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I am a bit confused, can you help?

So, let's say I have the task of selecting cards from a deck of playing cards. There are 52 cards total; 50% are red and 50% are black.

The probability of getting a red card is 50% The probability of subsequently selecting a black card is 26/51 (because 1 card is removed)

But the probability of a red card, given we've already got a black card, I think is: p(A∩B) = p(A) * p(B|A) - Which would be 0.5 * 0.51

But then I hear about a different formula for conditional probabiltiy, where I should divide the result by the probability of selecting a black.

I am confused, as I watched a youTube video which explained that the multiplication rule works like I have it & then read some blogs about dividing by the p(b).

Can someone help me figure out why we do that & whether my initial view was correct?

Thanks a lot

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  • $\begingroup$ You've set up the problem you want to ask about in a slightly jumbled way. The "task of selecting cards from a deck of playing cards" might be selecting two of them, or some other number of cards (or indeed a random number of them). Second paragraph talks about (a) the probability of (first) getting a red card, and (b) the probability of subsequently (to first getting a red card) selecting a black card. Your next paragraph drops the sequence of card selections, perhaps reversing it, but in any case confusing Readers as to whether conditional probability... $\endgroup$
    – hardmath
    Dec 27, 2020 at 20:40
  • $\begingroup$ or the probability of a conjunction (intersection) of events is meant. $\endgroup$
    – hardmath
    Dec 27, 2020 at 20:46
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    $\begingroup$ As for "then read some blogs about dividing by the p(b)" you may be referring to $\Pr(A\mid B) = \dfrac{\Pr(A\cap B)}{\Pr(B)}$ which is by definition and you will notice is equivalent to $\Pr(A\cap B) = \Pr(B)\Pr(A\mid B)$ like you had before, just a rearrangement of terms. Applying this again from the other event's perspective gives you that $\Pr(A)\Pr(B\mid A) =\Pr(A\cap B)=\Pr(B)\Pr(A\mid B)$ and dividing gives the relation $\Pr(A\mid B) = \dfrac{\Pr(B\mid A)\Pr(A)}{\Pr(B)}$ which is referred to as "Bayes' Theorem". $\endgroup$
    – JMoravitz
    Dec 27, 2020 at 20:59
  • $\begingroup$ As for why this might be useful to do, it depends on what values are given in a problem statement and/or are easy to calculate and what information you are looking for in the first place. $\endgroup$
    – JMoravitz
    Dec 27, 2020 at 21:00
  • $\begingroup$ Thanks @JMoravitz , so my formula is equivalent, because its just a reordered version of the same formula? Hence that formula is already bayes theorem? $\endgroup$
    – kikee1222
    Dec 28, 2020 at 6:57

1 Answer 1

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"...the probability of a red card, given we've already got a black card, is $P(A\cap B)=\dots$."

No. $P(A\cap B)$ is the probability of getting a red card and then getting a black card (or getting a black card and then getting a red card). The probability of getting a red card, given we've already got a black card is a conditional probability, denoted $P(A\mid B)$, and calculated by $$P(A\mid B)={P(A\cap B)\over P(B)}$$ In your example, $P(A\cap B)=(26/52)(26/51)=13/51$, while $P(A\mid B)=P(A\cap B)/P(B)=(13/51)/(26/52)=26/51$.

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  • $\begingroup$ Hey there, I am sorry I took some time to come back to you, I really appreciate the help. I think I am a bit clearer, I want to ask a follow up question, I'll mark this one as solved & hopefully the next one will fully clarify it in my mind :) $\endgroup$
    – kikee1222
    Dec 30, 2020 at 10:28

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