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.

Answer

N/a

Explanation

Let `E_(1)=` event that card lost is of diamond
`implies n(E_(1))=13`
`E_(2)=` event that lost card is not of diamond
`implies n(E_(2))=52-13=39`
and sample space `n(S)=52`
Then `E_(1)` and `E_(2)` are mutually exclusive and exhaustive event.
`:. P(E_(1))=13/52=1/4` and `P(E_(2))=39/52=3/4`
Let `E=` event that 2 cards drawn from remaining cards are of diamond. Now there will be 12 cards of diamond in the remaining 51 cards, when a lost card is of diamond. Now total number of ways of drawing 2 cards of diamond will be `.^(12)C_(2)` from 12 cards of diamond.
Similarly total number of ways of drawing 2 cards from 51 cards are `.^(51)C_(2)`.
`:.P(E/(E_(1)))=P` (when the lost card is of diamond then 2 cards drawn from remaining cards are of diamond)
`=(.^(12)C_(2))/(.^(51)C_(2))=((12xx11)/(1xx2))/((51xx50)/(1xx2))=(12xx11)/(51xx50)`
and `P(E/(E_(2)))=P` (when the lost card is not of diamond then 2 cards drawn from remaining cards are of diamond)
`=(.^(13)C_(2))/(.^(51)C_(2))=(13xx12)/(51xx50)`
From Bayes' theorem
`P((E_(1))/E)=(P(E_(1)).P(E/E_(1)))/(P(E_(1)).P(E/(E_(1)))_P(E_(2))P(E/(E_(2))))`
`implies (1/4xx(12xx11)/(51xx50))/(1/4xx(12xx11)/(51xx50)+3/4xx(13xx12)/(51xx50))`
`=(12xx11)/(12xx11+3xx13xx12)=11/(11+39)`
`=11/50`