Testbook-testseries-probability

Question

Comments

$\large \left ( \frac{2}{3} \right ) - \left (\frac{2}{3} \right )^2 = \frac{2}{9}$

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thank you

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@Mk Utkarsh , Can you post detailed solution for this?

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added below $\vee$

Answer 1

$P(Heads) =$$\large \frac{2}{3}$

$P(Tails) =$$\large \frac{1}{3}$

$X_i = 1, when \  heads$

$X_i = 0, when \  tails$

$E(X) = $ $\large 0\left ( \frac{1}{3} \right ) + 1\left ( \frac{2}{3} \right ) = \frac{2}{3}$

$E(X^2) = $$\large 0^2 \left ( \frac{1}{3} \right ) + 1^2\left ( \frac{2}{3} \right ) = \frac{2}{3}$

$Var(X) = E(X^2) - \left [E(X)  \right ]^2$

$Var(X) =$$\large \frac{2}{3} - \left [\frac{2}{3}  \right ]^2 = \frac{2}{9} $