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Question
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.
Solution
Let E1: the losing card is a diamond;
E2: The losing card is a heart.
E3: The losing card is a club.
E4: The losing card is a spade.
E: Event of drawing 2 diamonds from the remaining cards
Then P(E1) = P(E2) = P(E3) = P(E4) = `13/52 = 1/4`
P(E|E1) = Both cards are diamonds, action if the diamond card is lost.
= `(""^12C_2)/(""^51C_2)`
= `(12 xx11)/(51 xx 50)`
= `44/850`
= `22/425`
P(E|E2) = Both cards are diamonds, action if the heart card is lost.
= `(""^13C_2)/(""^51C_2)`
= `(13 xx 12)/(51 xx 50)`
= `26/(17 xx 25)`
= `26/425`
Similarly, P(E|E3) = `26/425`, P(E|E4) = `26/425`
Then, by Bayes' theorem, the lost card is a diamond, while the two cards drawn from the remaining cards are diamonds.
P(E1|E) = `(P(E_1) xx P(E|E_1))/(P(E_1) xx P(E|E_1) + P(E_2) xx P(E|E_2) + P(E_3) xx P(E|E_3) + P(E_4) xx P(E|E_4)`
= `(1/4 xx 22/425)/(1/4 xx 22/ 425 + 1/4 xx 26/425 + 1/4 xx 26/425 + 1/4 xx 26/425)`
= `22/(22 + 78)`
= `22/100`
= `11/50`
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