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A standard deck of cards is shuffled well. Two cards are drawn randomly, one at a time without replacement. Let A be the event that the first card is a heart. Let B be the event that the second card is red. Find P(A|B) and P(B|A).

Let’s start with P(B|A): If first card is a heart, then remaining cards consist of 25 red and 26 black(all of which are equally likely) Therefore the conditional probability of getting a red card is $\frac{25}{25+26}=25/51$

The book(introduction to probability) then goes on to state that it’s harder to find P(A|B) in this way.

My question is why is it harder to find P(A|B) using reasoning similar to above?

Why is the following incorrect: $P(A|B)=P(A|\text{red and heart on second})+P(A|\text{red but not heart on second draw})$

(partition event B)

$=12/51 +13/51 $ because if second draw is a heart, then there are 13-1 possible heart cards for the first draw, out of 51 equally likely cards, and then if the second draw is red but not a heart then there are 13 possible hearts for the first draw out of 51 equally-likely-to-be-chosen cards.

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    $\begingroup$ This is a minor variation of the question asked and answered at stats.stackexchange.com/questions/113306. $\endgroup$ – whuber Apr 19 at 12:45
  • $\begingroup$ @whuber - I’ve read the question and its great answer, however my question is different: I’m looking at why the system of reasoning used in the first part of my question doesn’t apply to the second part. Specifically, why doesn’t partitioning the event B work in this case? Or does it? $\endgroup$ – apprentice9 Apr 19 at 14:31
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    $\begingroup$ The hope was you could employ a comparable visualization to understand your problem. $\endgroup$ – whuber Apr 19 at 16:59
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Using intuition, knowing that the first card was red doesn't change the probability of choosing a heart that much, so it seems wrong that it could be almost double.

$P[A|B \text{ or } C]$ is not necessarily equal to $P[A|B]+P[A|C]$, even if $B$ and $C$ are disjoint.

Suppose $B$ and $C$ are disjoint, like they are in the example where $B$ means the first card is a heart and $C$ means the first card is a diamond.

$P[A|B \text{ or } C]=\frac{P[A \text{ and } \{ B \text{ or } C\}]}{P[B \text{ or } C]} =\frac{P[A \text{ and } B]+P[A \text{ and } C]}{P[B \text{ or } C]}$

But, $P[A|B]+P[A|C]=\frac{P[A \text{ and } B]}{P[B]}+\frac{P[A \text{ and } C]}{P[C]}$

So, these two things are not necessarily the same.

If you further assume that $P[B]=P[C]$, as in this example, then $P[B \text{ or } C]=2P[B]$ and then you will have the first is half as big as the second.

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The problem with your reasoning is that you don't take into account the probability of the second card to be a heart or not (p = 0.5).

P(A|B) = (12/51)x0.5 + (13/51)x0.5 = 25/102.

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  • $\begingroup$ I see. Thanks. But why don’t we do the same in the first example - that is, why don’t we times by the probability of the conditioning event? $\endgroup$ – apprentice9 Apr 20 at 7:51
  • $\begingroup$ That is actually my mistake. P(A|red and heart on second) + P(A|red but not heart on second draw) = 12/51 + 13/51 as you correctly said, but P(A|B) is not equal to that sum. $\endgroup$ – Clabis Apr 21 at 12:33
  • $\begingroup$ Fixed the answer also since it was misleading. $\endgroup$ – Clabis Apr 21 at 12:34

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