 )


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problem 6 and problem 7 of bayes theorem

There are three coins. One is a two headed coin (having head on both faces),another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ?


Let E1 be the event that the selected coin is the two headed coin .

let E2 be the event that the selected coin is the biased coin that comes up heads 75% of the time.

let E3 be the event that the selected coin is the unbiased coin.

let A be the event that the  toss of the selected coin resulted in a head.

Assuming E1 , E2 , E3 are equally likely

P( E1 ) = ( 1 / 3 )

P( E2 ) = ( 1 / 3 )

P( E3 ) = ( 1 / 3 )

P( A / E1 ) = 1 { since the coin is two headed  in the event of E1 }

P( A / E2 ) = ( 75 / 100 ) = ( 3/4 )  { since the biased coin comes up heads 75% of the time in E2 }

P( A / E3 ) = ( 1 / 2 )  { since the  coin is unbiased in E3 }

Required probability = P [ a person that the two headed coin was selected given that the toss resulted in a head ]

Required probability = P [ E1 / A ]


              



P ( E1 / A ) = [ ( 1 / 3 )( 1 )] / { [( 1 / 3 )( 1 )] + [( 1 / 3 )( 3/4 ) ] + [( 1 / 3 )( 1/2 ) ] }


P ( E1 / A ) =  [1] / { [1]+[3/4]+[1/2] } =  ( 4 /9 )

index of more problems on baye's theorem for ncert cbse mathematics

problem 7

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Let E1 be the event that the person is a scooter driver .

let E2 be the event that  the person is a car driver.

let E3 be the event that the person is a truck driver.

let A be the event that the  person met with an accident.

total number of vehicles = 2000 + 4000 + 6000 = 12000

P( E1 ) = ( 2000 / 12000 )

P( E2 ) = ( 4000 / 12000 )

P( E3 ) = ( 6000 / 12000 )

P( A / E1 ) = 0.01

P( A / E2 ) = 0.03

P( A / E3 ) = 0.15

Required probability = P [ person is a scooter driver given that the person met with an accident ]

Required probability = P [ E1 / A ]


         


P ( E1 / A ) = [(2000 / 12000 ) ( 0.01 )] / { [(2000 / 12000) (0.01)] + [(4000 / 12000) (0.03)] + [(6000 / 12000) (0.15)] }


P ( E1 / A ) =  [20] / { [20]+[120]+[900] } =  ( 20 / 1040 ) = ( 1 / 52 )

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problem 4 and 5 of bayes theorem

problem 4

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let ( 3/4 ) be the probability that he knows the answer and ( 1/4) be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability ( 1/4 ). What is the probability that the student knows the answer given that he answered it correctly?


Let E1 be the event that the  student knows the answer.

let E2 be the event that the student guesses the answer.

let A be the event that the answer is correct.

P( E1 ) = ( 3 / 4 )

P( E2 ) = ( 1 / 4 )

P( A / E1 ) = 1 [ E1  means student knows the answer and hence the answer will be correct ]

P( A / E2 ) = ( 1 / 4 )  [ given in the question  ]

Required probability = P [ student knows the answer given that he answered it correctly ]

Required probability = P [ E1 / A ]


 


P ( E1 / A ) = [ ( 3 / 4 )( 1 )] / { [ ( 3 / 4 )( 1 ) ] + [ ( 1 / 4 )( 1 / 4 ) ] }


P ( E1 / A ) =  [3] / { 3 + ( 1/4 )} =  ( 12 / 13 )

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problem 5

A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive ?


Let E1 be the event that the person has the disease .

let E2 be the event that the person does not have the disease.

let A be the event that the test result of the person is positive.

P( E1 ) = ( 0.1 / 100 ) [0.1 percent of the population actually has the disease]

P( E2 ) = ( 99.9 / 100 ) [ complement, 100 - 0.1 = 99.9 % of the population does not have the disease ]

P( A / E1 ) = ( 99 / 100 ) [ test is 99% effective in detecting a certain disease when it is in fact, present ]

P( A / E2 ) = ( 0.5 / 100 )  [ the test also yields a false positive result for 0.5% of the healthy person tested ]

Required probability = P [ a person has the disease given that his test result is positive ]

Required probability = P [ E1 / A ]





P ( E1 / A ) = [ ( 0.1 / 100 )( 99 / 100 )] / { [( 0.1 / 100 )( 99 / 100 )] + [ ( 99.9 / 100 )( 0.5 / 100 ) ] }


P ( E1 / A ) =  [ 9.9 ] / { 9.9 + 49.95 } =  ( 9.9 / 59.85 ) = ( 990 / 5985) = ( 22 / 133 )


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bayes theorem problem 2 and 3

A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Let E1 be the event that the first bag is selected.

let E2 be the event that the second bag is selected.

let A be the event of drawing a red ball from the selected bag.

