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The question reads:

"Three cards are randomly selected, without replacement, from an ordinary deck of $52$ playing cards. Compute the conditional probability that the first card selected is a spade given that the sec- ond and third cards are spades."

This is my attempt, which I know is incorrect. I would like an explanation for why the reasoning was incorrect

$P(A)=$ First card is a spade

$P(B)=$ Second card is a spade

$P(C)=$ Third card is a spade

So, we want to find $P\left(A \vert B\cap C\right)$ $$P\left(A \vert B\cap C\right)= \frac{P\left(A \cap B\cap C\right)}{P\left(B\cap C\right)}$$

For the numerator we choose $3$ out $13$ cards out of a possible ${52 \choose 3}$.

For the denominator we choose $2$ out of a remaining $12$ cards out of a possible ${51 \choose 2}$

Our final answer should be $\frac{\frac{{13 \choose 3}}{{52 \choose 3}}}{\frac{{12 \choose 2}}{{51 \choose 2}}}= 0.25$

The given answer is $0.22$. Their solution includes using a lot of properties of probability conversions. If that is the only correct way to solve this problem, there is no reason to explain it since I have the explanation.

I was wondering what's wrong with my approach, and if it's not a simple numerical error, why this approach is incorrect.

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    $\begingroup$ Are you able to answer if the question read instead "What is the probability that the third card is a spade given that the first two cards are spades?" Are you able to reason that the answer to this different question should be the same as the answer to the first? $\endgroup$ – JMoravitz Oct 18 '17 at 0:19
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    $\begingroup$ As for your attempted answer, you do have an error. Your denominator you used $\dfrac{\binom{12}{2}}{\binom{51}{2}}$. Why? It should have been $\dfrac{\binom{13}{2}}{\binom{52}{2}}$. $\endgroup$ – JMoravitz Oct 18 '17 at 0:21
  • $\begingroup$ I see where my problem is now. I was using $P\left(B \cap C\vert A\right)$ instead of $P\left(B \cap C\right)$, correct? $\endgroup$ – Allan Oct 18 '17 at 0:26
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    $\begingroup$ That is correct. Once you simplify your fraction, you'll arrive at an answer of $\frac{11}{50}$, which is as alluded to the same answer to the question I pose in my first comment. In general, you can draw your cards in whatever order you want: third card first then first card then second card or what have you. There is a proper formal argument you can make as to why this is involving bijections between ways of drawing in the standard order versus drawing in an unusual order. $\endgroup$ – JMoravitz Oct 18 '17 at 0:28
  • $\begingroup$ Thank you for the help $\endgroup$ – Allan Oct 18 '17 at 0:29
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What is the probability that a spade is drawn from a pack of 50 cards with 11 spades? $11/50$.

Or by the method you attempted:

$$\mathsf P(A\mid B\cap C) ~{= \dfrac{\mathsf P(A\cap B\cap C)}{\mathsf P(B\cap C)} \\= \dfrac{\binom{13}3/\binom{52}3}{\binom{13}2/\binom{52}{2}} \\ = \dfrac{11}{50}}$$

Reason: $\{B\cap C\}$ is the event of drawing two spades when drawing two cards from the full deck .   That the position of these cards is second and third is irrelevant.

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