We split 8 colored (and distinguishable from each other[each ball is unique]) balls to 4 kids, 2 balls for each kid.
There are 2 blue balls, 2 red balls, 2 yellow balls, 2 green balls [still each ball is unique]

A) It is known that Amy got at least 1 red ball, what is the probability that also John got at least 1 red ball?

B) It is known that Amy got balls of different colors, what is the probability that also John got balls of different colors?

What I have done is as follows:

A) let A be "Amy got at least 1 red ball"(B the same for John) hence $P(A)=1-{6\over8}*{5\over7}={13\over28}$ (all the cases subtract the case in which the first ball isn't red and also the 2nd isn't red)
So $P(A|B)={{{(2*{2\over8}*{6\over7})}^2}\over{13\over28}}=0.395$
This is because there is $2\over8$ chance to get red ball first and then $6\over7$ chance of getting non red ball, this chance is multiplied by 2 since we could do it the other way around(1st non red 2nd yes red) and then I square it all since the same chance applies to B(John).
I feel like I may have done it too complicated? is it even the right answer? not sure.

B)first ball can be any color $8\over8$ the 2nd ball has to be different than the first so $6\over7$ chance of that.
But now when I tried to find the new(A is now different color balls so is B) $P(A|B)$ I got something weird like that: ${({6\over7}*{6\over7})\over{6\over7}}={6\over7}$ which I really dont feel good about


We solve the first problem. Let $A$ be the event Amy got at least one red, and $B$ the event John got at least one red. We are asked to find $\Pr(B\mid A)$, which by definition is equal to $\Pr(A\cap B)/\Pr(A)$.

We compute the two required probabilities. You found $\Pr(A)$ correctly. Now we need $\Pr(A\cap B)$. This is the probability Amy and John each got one red.

Imagine that Amy drew a ball, then another, then John drew a ball, then another. The probability Amy got exactly one red is $\frac{2}{8}\cdot \frac{6}{7}+\frac{6}{8}\cdot\frac{2}{7}$, that is, $\frac{24}{56}$.

Given that Amy got exactly one red, the probability John got a red is $\frac{1}{6}\cdot \frac{5}{5}+\frac{5}{6}\cdot \frac{1}{5}$, that is, $\frac{10}{30}$.

Thus $\Pr(A\cap B)=\frac{24}{56}\cdot \frac{10}{30}$. Now we can find $\Pr(B\mid A)$.

It might be a little smoother to use binomial coefficients. For example, the probability that Amy gets exactly one red is $\frac{\binom{2}{1}\binom{6}{1}}{\binom{8}{2}}$.

The second problem is in a sense somewhat easier than the first. Change the meanings of $A$ and $B$ in the obvious way. You can find $\Pr(B\mid A)$ directly, without finding $\Pr(A)$ and $\Pr(A\cap B)$.

  • $\begingroup$ Thanks I was sure that because we are using the conditional formula I shall ignore the fact that something is given when I calculate $P(A \ and \ B)$ because in a sense if I do consider what is given then $P(A \ and \ B)$ is itself what I was looking for without the long formula isnt it? kind of weird when I think about it [we have used the given information inside the intersection already so why shall we divide by $P(A)$ now] $\endgroup$ – The One Apr 14 '16 at 15:10
  • $\begingroup$ You are welcome. If the question were: Given Amy got exactly one red, what is the probability John got exactly one red, we could find this (different) conditional probability directly. One red and one non-red are gone, say a blue, and now we can compute for John. (This is an additional hint for the second problem!) What complicates things in the first problem is the possibility that Amy got two red. This kind of forces us to go through the conditional probability $\Pr(A\cap B)/\Pr(A)$ machinery. $\endgroup$ – André Nicolas Apr 14 '16 at 15:17
  • $\begingroup$ Oh now I see, our intersection says something different indeed. about the 2nd question I thought to say that if Amy got 2 different color balls we now only have 2 sets of same color balls so the only way for John to fail is to get 1 ball from the duplicated set (there are 2 of those) and to take the pair of that ball so its $1-{2\over6}*{1\over5}*2={13\over15}$ is that the shortcut? $\endgroup$ – The One Apr 14 '16 at 15:27
  • $\begingroup$ Thanks I at the start did both 4/6 and multiplication by 2 then I thought that its weird so I changed to 2/6 and multiplication by 2 but now I understand how I could use my first 4/6 way (which clicks better in my head) $\endgroup$ – The One Apr 14 '16 at 15:46
  • $\begingroup$ I will delete my previous comment, since while I was typing you changed the $4$ to a $2$, and you understand the issues. In the longer run, you will need to use the "binomial coefficient" approach that I mentioned, since tracing all patterns would be too painful. $\endgroup$ – André Nicolas Apr 14 '16 at 15:52

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