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 )

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

**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 )

**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

## No comments:

## Post a Comment

please leave your comments