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  • A six faced die is biased such that <img alt="\mathrm{3 \times \mathrm{P}( a \: prime\: number )=6 \times \mathrm{P}(a\: composite\: number )=2 \times \mathrm{P}(1)}" src="/latex-image/?%5Cm

A six faced die is biased such that
\mathrm{3 \times \mathrm{P}( a \: prime\: number )=6 \times \mathrm{P}(a\: composite\: number )=2 \times \mathrm{P}(1)}.
Let \mathrm{X} be a random variable that counts the number of times one gets a perfect square on  some throws of this die. If the die is thrown twice, then the mean of \mathrm{X} is

Option: 1

\frac{3}{11}


Option: 2

\frac{5}{11}


Option: 3

\frac{7}{11}


Option: 4

\frac{8}{11}


Answers (1)

best_answer

\begin{aligned} &\text { Let }\mathrm{ P(1)=k} \\ &\therefore \mathrm{\quad P(2)=P(3)=P(5)=\frac{2 k}{3} }\\ &\text { and } \mathrm{P(4)=P(6)=\frac{k}{3} }\\ &\text { As Sum=1} \\ &\quad \mathrm{k+(2 k)+\frac{2 k}{3}=1} \\ &\quad \Rightarrow \quad\mathrm{ k=\frac{3}{11} }\\ &\text { Now } \mathrm{n=2} \\ &\text { and } \mathrm{P=P(1)+P(4)=k+\frac{k}{3}=\frac{4 k}{3}} \\ &=\frac{4}{11}\\ &\therefore \text { Mean }=\mathrm{np=2 \times \frac{4}{11}=\frac{8}{11}}\\ &\therefore \text{ Option(D)} \end{aligned}

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Ajit Kumar Dubey

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