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I'm reading Introduction to Probability (Blitzstein and Hwang) and cannot wrap my head around an example they are using to provide intuition behind the formula for conditional probability.

In essence, we have a standard deck of shuffled cards and draw two cards randomly (one at a time, without replacement). We define A to be the event that the first card is a heart, and B the event that the second card is red. Then we are asked to find P(A|B) and P(B|A).

Where my confusion lies is in two calculations: how they calculated P(A ∩ B) and how they calculated P(B). They say "By the Naive definition of probability and the multiplication rule, P(A ∩ B) = (13/52) x (25/51)." Why is this valid when A and B are clearly not independent? Logically, I understand the math. But P(A)P(B) is clearly not equal to P(A ∩ B). How can you explain why this is valid, and is there a more clear way to calculate P(A ∩ B) (maybe more conditioning)?

I also don't know how you would go about calculating P(B) since it is not the first draw - any thoughts on the intuition / methodology to go about calculating this?

Much appreciated in advance!

Ryan
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    Hi: you are correct, they are not independent. So, P(B|A) P(A) = P(A $\cap$ B) = 25/51 $\times$ 13/52. – mark leeds Oct 19 '23 at 17:37
  • Note that $25/51$ is not equal to $\Pr(B)$. That is $\Pr(B\mid A)$. Rather, $\Pr(B)=26/52$ – JMoravitz Oct 19 '23 at 17:42
  • With regards to calculating $\Pr(B)$, see If you draw two cards, what is the probability the second card is a Queen?. With regards to calculating $\Pr(A\cap B)$... use counting arguments, noting there are $52\times 51$ ways to draw two cards in sequence, $13\times 25$ ways to pick a heart followed by a red card (where the second card is not the same card as the first) to get $\dfrac{13\times 25}{52\times 51}$. Split this into two fractions is a stylistic choice – JMoravitz Oct 19 '23 at 17:44
  • With regards to finding $\Pr(A\mid B)$... use Bayes' theorem. – JMoravitz Oct 19 '23 at 17:45
  • Do you imagine that for some reason, red cards tend to be more likely to show up at the exact top of the deck rather than just under the top card? – David K Oct 19 '23 at 17:54
  • @JMoravitz, thanks for the clarification. I might be getting lost in the weeds here, but assuming we know events are not independent, how would you generally go about calculating Pr(∩)? Is it just something you have to reason out using combinatorics like you did? Moreover, splitting it into fractions so it reads (13/52) * (25/51) is misleading because it almost reads as P(A) * P(B), when clearly (25/51) is not P(B), right? – Ryan Oct 19 '23 at 18:37
  • Splitting it into fractions can be seen as $\Pr(A)\times \Pr(B\mid A)$ is perfectly fine. How would you find $\Pr(A\cap B)$ typically? You should be able to reason how to find $\Pr(B\mid A)$ in situations like this by considering "What if we were actually in this scenario, what is the probability then" such as here "What if we had a deck of cards but it was missing a heart? What is the probability of drawing a red card?" That question should be intuitive how to answer, even if it is informal to get to the answer. – JMoravitz Oct 19 '23 at 18:40
  • Alternatively, if you wish to be more formal, yes... counting arguments so long as you are working in an equiprobable sample space like this is good, but recalling your multiplication principle is often better so long as you are given the conditional probability since not every question involves equiprobable sample spaces. If we don't know any additional information beyond $\Pr(A)$ and $\Pr(B)$ then there is not enough information to calculate $\Pr(A\cap B)$, but here we know the context and that informs us how we can proceed. – JMoravitz Oct 19 '23 at 18:43
  • Got it, thank you so much. If I'm asked to find Pr(∩), I would typically ask myself first if the events are independent. If they are, that makes it a walk in the park. Likewise if we know / reason P(B|A), it's not too difficult to rearrange the equation. I initially got confused with the fractional part and mistakenly thought (25/51) was Pr(B) and so I couldn't follow the logic of multiplying that by P(A) (13/52) when they clearly weren't independent. But your reasoning clarified that for me so this all makes sense now. – Ryan Oct 19 '23 at 18:53

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