Conditional Probability lost card problem.

Question

Q : A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn. What is the probability that they both are diamonds?

My Approach : I calculated the probability of selecting 2 diamonds when A Diamond card is lost and added it to the probability of selecting 2 diamonds when a diamond card is not lost.

P($\frac{D}{LD})$ + P($\frac{D}{NLD}$) $= \frac{^{12}C_2}{^{51}C_2} + \frac{^{13}C_2}{^{51}C_2}$

$=\frac{48}{425}$

But the answer given in the book is $\frac{1}{17}$. Where am I going wrong?

Answer 1

Alternatively, intuitively, it's like saying, instead of drawing the top two cards after shuffling, move the top one to the bottom and reveal card 2 and 3. What's the probability that these two are diamonds? $$\frac{\binom{13}{2}}{\binom{52}{2}} = \frac{1}{17}.$$ This is symmetry.

Answer 2

We use your analysis, with some correction. Let $T$ be the event $2$ diamonds, and let $L$ be the event it was a diamond that was lost. Then $$\Pr(T)=\Pr(T\mid L)\Pr(L)+\Pr(T\mid L^c)\Pr(L^c).$$ Note that $\Pr(L)=1/4$ and $\Pr(L^c)=3/4$.

Your expressions for $\Pr(T\mid L)$ and $\Pr(T\mid L^c)$ are correct.

Answer 3

I partitioned the cases and then used the Law of Total probability instead of using combinations. Apologies if you need an approach that uses combinations.

The approach:

Let $P(X)$ be the probability of the lost card being a diamond ($=1/4$)

Let $P(Y)$ be the probability of the lost card not being a diamond ($=3/4$)

In the case of event $X$ (the lost card is a diamond), the probability of selecting 2 cards and having them both be a diamond under condition $X$ is then:

$P(2D | X) = (12/51)*(11/50)=(22/425)$

Under condition $Y$:

$P(2D | Y) = (13/51)*(12/50)=(26/425)$

With the four aforementioned probabilities, we can use the Law of Total probability to obtain the textbook's answer:

$P(2D) = P(2D | X)P(X) + P(2D | Y)P(Y) = (22/425)(1/4)+(26/425)(3/4) = 1/17$

(According to Wolfram Alpha).