Find the probability distribution of the number of sixes in throwing three dice once.
Let $X$ denote a random variable which is the number of sixes obtained when three dice are thrown at once.
$X$ can take values $0, \; 1, \; 2, \; 3$
Probability of getting a $6$ when a dice is thrown $= \dfrac{1}{6}$
Probability of NOT getting a $6$ when a dice is thrown $= \dfrac{5}{6}$
Now,
$P \left(X = 0\right) = P \left(\text{No } 6 \text{ is obtained in any throw}\right) = \left(\dfrac{5}{6}\right)^3 = \dfrac{125}{216}$
$P \left(X = 1\right) = P \left(\text{One } 6 \text{ is obtained in } 3 \text{ throws}\right) = 3 \times \dfrac{1}{6} \times \left(\dfrac{5}{6}\right)^2 = \dfrac{75}{216}$
$P \left(X = 2\right) = P \left(\text{Two } 6's \text{ are obtained in } 3 \text{ throws} \right) = 3 \times \left(\dfrac{1}{6}\right)^2 \times \dfrac{5}{6} = \dfrac{15}{216}$
$P \left(X = 3\right) = P \left(6's \text{ are obtained in all throws}\right) = \left(\dfrac{1}{6}\right)^3 = \dfrac{1}{216}$
Hence the probability distribution of the random variable $X$ is given by
| $X$ | $0$ | $1$ | $2$ | $3$ |
|---|---|---|---|---|
| $P \left(X\right)$ | $\dfrac{125}{216}$ | $\dfrac{75}{216}$ | $\dfrac{15}{216}$ | $\dfrac{1}{216}$ |