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A die is tossed twice. Getting a number greater than 4 is considered a success. Then the variance of the probability distribution of the number of successes is
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The correct answer is:
${ }^{\frac{4}{9}}$
Obviously, $p=\frac{2}{6}=\frac{1}{3} \Rightarrow$ $q=1-\frac{1}{3}=\frac{2}{3}$
also $n=2$. Therefore, variance $=n p q=2 \times \frac{1}{3} \times \frac{2}{3}=\frac{4}{9}$.
also $n=2$. Therefore, variance $=n p q=2 \times \frac{1}{3} \times \frac{2}{3}=\frac{4}{9}$.
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