Finding expected number of white balls drawn before any red ball [closed]

A box contains 12 balls: 6 white, 4 black, and 2 red. Draws are made without replacement. Find the expected number of white balls drawn before any red ball is drawn.

Similarly, how do I find the probability that all white balls are drawn before any red ball?

I have a hunch that this might be easily solved with indicator variables, but I am stuck on how to set up this problem.

closed as off-topic by Did, Najib Idrissi, Przemysław Scherwentke, Venus, Yiorgos S. SmyrlisMar 5 '15 at 12:29

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The black balls can be ignored. Label the white balls $W_1$ to $W_6$.

Let $X_i=1$ if white ball $W_i$ is drawn before any red, and $0$ otherwise. Then the number of white balls drawn before any red is $X_1+\cdots+X_6$.

We have $\Pr(X_i=1)=\frac{1}{3}$, and therefore $E(X_i)=\frac{1}{3}$. This is because if we look at $W_i$ and the two reds, it is equally likely that $W_i$ is in front of the two reds, somewhere in between, or after.

Thus the expected number of whites drawn before any red is $\frac{6}{3}$.

The probability that all whites are drawn before any red is the probability that the reds occupy the last $2$ positions. There are $\binom{8}{2}$ equally likely ways to choose positions for the reds, so the required probability is $\frac{1}{\binom{8}{2}}$.

• I just tried to solve it by noting that there are 7 different ways that 6 whites and 1 red can be arranged. Only one of these combinations has the 6 whites before the 1 red. So shouldn't $Pr(X_i=1)$ = 1/7? – Luchia Dec 5 '13 at 1:03
• Definitely $1/3$. We are just concerned with whether the particular ball $W_i$ is before any of the two reds. There are only $3$ balls relevant to the calculation. – André Nicolas Dec 5 '13 at 1:39

Hint: you can ignore the black balls. Your sample space is $R, WR, WWR, WWWR, \dots WWWWWWR$ what is the probability of each? Now $0P(R)+1P(WR)+2P(WWR)+\dots6P(WWWWWWR)$

• Thanks. Do you possibly know of a way I can use indicator variables to solve this? – Luchia Dec 5 '13 at 0:44
• I haven't seen one, but that is not something I often work with. – Ross Millikan Dec 5 '13 at 0:45
• This loses all the symmetry of the setting. – Did Dec 17 '13 at 22:04