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.
Best Answer
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.