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.

 

show that 9^(n+1) - 8n -9 is divisible by 64 whenever n is a positive integer

 ncert cbse chapter 8 binomial theorem exercise 8.1

13.show that 9^(n+1) - 8n -9 is divisible by 64 whenever n is a positive integer 

(x+1)^(n+1) = [x^(n+1)] +C[(n+1),1][x^(n)]+...+C[(n+1),(n-1)][x^2] +C[(n+1),n][x] + 1

 

C[(n+1),n] =C[(n+1),1] = (n+1)

C[(n+1),(n-1)]= C[(n+1),2] = (n+1)(n)/2

(x+1)^(n+1) = [x^(n+1)] +(n+1)[x^(n)]+...+[(n+1)(n)/2][x^2]+(n+1)[x] + 1

put x=8

 (8+1)^(n+1) = [8^(n+1)] +(n+1)[8^(n)]+...+[(n+1)(n)/2][8^2]+(n+1)[8] + 1

concentrating on the last 2 terms only

9^(n+1) =  { [8^(n+1)] +(n+1)[8^(n)]+...+[(n+1)(n)/2][8^2]} + [8n+8]+1

9^(n+1) =  { [8^(n+1)] +(n+1)[8^(n)]+...+[(n+1)(n)/2][8^2]} + 8n+9

re-arranging

 9^(n+1) - 8n - 9 ={ [8^(n+1)] +(n+1)[8^(n)]+...+[(n+1)(n)/2][8^2]}

note that C[(n+1),(n-1)]= C[(n+1),2] = (n+1)(n)/2 is an integer

so after taking factor of (8^2) = 64 out,

9^(n+1) - 8n - 9 = 64*[8^(n-1)+ ...+[(n+1)(n)/2] ] =divisible by 64

8. evaluate (101)^4

using expansion with 

C(4,1) =4=C(4,3)

C(4,2)=6

(x+y)^4 = (x^4)+4(x^3)(y)+6(x^2)(y^2)+4(x)(y^3)+(y^4)

x=100

y=1

(100+1)^4 = (100^4)+4(100^3)(1)+6(100^2)(1^2)+4(100^1)(1^3)+(1^4)

(101)^4 =100000000+4000000+60000+400+1

(101)^4 =104060401

 

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

 

5.evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

solution 

 6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4 

solution

 

7.find an approximate value of (0.99^5) using the first three terms of its expansion

solution  

 

8.find n if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of [(fourth root of 2) + {1/(fourth root of 3)}]^n is (sqrt6):1 

solution

 

exercise 8.2

 

 5. find the 4th term in the expansion of (x-2y)^12

solution

7. Find the middle terms in the expansion of [3 - ((x^3) / 6)]^7

solution

Q8) Find the  middle terms in the expansion of [(x/3)+9y)]^10

solution

 

 10.The coefficients of the (r-1)th, rth, (r+1)th  terms in the expansion of [(x+1)^n] is in the ratio 1:3:5. Find n and r.

solution 

 

exercise 8.1

8. evaluate (101)^4

solution

13.show that 9^(n+1) - 8n -9 is divisible by 64 whenever n is a positive integer 

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

find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4

 ncert cbse chapter 8 binomial theorem miscellaneous exercise    

 6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4 

using expansion with 

C(4,1) =4=C(4,3)

C(4,2)=6

(x+y)^4 = (x^4)+4(x^3)(y)+6(x^2)(y^2)+4(x)(y^3)+(y^4)

(x-y)^4 = (x^4)-4(x^3)(y)+6(x^2)(y^2)-4(x)(y^3)+(y^4)

adding the two expansions 

(x+y)^4 +(x-y)^4= 2[(x^4) +6(x^2)(y^2) +(y^4) ]

use x =(a^2) y = sqrt{(a^2)-1}

[(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4

 = 2[{(a^2)^4}+6{(a^2)^2}{[ sqrt{(a^2)-1}]^2} +{[ sqrt{(a^2)-1}]^4}]

 =2[ (a^8) +6(a^4)((a^2)-1) +{(a^2)-1}^2 ]

=2 [ (a^8) +6(a^6) -6(a^4) +{(a^4) -2(a^2)+1} ]

=2[ (a^8) +6(a^6) -5(a^4)-2(a^2)+1  ]

=2(a^8) +12(a^6) -10(a^4)- 4(a^2)+2

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

 

5.evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

solution 

 6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4 

solution

 

7.find an approximate value of (0.99^5) using the first three terms of its expansion

solution  

exercise 8.2

Q8) Find the  middle terms in the expansion of [(x/3)+9y)]^10

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

 

 

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

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

 

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

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

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

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

ncert cbse 7th chapter permutations miscellaneous exercise

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

The vowels are A,U,E

We have to select 2 out of the 3 vowels in C(3,2)= [3*2]/[1*2]= 3 ways

The consonants are D,G,H,T,R

We have to select 3 out of the 5 vowels in C(5,3)= [5*4] / [1*2] =10ways

because C(n,r) = C[n,(n-r)]  gives C(5,3)=C(5,2)

now the 2+3= 5 selected letters can be arranged among themselves in

5! =5*4*3*2*1=120 ways

Required number of words = 3*10*120 =3600

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

 EQUATION  so that the vowels and consonants occur together?