We can assume that E1 and E2 are equally likely so that

P( E1 ) = 1 / 2

P( E2 ) = 1 / 2

P( A / E1 ) = 4 / ( 4 + 4 )  = 4 / 8 [ 4 red balls out of a total of 4 + 4 = 8 balls in the first bag ]

P( A / E2 ) = 2 / ( 2 + 6 ) = 2 / 8  [ 2 red balls out of a total of 2 + 6 = 8 balls  in the second bag]

By bayes theorem ,

 probability that the ball is drawn from the first bag = P ( E1 / A )

                         
               



P ( E1 / A ) =  { (1/2)(4/8) } / { (1/2)(4/8) + (1/2)(2/8) }


P ( E1 / A ) = 4 / 6 =  2 / 3

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problem 3

Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostelier?


Let E1 be the event that the chosen student is a hostelier.

let E2 be the event that the chosen student is a day scholar.

let A be the event that the chosen student has a A grade in the annual exam.

P( E1 ) = 60 / 100

P( E2 ) = 40 / 100

P( A / E1 ) = 30 / 100 [ Previous year results report that 30% of all students who reside in hostel attain A grade ]

P( A / E2 ) = 20 / 100  [ Previous year results report that 20% of day scholars attain A grade  ]

Required probability = P [ chosen student is a hostelier given that he has an A grade ]

Required probability = P [ E1 / A ]

  



P ( E1 / A ) = [ (60 / 100 ) ( 30 / 100 )] / { [ (60 / 100 ) ( 30 / 100 ) ] + [ (40 / 100 ) ( 20 / 100 ) ] }


P ( E1 / A ) = ( 18 / { 18 + 8 } ) =  ( 18 / 26 ) = ( 9 / 13 )

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bayes theorem is a topic for cbse / ncert  / scert of 12th standard in India. These are some of the important questions and their solutions from the topic of bayes theorem from various cbse ncert textbooks and old question papers.

bayes theorem of probability

total probability

bayes theorem of probability

Let E1, E2, . . . En be a partition of the sample space S, where E1, E2, . . . En are pairwise disjoint , non empty events.Let A be an event of non zero probability then

total probability

problem 1

An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is theprobability that the second ball is red?

Let E1 be the event that the ball taken out from the first 10 balls is red

let E2 be the event that the ball taken out from the first 10 balls is black.

Let A be the event that the ball taken out from the 12 balls ( after addition of 2 more balls ) is red.

If E1 occurs, then two more red balls will be added to make up seven red balls and five black balls with a total of twelve balls.

If E2 occurs, then two more black balls will be added to make up five red balls and seven black balls with a total of twelve balls.


P( E1 ) = 5 / 10

P( E2 ) = 5 / 10

P( A / E1 ) = 7 / 12 ( if red ball was drawn out first, two more red balls will be added to give 5+2 =7 red balls )

P( A / E2 ) = 5 / 12 ( if black ball was drawn out first, two more black balls will be added and number of red balls will not increase )


P(A) = P( E1 )  P( A / E1 )  +  P( E2 ) P( A / E2 ) ( total probablity )

P(A) = [ 5 / 10 ] [ 7 / 12 ]  +  [ 5 / 10 ] [ 5 / 12 ]

P(A) =  [ 60 / 120 ] =  ( 1 / 2 )




problem 2
A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

solution to problem 2 of  bayes theorem

problem 3
Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostelier?

solution to problem 3 of  bayes theorem


problem 4
In answering a question on a multiple choice test, a student either knows the answer or guesses. Let ( 3/4 ) be the probability that he knows the answer and ( 1/4) be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability ( 1/4 ). What is the probability that the student knows the answer given that he answered it correctly?
solution to bayes theorem problem 4 for ncert cbse 12th mathematics

problem 5

A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive ?
solution to bayes theorem problem 5 for ncert cbse 12th mathematics

problem 6

There are three coins. One is a two headed coin (having head on both faces),another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ?

solution to problem 6 of bayes theorem for cbse ncert mathematics

problem 7

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

solution to problem 7 of bayes theorem for cbse ncert mathematics 

problem 8
 A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further,2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?
solution of problem 8 on bayes theorem for cbse mathematics 

problem 9
Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.
solution of problem 9 on bayes theorem for cbse mathematics

problem 10
Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
solution of problem10 on bayes theorem for ncert cbse mathematics

problem 11
A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?
solution of problem11 on bayes theorem for ncert cbse mathematics

problem 12
A card from a pack of 52 cards is lost. From the remaining cards of the pack,two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
 solution to bayes theorem problem 12 for ncert cbse mathematics probability

problem 13
Probability that a man speaks truth is (  4 / 5 ). A coin is tossed and the man reports that a head appeared.Find the probability that actually there was a head.
 solution to bayes theorem problem 13for ncert cbse mathematics probability


miscellaneous exercise problem 14
Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of a certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?
solution to miscellaneous exercise problem 14 on bayes theorem in  ncert cbse 12th mathematics

miscellaneous exercise problem 15
Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls.One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
solution to miscellaneous exercise problem 15on bayes theorem in  ncert cbse 12th mathematics


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bayes theorem is a topic for cbse / ncert  / scert of 12th standard in India. These are some of the important questions from the topic of bayes theorem from various cbse ncert textbooks and old question papers.