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.

  • 1
    $\begingroup$ This is a minor variation of the question asked and answered at stats.stackexchange.com/questions/113306. $\endgroup$
    – whuber
    Commented Apr 19, 2021 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? $\endgroup$ Commented Apr 19, 2021 at 14:31
  • 1
    $\begingroup$ The hope was you could employ a comparable visualization to understand your problem. $\endgroup$
    – whuber
    Commented Apr 19, 2021 at 16:59

2 Answers 2


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.


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.

  • $\begingroup$ 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? $\endgroup$ Commented Apr 20, 2021 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
    Commented Apr 21, 2021 at 12:33
  • $\begingroup$ Fixed the answer also since it was misleading. $\endgroup$
    – Clabis
    Commented Apr 21, 2021 at 12:34

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