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  • Two biased dice are thrown together. For the first die P (6) = 1/2. the other scores being equally likely while for the second die, P (1) = 2/5 and the other scores are equally likely. Find the probability distribution of ‘the number of ones seen’.

Two biased dice are thrown together. For the first die P (6) = 1/2. the other scores being equally likely while for the second die, P (1) = 2/5 and the other scores are equally likely. Find the probability distribution of ‘the number of ones seen’.

Answers (1)

Given-

\begin{aligned} &\text { For first die, }\\ &\mathrm{P}(6)=\frac{1}{2} \text { and } \mathrm{P}\left(6^{\prime}\right)=\frac{1}{2}\\ &\Rightarrow P(1)+P(2)+P(3)+P(4)+P(5)=\frac{1}{2}\\ &[\because \mathrm{P}(1)=\mathrm{P}(2)=\mathrm{P}(3)=\mathrm{P}(4)=\mathrm{P}(5)]\\ &\Rightarrow 5 \mathrm{P}(1)=\frac{1}{2}\\ &\Rightarrow \mathrm{P}(1)=\frac{1}{10}\\ &\Rightarrow \mathrm{P}\left(1^{\prime}\right)=1-\mathrm{P}(1)\\ &=1-\frac{1}{10}=\frac{9}{10} \end{aligned}

For the second die
P(1)=\frac{2}{5}$ and $P\left(1^{\prime}\right)=1-\frac{2}{5}=\frac{3}{5}
Let x  be the number of ones seen

For X=0,

\\\mathrm{P}(\mathrm{X}=0)=\mathrm{P}\left(1^{\prime}\right) \cdot\mathrm{P}\left(1^{\prime}\right)=\frac{9}{10} \times \frac{3}{5}=\frac{27}{50}=0.54$ \\$\mathrm{P}(\mathrm{X}=1)=\frac{9}{10} \times \frac{2}{5}+\frac{1}{10} \times \frac{3}{5}$ \\$=\frac{18}{50}+\frac{3}{50}=\frac{21}{50}$ \\$=0.42$ \\$\mathrm{P}(\mathrm{X}=2)=\mathrm{P}(1) \cdot \mathrm{P}(1)

\begin{aligned} &=\frac{1}{10} \times \frac{2}{5}\\ &=0.04\\ &\text { Hence, the required probability distribution is as follows }\\ &\begin{array}{|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 \\ \hline \mathrm{P}(\mathrm{X}) & 0.54 & 0.42 & 0.04 \\ \hline \end{array} \end{aligned}

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