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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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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