# Find the conditional probability of 6 appearing exactly once in $3$ rolls of a die given that it appeared at least once.

Did I find the correct probability?

A fair die is rolled $3$ times. The conditional probability of 6 appearing exactly once, given that it appeared at least once.

So,the combined probability that 6 appeared exactly once and it appeared at least once is $$3\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{2}$$ since, the case is that the 6 appears only once in the $3$ throws and it can appear at any throw: 1, 2 or 3. And the probability that 6 appeared at least once is $$1-\left(\frac{5}{6}\right)^3$$

So the required probability is $$\frac{3\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{2}}{1-\left(\frac{5}{6}\right)^3}$$ P.S.I don't have anything to check the answer, so I posted it here to verify. Please don't mind.

• Your solution is correct. Jan 24, 2016 at 11:30

Perhaps just a better way to rephrase all this:

The probability of 6 appearing exactly $\color\red1$ out of $\color\green3$ times:

$$\binom{\color\green3}{\color\red1}\cdot\left(\frac16\right)^{\color\red1}\cdot\left(1-\frac16\right)^{\color\green3-\color\red1}=\frac{75}{216}$$

The probability of 6 appearing at least $\color\red1$ out of $\color\green3$ times:

$$\sum\limits_{n=\color\red1}^{\color\green3}\binom{\color\green3}{n}\cdot\left(\frac16\right)^{n}\cdot\left(1-\frac16\right)^{\color\green3-n}=\frac{91}{216}$$

So the conditional probability is $\dfrac{\frac{75}{216}}{\frac{91}{216}}=\dfrac{75}{91}$.