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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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 )

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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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