(a) You have three fair dice: one blue, one red and one green. You roll the blue die six times, the red die five times and the green die four times. What is the probability that you roll a one exactly three times in total?

(b) If you rolled 5 ones, what is the probability that two of them were rolled on the blue die?

For (a): $$P(Y=3)=\binom{15}3\left(\frac16\right)^3\left(\frac56\right)^{12}=0.2363$$ But for (b) I don't seem to get the concept: $$P(XB=2\mid Y=5)=\frac{\binom62\binom93}{\binom{15}5}$$ I'm not getting the same answer as the book.

  • 1
    $\begingroup$ What was the book's answer? $\endgroup$ Dec 12, 2017 at 4:27
  • $\begingroup$ the book used P(XB=3|Y=5) and i’m losing my mind over it $\endgroup$ Dec 12, 2017 at 4:29
  • $\begingroup$ Do they provide a numerical answer? $\endgroup$
    – Remy
    Dec 12, 2017 at 4:29
  • $\begingroup$ I think XB corresponds to number of 1s from non blue die $\endgroup$ Dec 12, 2017 at 4:31
  • $\begingroup$ they found as final answer in the book 0.2398 $\endgroup$ Dec 12, 2017 at 4:33

2 Answers 2


The first question is easy : since colour does not matter, it is simply the probability of rolling exactly $3$ ones in $15$ attempts. This is given by a binomial distribution with parameters $15$(number of rolls) and probability parameter $\frac 16$, so the answer is just $\binom{15}{3} \left(\frac 16\right)^3 \left(\frac 56\right)^{12}$, as expected.

For the second one, well, what we do know is that there were $5$ ones rolled. The probability of this event, is given again by a similar binomial expression to above, namely $\binom{15}{5} \left(\frac 16\right)^5 \left(\frac 56\right)^{10}$.

Now, suppose two $1$s came on the blue die, and three on the others. What is the probability of getting two ones when rolling a dice six times? It is just $\binom{6}{2} \left(\frac 16\right)^2 \left(\frac 56\right)^{4}$.

Then, three ones must come on the other rolls, but we do not care how many come on which coloured dice. So the answer here is just $\binom{9}{3} \left(\frac 16\right)^3 \left(\frac 56\right)^{6}$.

Putting these together gives you the answer $\frac{\binom 62 \binom 93}{\binom {15}5}$, which is exactly what you have written, and which if you like is $0.41958...$ or $\frac{60}{143}$. So you have your concept absolutely right, if anything the textbook has got it wrong.

  • $\begingroup$ You have a small typo: It's 0.41958 $\endgroup$
    – Remy
    Dec 12, 2017 at 4:49
  • $\begingroup$ @Remy Yes, thank you for pointing this out. $\endgroup$ Dec 12, 2017 at 4:49

Let $Y_B$ denote the number of blue ones and let $Y$ denote the number of total ones. Then

$$\begin{align*} P(Y_B = 2 | Y=5) &=\frac{P(Y_B =2 \cap Y=5)}{P(Y=5)}\\\\ &=\frac{{6\choose2}\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^4 {9\choose{3}}\left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^6}{{15\choose{5}}\left(\frac{1}{6}\right)^5\left(\frac{5}{6}\right)^{10}}\\\\ &\approx0.4195804 \end{align*}$$

which agrees with the other answer.

The answer in the back of the book is surely wrong.


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