In a game, a man wins Rs. 5 for getting a number greater than 4 and loses Rs. 1 otherwise, when a fair dice is thrown. The man decided to throw a die thrice but to quit as and when he gets a number greater than 4. The expected value of the amount (in Rs.) he wins or loses is:

19/9

The correct answer is Option (2) → 19/9

Let X be the amount won by the man.

Winning Rs. 5 if die shows 5 or 6 (probability = 2/6 = 1/3), losing Rs. 1 otherwise (probability = 4/6 = 2/3).

The man throws the die up to 3 times, stopping immediately if he gets a number greater than 4.

Compute expected value using probability tree:

Case 1: Wins on first throw:

Probability = $\frac{1}{3}$, Amount = 5

Contribution = $\frac{1}{3} \cdot 5 = \frac{5}{3}$

Case 2: First throw loses, second throw wins:

Probability = $\frac{2}{3} \cdot \frac{1}{3} = \frac{2}{9}$, Amount = -1 + 5 = 4

Contribution = $\frac{2}{9} \cdot 4 = \frac{8}{9}$

Case 3: First two throws lose, third throw wins:

Probability = $\frac{2}{3} \cdot \frac{2}{3} \cdot \frac{1}{3} = \frac{4}{27}$, Amount = -1 -1 + 5 = 3

Contribution = $\frac{4}{27} \cdot 3 = \frac{12}{27} = \frac{4}{9}$

Case 4: All three throws lose:

Probability = $(\frac{2}{3})^3 = \frac{8}{27}$, Amount = -1 -1 -1 = -3

Contribution = $\frac{8}{27} \cdot (-3) = -\frac{24}{27} = -\frac{8}{9}$

Expected value:

E(X) = $\frac{5}{3} + \frac{8}{9} + \frac{4}{9} - \frac{8}{9} = \frac{5}{3} + \frac{4}{9} = \frac{15}{9} + \frac{4}{9} = \frac{19}{9}$

E(X) = $\frac{19}{9} \approx 2.11$ Rs.