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Home  >>  CBSE XII  >>  Math  >>  Probability
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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

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  • Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: \(\;P(E_i/A)\)=\(\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}\)
Let $E_1$ be the event that the lost card is a diamond. Given that there are 13 diamonds in the deck, P($E_1$) = $\large\frac{13}{52} = \frac{1}{4}$
Let $E_2$ be the event that the lost card is not a diamond. P($E_2$) =1 - $\large\frac{13}{52} = \frac{39}{52} = \frac{3}{4}$
Let A be the event that the two cards drawn are found to be both diamonds.
We need to calculate P (2 cards are both diamond w the lost card being a diamond) next.
Given 13 diamond cards, if one is lost, then we have 12 remaining diamonds out of 51 total cards now.
The two cards that are both diamond can be drawn in $^{12}C_2 = \large\frac{12 \times 11}{1 \times 2} = \frac{132}{2} $ = 66 ways.
Likewise, the two diamonds can be drawn from the pack of 51 remaining cards in $^{51}C_2 = \large\frac{51 \times 50}{1 \times 2} = \frac{2550}{2} $ = 1275 ways.
Therefore, the P (2 cards drwan are diamond given one is lost) = P (A|$E_1$) = $\large\frac{66}{1275}$
Now, consider the event where the two cards drawn are both diamonds but the lost card is not a diamond.
The two cards that are both diamond can be drawn in $^{13}C_2 = \large\frac{13 \times 12}{1 \times 2} = \frac{156}{2} $ = 78ways.
Likewise, the two diamonds can be drawn from the pack of 51 remaining cards in $^{51}C_2 = \large\frac{51 \times 50}{1 \times 2} = \frac{2550}{2} $ = 1275 ways.
Therefore, the P (2 cards drwan are diamond given one card which is not a diamond is lost) = P (A|$E_1$) = $\large\frac{78}{1275}$
We need to calcuate the probability that the lost card is a diamond. P ($E_1$|A).
We can use Baye's theorem, according to which $P(E_1|A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$
P($E_1$|A) = $\large\frac{\frac{66}{1275}.\frac{1}{4}}{\frac{66}{1275}.\frac{1}{4} + \frac{78}{1275}.\frac{3}{4}} = \frac{66}{66+78 \times 3} = \frac{66}{300} = \frac{11}{50}$
answered Jun 19, 2013 by balaji.thirumalai
 

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