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Let's say I have an urn with 10 white balls and 20 black ones. If I draw 8, what is the probability that I get 1 white in the last 3 if I drew 1 white in the first 5?

The most straightforward way is to say that after the five draws I only have 9 W and 16 B left so

$$P(\text{1 W in last 3}) = \frac{9\times {16\choose2}}{25\choose3}$$

But if I want to use P(A|B) = $\frac{P(A \cap B)}{P(B)}$ how would I calculate $P(A \cap B)$?

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For any particular ordering of $2$ White, $8$ Black, you'll have the following probability: $$p=\frac{P(10,2)P(20,6)}{P(30,8)}$$ i.e. think about the case where the first $2$ draws are white and the rest is black and then write the probabilities one by one. We also need to count the number of places we can distribute these two White balls, e.g. instead of the first two places. Choose one from the first five, and one from the last three, i.e. $15$ ways, which makes the probability you seek for, i.e. $P(A\cap B)=15p$.

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  • $\begingroup$ I've never seen that comma notation before. What does it mean? $\endgroup$ – planarian Jan 11 at 15:56
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    $\begingroup$ $P(10,2)$ means choose $2$ from $10$ and order, i.e. $$P(n,k)={n\choose k}k!$$ $\endgroup$ – gunes Jan 11 at 15:58
  • $\begingroup$ @haventureinstatedmonicayet in a Bayesian context the comma is sometimes used as a shorthand in joint probabilities. Since Probability and Permutations can be notated with a 'P', it's easy to assume the wrong context. $\endgroup$ – jkm Jan 11 at 18:39
  • $\begingroup$ @jkm right, abuse of notations can easily cause the confusion. I'd insist on using $p_{xy}(10,2)$ or $P(X=10,Y=2)$ instead of $P(10,2)$ in every contexts if we're referring to joint probabilities/distributions. planarian, this notation in permutation is called as k-permutations of n: en.wikipedia.org/wiki/Permutation#/k-permutations_of_n $\endgroup$ – gunes Jan 11 at 22:54

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