Let x denotes the number of doublets in three throws of a pair of dice.
\(\therefore\) Possible values of X are 0, 1, 2 and 3 .
In a toss of pair of dice, possible doublets are (1, 1), (2, 2),(3, 3), (4, 4)(5, 5) and (6, 6).
\(\therefore\) Probability of getting a doublet \(=p=\frac{6}{36}=\frac{1}{6}\)
\(\therefore\) Probability of not getting a doublet \(=q=1-\frac{1}{6}=\frac{5}{6}\)
\(\therefore\) \(\mathrm{P}(\mathrm{X}=0)=\mathrm{P}( no \ doublet )=q \times q \times q=\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}=\frac{125}{216}\)
\(\mathrm{P}(\mathrm{X}=1)=\mathrm{P}( one \ doublet )=p q q+q p q+q q p =\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6}+\frac{5}{6} \times \frac{1}{6} \times \frac{5}{6}+\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}=\frac{75}{216}\)
\(\mathrm{P}(\mathrm{X}=2)=\mathrm{P}( two \ doublets )=p p q+p q p+q p p =\frac{1}{6} \times \frac{1}{6} \times \frac{5}{6}+\frac{1}{6} \times \frac{5}{6} \times \frac{1}{6}+\frac{5}{6} \times \frac{1}{6} \times \frac{1}{6}=\frac{15}{216}\)
\(\mathrm{P}(\mathrm{X}=3)=\mathrm{P} (three\ doublets )=p \times p \times p=\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}=\frac{1}{216}\)
\(\therefore\) Probability distribution of $X$ is as follows:
| X |
0 |
1 |
2 |
3 |
| P(X = x) |
\(\frac{125}{216}\) |
\(\frac{75}{216}\) |
\(\frac{15}{216}\) |
\(\frac{1}{216}\) |
