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 )

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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)

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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)

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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 

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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

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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

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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

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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]

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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
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miscellaneous problem 4 on binomial distribution

miscellaneous problem 4 on binomial distribution for cbse xii probability

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?

Let X be the number of people who are right handed out of a sample of 10 people.

Assuming X follows binomial distribution with
n = 10
p = (90/100) = (0.9) [given that 90% of people are right handed]
q = 1 - p = (0.1)

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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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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 )

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miscellaneous problem on bayes theorem 14 and 15

miscellaneous 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?

Let E1 be the event that the patient follows a course of meditation and yoga.

let E2 be the event that the patient follows the prescription of a certain drug.

let A be the event that the patient sufferes a heart attack.


P( E1 ) = ( 1 / 2 )  { given a patient can choose any one of the two options with equal probabilities}

P( E2 ) = ( 1 / 2 )  { given a patient can choose any one of the two options with equal probabilities}

P( A / E1 ) = ( 70 / 100 )( 40 / 100 ) { If E1 occurs, risk of heart attack is less by 30% }

P( A / E2 ) = ( 75 / 100 )( 40 / 100 ) { If E2 occurs, risk of heart attack is less by 25% }


Required probability = P [ the patient follows a course of meditation and yoga given that the patient sufferes a heart attack ]

Required probability = P [ E1 / A ]



P(E1 / A)=[( 1 / 2 )( 70 / 100 )( 40 / 100 ) ] / {[(1 / 2)(70/100)(40 / 100)] + [(1 / 2)(75/100)(40 / 100)]  }


P ( E1 / A ) =  [70] / { [70] + [75] } = ( 14 / 29 )
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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.

Let E1 be the event that the ball transferred from Bag I to Bag II is black.

let E2 be the event that the ball transferred from Bag I to Bag II is red.

let A be the event that the ball drawn from Bag II after the transfer is red.


P( E1 ) = ( 4 / 7)  { 4 black in Bag I out of a total of 3 + 4 = 7  }

P( E2 ) = ( 3 / 7 )  { 3 red in Bag I out of a total of 3 + 4 = 7}

P( A / E1 ) = ( 4 / 10 ) { If E1 occurs, 1 more black makes 4R and 6B in Bag II }

P( A / E2 ) = ( 5 / 10 ) { If E2 occurs, 1 more red makes 5R and 5B in Bag II }


Required probability = P [ transferred ball is black given that ball drawn from Bag II after the transfer is red ]

Required probability = P [ E1 / A ]



P(E1 / A)=[( 4 / 7 )( 4 / 10 ) ] / {[( 4 / 7 )( 4 / 10 ) ] + [( 3 / 7 )( 5 / 10 ) ]  }


P ( E1 / A ) =  [16] / { [16] + [15] } = ( 16 / 31 )

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problem 12 and problem 13 of bayes theorem

 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.

Let E1 be the event that the lost card is a diamond.

let E2 be the event that the lost card is not a diamond.

let A be the event that the two cards selected from the remaining 51 cards are both diamonds .


P( E1 ) = ( 13 / 52 )  { 13 diamonds in a pack of 52 cards }

P( E2 ) = ( 39 / 52 )  { 52 - 13 = 39 non-diamonds in a pack of 52 cards }

P( A / E1 ) = ( C (12,2) / C(51,2) ) { If E1 occurs, there are only 12 more diamonds among the remaining 51 cards}

P( A / E2 ) = (  C (13,2) / C(51,2) ) { If E2 occurs, there are  13 diamonds among the remaining 51 cards}

where C(n,r) = number of combinations of n things taken r at a time.



Required probability = P [ lost card is a diamond given that the two cards drawn from the remaining 51 cards are both diamonds ]

Required probability = P [ E1 / A ]




P(E1 / A)=[( 13/52 )( C (12,2)/C(51,2))] / {[( 13/52 )( C (12,2)/C(51,2))]+[( 39/52)( C (13,2)/C(51,2))]  }


P ( E1 / A ) =  [(13) * C (12,2) ] / { [(13) * C (12,2) ] + [(39) * C (13,2) ] }

P ( E1 / A ) =  [ C (12,2) ] / { [ C (12,2) ] + [(3) * C (13,2) ] }  = ( 11 / 50 )

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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.

Let E1 be the event that the coin toss actually resulted in a head.

let E2 be the event that the coin toss did not result in a head.

let A be the event that the man reports that a head appeared in the toss.


P( E1 ) = ( 1 / 2 ) 

P( E2 ) = ( 1 / 2 ) 

P( A / E1 ) = ( 4 / 5 ) { If E1 occurs, head has occured and the man is speaking the truth }

P( A / E2 ) = (  1 / 5 ) { If E2 occurs, head has not occured and the man is lying  hence [ 1 - (1/5)] }


Required probability = P [ the coin toss actually resulted in a head given that the man reports a head ]

Required probability = P [ E1 / A ]


 


P(E1 / A)=[( 1 / 2 )( 4 / 5 ) ] / {[( 1 / 2 )( 4 / 5 ) ] + [( 1 / 2 )( 1 / 5 ) ]  }


P ( E1 / A ) =  [4] / { [4] + [1] } = ( 4 / 5 )

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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 )
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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 )

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


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 )


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

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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

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


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


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


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.