1.

A card from a pack of 52 cards is lost. From the remaining cardsof the pack, two cards are drawn and are found to be both spades. Find the probability of the lost card being a spade.

Answer»

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Solution :Let `E_1,E_2,E_3and E_4` be the events of losing a card of SPADES, clubs, hearts and diamonds repectively.
Then, `P(E_1)=P(E_2)=P(E_3)=P(E_4)=13/52=1/4.`
Let E be the event of drawing 2 spades from the remaining 51 cards. Then,
`P(E//E_1)`= PRBABILITY of drawing 2 spades, given that a card of spades is missing
`(.^12C_2)/(.^51C_2)=((12xx11))/(2!)XX(2!)/((51xx50))=22/425`
`P(E//E_2)`=probability of drawing 2 spades, giventhat a card of clubs is missing
`(.^13C_2)/(.^51C_2)=((13xx12))/(2!)xx(2!)/((51xx50))=26/425`
`P(E//E_3)`= probability of drawing 2 spades, given that a card of hearts is missing
`=(.^13C_2)/(.^51C_2)=26/425`.
`P(E//E_4)`= probability of drawing 2 spades, given that a card of diamonds is missing
`=(.^13C_2)/(.^51C_2)=26/425`.
`:. P(E_1//E)`= probability of the lost card being a spade, given that 2 spades are drawn from the remaining 51 cards
`(P(E_1).P(E//E_1))/(P(E_1).P(E//E_1)+P(E_2).P(E//E_2)+P(E_3).P(E//E_3)+P(E_4).P(E//E_4))`
`=((1/4xx22/425))/((1/4xx22/425)+(1/4xx26/425)+(1/4xx26/425)+(1/4xx26/425))`
Hence, the required probability 0.22.


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