This is in reference to an already asked and solved question.

I have the same problem statement:

Bowl 1 contains six red chips and four blue chips. Five of these 10 chips are selected at random and without replacement and put in Bowl 2, which was originally empty. One chip is then drawn at random from Bowl 2. Given that this chip is blue, find the conditional probability that two red chips and three blue chips are transferred from Bowl 1 to Bowl 2. Solve using Bayes Theorem

Bayes Theorem clearly states that:

$$P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)}$$

My question that specializes my post and separates it from the linked post, is an explanation of how P(A) is found.

According to the proposed solution in the linked question, P(B) is equal to 3/5, which makes sense as there are 3 blue chips and 5 total.

How does the following make sense?

$$\begin{align} &P(A) = \frac {\binom 6 2 \cdot \binom 4 3}{\binom {10} 5}\\ \end{align}$$

And later, how does that translate to: $$\begin{align} P(A|B) = \left(\dfrac 3 5\right)\cdot \dfrac{ \binom 6 2 \binom 4 3}{\binom {10} 5\cdot \dfrac{4}{10}} \end{align}$$

  • $\begingroup$ Read it again. $\mathsf P(B)$ is not $3/5$. That is $\mathsf P(B\mid A)$. $\endgroup$ Feb 5, 2020 at 1:52

1 Answer 1


The event $A$ is that 2 from 6 red chips and 3 from 4 blue chips in the first bowl are transfered.

The event $B$ is that 1 from the blue chips transfered into the second bowl are then drawn from it.

The probabilities are thus:

  • $\mathsf P(A)=\binom 62\binom 43/\binom{10}5$ is the probability for transferring 2 from 6 red chips and 3 from 4 blue chips, when selecting 5 from 10 chips in the first bowl.

  • $\mathsf P(B\mid A)=3/5$ is the probability for selecting one from the blue chips transfered to the second bowl when given that 3 blue chips were among the 5 transfered to the second bowl.

  • $\mathsf P(B) = 4/10$ is the probability for selecting, from the second bowl, one from the 4 blue chips that were (originally) in the first bowl (since each from the 10 chips had an equal chance of being transferred and selected; 4 from these 10 chips are blue).

    Hence the probability for transferring 2 red chips (and 3 blue) when given that a blue chip was selected from the second bowl is:

$$\begin{align}\mathsf P(A\mid B)&=\dfrac{\mathsf P(B\mid A)\cdot\mathsf P(A)}{\mathsf P(B)}\\[2ex]&=\cfrac{\dfrac 35\cdot\left.\dbinom 62\dbinom 43\middle/\dbinom{10}5\right.}{\dfrac 4{10}}\\[2ex]&=\dfrac{5}{14}\end{align}$$

  • $\begingroup$ Thanks, @GrahamKemp. What probability rule or law allows you to find P(A)? I understand it logically when you write it out, but I am trying to tie this back into some lecture material. $\endgroup$
    – user615962
    Feb 5, 2020 at 2:16
  • $\begingroup$ @wundermahn It is just the count for outcomes in the event divided by the count for outcomes in the space (where all outcomes are equiprobable). $\endgroup$ Feb 5, 2020 at 2:44
  • $\begingroup$ Thanks @GrahamKemp. Lastly, why in your second to last step is it (10 2) instead of (10 5). Apologies for the questions $\endgroup$
    – user615962
    Feb 5, 2020 at 2:50
  • 1
    $\begingroup$ Ah, that is a typo. Thanks for pointing it ouy. Corrected $\endgroup$ Feb 5, 2020 at 3:01

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