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Question
Find the probability distribution of the number of doublets in four throws of a pair of dice. Also find the mean and variance of this distribution.
Solution
Let p denote the probability of getting a doublet in a single throw of a pair of dice. Then
\[p = \frac{6}{36} = \frac{1}{6}\]
\[\text { Therefore,} q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}\]
Let X denote the number of doublets in four throws of a pair of dice. Clearly, X has the binomial distribution with n = 4, \[p = \frac{1}{6}\] and \[q = \frac{5}{6}\] .
\[\therefore P\left( X = r \right) =^{4}{}{C}_r \left( \frac{5}{6} \right)^{4 - r} \left( \frac{1}{6} \right)^r\] where r = 0, 1, 2, 3 and 4
\[\Rightarrow P\left( X = r \right) =^{4}{}{C}_r \frac{5^{4 - r}}{6^4}\]
∴ P(0 success)
\[=\] P(X = 0)
\[=^{4}{}{C}_0 \cdot \frac{5^{4 - 0}}{6^4}\]
\[ = \frac{5^4}{6^4}\]
\[ = \frac{625}{1296}\]
P(1 success)
\[=\] P(X = 1)
\[=^{4}{}{C}_1 \cdot \frac{5^{4 - 1}}{6^4}\]
\[ = 4 \cdot \frac{5^3}{6^4}\]
\[ = \frac{125}{324}\]
P(2 successes)
\[=\] P(X = 2)
`= ^4C_2 5^(4-2)/6^4`
`= 6 25/1296`
`= 25/216`
P(3 successes)
\[=\] P(X = 3)
\[=^{4}{}{C}_3 \cdot \frac{5^{4 - 3}}{6^4}\]
\[ = 4 \cdot \frac{5}{6^4}\]
\[ = \frac{5}{324}\]
P(4 successes)
\[=\] P(X = 4)
\[= ^{4}{}{C}_4 \cdot \frac{5^{4 - 4}}{6^4}\]
\[ = 1 \cdot \frac{1}{6^4}\]
\[ = \frac{1}{1296}\]
Thus, the probability distribution of X is given by
0 \[\frac{625}{1296}\]
1 \[\frac{125}{324}\]
2 \[\frac{25}{216}\]
3 \[\frac{5}{324}\]
4 \[\frac{1}{1296}\]
\[\therefore \text { Mean of } X\]
\[ = \sum_{} x_i p_i \]
\[ = 0 \times \frac{625}{1296} + 1 \times \frac{125}{324} + 2 \times \frac{25}{216} + 3 \times \frac{5}{324} + 4 \times \frac{1}{1296}\]
\[ = \frac{125}{324} + \frac{50}{216} + \frac{15}{324} + \frac{1}{324}\]
\[ = \frac{2}{3}\]
Now,
\[\sum_{} {x_i}^2 p_i = 0^2 \times \frac{625}{1296} + 1^2 \times \frac{125}{324} + 2^2 \times \frac{25}{216} + 3^2 \times \frac{5}{324} + 4^2 \times \frac{1}{1296}\]
\[ = \frac{125}{324} + \frac{100}{216} + \frac{45}{324} + \frac{4}{324}\]
\[ = 1\]
\[\therefore \text { Var }\left( X \right) = \sum_{} {x_i}^2 p_i - \left( \sum_{} x_i p_i \right)^2 \]
\[ = 1 - \left( \frac{2}{3} \right)^2 \]
\[ = \frac{5}{9}\]
Thus, the mean and the variance of the distribution are
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