Vowels are E,U,A,I,O

Treat the 5 vowels as a single unit. Inside this the 5 vowels can be arranged

among themselves in 5!=5*4*3*2*1=120 ways

Consonants are Q,T,N

Treat the 3 consonants as a single unit. Inside this the 3 consonants can be arranged among themselves in 3!=3*2*1=6 ways

Now the two units can be arranged among themselves in 2!=2*1 = 2 ways

Required number of words =120*6*2=1440 words

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

integral ∫ { cosx/ [ sinx +cosx] } dx

integral  ∫  { cosx / [ sinx +cosx]  } dx

first express the numerator in terms of the denominator and its derivative
split into two integrals ;one  of them is of the form u' / u which will integrate to ln(u)

 PAGE 1 BASIC INTEGRATION

PAGE 2 INTEGRATION BY SUBSTITUTION 

PAGE 3 INTEGRATION BY COMPLETION OF SQUARES

PAGE 4 INTEGRATION BY PARTS

PAGE 5 INTEGRATION BY MANIPULATION OF NUMERATOR IN TERMS OF DENOMINATOR

 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

some more problems from binomial distribution for cbse ncert class xii mathematics probability

some more problems from binomial distribution for cbse ncert class xii  mathematics probability

4.Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that (i) all the five cards are spades (ii) only 3 cards are spades and (iii) none is a spade?

Let X be the number of spades among the five cards.

Assume X follows Binomial distribution with

n = 5

p = (13/52) = ( 1 / 4 ) [13 spades among the 52 cards ]

q =1 -p

q = (3 / 4)

P[X=r] = _nC_r p^rq^(n-r) , r = 0,1,2,...,n

P[X=r] = ₅C_r (1/4)^r(3/4)^(5-r) , r = 0,1,2,...,5

P[all the five cards are spades ] = P[X=5] =₅C₅(1/4)⁵(3/4)^(5-5) =( 1 / 1024 )

P[only three cards are spades ] = P[X=3] =₅C₃(1/4)³(3/4)^(5-3) =(90/1024)=(45/512)

P[none is a spade ] = P[X=0] =₅C₀(1/4)⁰(3/4)^(5-0) =(243/1024)

9.On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing ?

Let X be the number of questions he answered correctly out of the 5 questions. just by guessing

Assume X follows Binomial distribution with

n =5

p = ( 1 / 3 ) [one out the three possible answers is correct and the candidate is guessing ]

q = 1 – p = ( 2 / 3 )

P[X=r] = _nC_r p^rq^(n-r) , r = 0,1,2,...,n

P[X=r] = ₅C_r ( 1 / 3 )^r( 2 / 3 )^{(5 - r)} r = 0, 1 ,..., 5

P[ candidate would get four or more correct answers just by guessing ] = P[X=4] + P[X=5]

= ₅C₄ ( 1 / 3 )⁴( 2 / 3 )^{(5 - 4)} + ₅C₅ ( 1 / 3 )⁵( 2 / 3 )^{(5 - 5)} = ( 11/243 )

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

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

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

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

 

binomial distribution problem for ncert cbse 12th mathematics probability

binomial distribution problem for ncert cbse 12th mathematics probability

1.A die is thrown 6 times. If getting an odd number is a success, what is the probability of (i) 5 successes (ii) at least 5 successes (iii) at most 5 successes?

Let X be the number of successes out of 6 throws

Assume X follows binomial distribution with

n = 6,

p = (3/6) = ( 1 / 2 )

q = 1 -p

q = ( 1 / 2 )

P[X=r] = _nC_r p^rq^(n-r) , r = 0,1,2,...,n

P[X=r] = ₆C_r ( 1 / 2 )^r( 1 / 2 )^{(6 - r)}

P[X=r] = ₆C_r ( 1 / 2 )⁶

P[ 5 successes ] =P[X=5] = ₆C₅ ( 1 / 2 )⁶ = (3 /32)

P[ at least 5 successes ] =P[X=>5]= P[X=5] +P[X=6]= (7/64)

P[ at most 5 successes ] =P[X<=5] =1 - P[X=6]= 1 - ₆C₆ ( 1 / 2 )⁶ = (63 /64)

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

>

2.A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of

two successes.

Let X be the number of doublets out of 4 tosses of a pair of dice.

Assume X follows binomial distribution with

n = 4,

p = (6 / 36) [6 doublets out of 36 possible outcomes in one toss of a pair of dice]

p = ( 1 / 6)

q= 1 -p

q = (5 /6)

P[X=r] = _nC_r p^rq^(n-r) , r = 0,1,2,...,n

P[X=r] = ₄C_r (1/6)^r(5/6)^{(4 - r)} , r = 0,1,2,3,4

P[two successes] = P[X = 2] =₄C₂ (1/6)²(5/6)^{(4 – 2)} = (25/216)

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

 

 

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

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

This problem and the answer is from binomial distribution in the chapter on probability for class xii of cbse ncert 12^th mathematics and is useful for the students preparing for the board examination 

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

finding n of binomial distribution given the probability

 finding n of binomial distribution given the probability from miscellaneous 12th cbse ncert mathematics

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

Let n be the required number of times the man must toss a fair coin so that the probability of having at least one head is more than 90%

Let X be the number of heads obtained when the coin is tossed n times.

Assuming X follows binomial distribution with

n=n

p = (½)

q = 1 – p = (½)

P[X=r] = _nC_r p^rq^(n-r) , r = 0,1,2,...,n

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

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

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

hurdle problem from miscellaneous cbse ncert mathematics probability

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?

Let X be the number of hurdles the player knocks down out of 10 hurdles

Assume X follows binomial distribution with

n=10

p = 1-(5/6) [because we defined X in terms of the hurdles knocked down ]

p=(1 /6)

q = 1 -p

q =(5/6)

P[X=r] = _nC_r p^rq^(n-r) , r = 0,1,2,...,n

P[X=r] = ₁₀C_r (1/6)^r(5/6)^{(10 - r)} , r = 0,1,2,...,10

P[the player will knock down fewer than two hurdles] = P[ X < 2 ]

P[ X < 2 ] = P[X=0] + P[X=1]

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

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