A six faced die is biased such that
$3 \times \mathrm{P}($a prime number$)\,=6 \times \mathrm{P}($a composite number$)\,=2 \times \mathrm{P}(1)$.
Let 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 X is :
Solution
<p>Let P(a prime number) = $\alpha$</p>
<p>P(a composite number) = $\beta$</p>
<p>and P(1) = $\gamma$</p>
<p>$\because$ $3\alpha = 6\beta = 2\gamma = k$ (say)</p>
<p>and $3\alpha + 2\beta + \gamma = 1$</p>
<p>$\Rightarrow k + {k \over 3} + {k \over 2} = 1 \Rightarrow k = {6 \over {11}}$</p>
<p>Mean = np where n = 2</p>
<p>and p = probability of getting perfect square</p>
<p>$= P(1) + P(4) = {k \over 2} + {k \over 6} = {4 \over {11}}$</p>
<p>So, mean $= 2\,.\,\left( {{4 \over {11}}} \right) = {8 \over {11}}$</p>
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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