In a game, a man wins ₹5 for getting a number greater than 4 and lose…

The correct answer is Option (3) → $2\frac{1}{9}$

Let X (in ₹) denote the amount which the man wins or loses.

As the man throws a die atmost thrice and quits the game as and when he gets a number greater than 4, the sample space of the random experiment is

S = {W, LW, LLW, LLL} where W denotes win and L denotes loss.

The possible values of X are:

$X = 5$ when he gets a number greater than 4 in the first throw of die,

$X = 4$ when he gets a number greater than 4 in the second throw of die,

$X=3$ when he gets a number greater than 4 in the third throw of die,

$X=-3$ when he does not get a number greater than 4 in any of three throws of die.

Here, -3 indicates the loss of 3.

P(getting a number greater than 4) = P(getting 5 or 6) = $\frac{2}{6}=\frac{1}{3}$

∴ P(not getting a number greater than 4) = $1-\frac{1}{3}=\frac{2}{3}$

$P(X=5)=\frac{1}{3},P(X=4)=\frac{2}{3}×\frac{1}{3}=\frac{2}{9}$,

$P(X=3)=\frac{2}{3}×\frac{2}{3}×\frac{1}{3}=\frac{4}{27},P(X=-3)=\frac{2}{3}×\frac{2}{3}×\frac{2}{3}=\frac{8}{27}$.

$∴ E(X)=∑p_ix_i=\frac{1}{3}×5+\frac{2}{9}×4+\frac{4}{27}×3+\frac{8}{27}×(-3)$

$=\frac{5}{8}+\frac{8}{9}+\frac{4}{9}-\frac{8}{9}=\frac{19}{9}$.

∴ The expected amount he wins = $₹\frac{19}{9}=₹2\frac{1}{9}$