The probability distribution of the number of doublets in four throws of a pair of dice is given by
The probability distribution of the number of doublets in four throws of a pair of dice is given by
- \begin{array}{|l|l|l|l|l|l|}
\hline \mathrm{X}: & 0 & 1 & 2 & 3 & 4 \\
\hline \mathrm{P}(\mathrm{X}): & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} \\
\hline
\end{array}
- \begin{array}{|l|l|l|l|l|}
\hline \mathrm{X}: & 0 & 1 & 2 & 3 \\
\hline \mathrm{P}(\mathrm{X}): & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\
\hline
\end{array}
- \begin{array}{|l|l|l|l|l|l|}
\hline \mathrm{X}: & 0 & 1 & 2 & 3 & 4 \\
\hline \mathrm{P}(\mathrm{X}): & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\
\hline
\end{array}
- \begin{array}{|l|l|l|l|l|l|}
\hline \mathrm{X}: & 0 & 1 & 2 & 3 & 4 \\
\hline \mathrm{P}(\mathrm{X}): & \frac{625}{1296} & \frac{125}{324} & \frac{25}{216} & \frac{5}{324} & \frac{1}{1296} \\
\hline
\end{array}
Solution
Let $\mathrm{p}=$ Probability of getting a doublet in a throw of pair of dice.
$\mathrm{p}=\frac{6}{36}=\frac{1}{6} \text { and } \mathrm{q}=1-\frac{1}{6}=\frac{5}{6}$
Dice are thrown 4 times.
Let $\mathrm{X}=0,1,2,3,4$ denote number of times a doublet is obtained.
When $\mathrm{X}=0, \mathrm{p}=\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}=\frac{625}{1296}$
When $\mathrm{X}=1, \mathrm{p}=\left({ }^4 \mathrm{C}_1\right) \times \frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}=\frac{125}{324}$
$\begin{aligned}
& X=2, p=\left({ }^4 C_2\right) \times \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} \times \frac{5}{6}=\frac{25}{216} \\
& X=3, p=\left({ }^4 C_3\right) \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6}=\frac{5}{324} \\
& X=4, p=\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}=\frac{1}{1296}
\end{aligned}$
Asked in: MHT CET 2021 (23 Sep Shift 2)
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