Q. Two dice are thrown, simultaneously. If $X$ denotes the number of sixes, then the expected value of $X$ is
Probability - Part 2
Solution:
$X$ can take values $0,1,2$.
$P ( X =0)=$ Probahility of not getting six on any dice $=\frac{25}{36}$
$P ( X =1)=$ Probability of getting one six $=\frac{10}{36}$
$P ( X =2)=$ Probability of getting two six $=\frac{1}{36}$
Thus, the probability distribution is
X
0
1
2
P(X)
$\frac{25}{36}$
$\frac{10}{36}$
$\frac{1}{36}$
$\therefore E ( X )=0 \times \frac{25}{36}+1 \times \frac{10}{36}+2 \times \frac{1}{36}$
$=\frac{10}{36}+\frac{2}{36}=\frac{1}{3}$
| X | 0 | 1 | 2 |
| P(X) | $\frac{25}{36}$ | $\frac{10}{36}$ | $\frac{1}{36}$ |