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:

For a fair dice:

P(getting > 4) = P(getting 5 or 6) = $\frac{2}{6} = \frac{1}{3}$

P(getting ≤ 4) = P(getting 1, 2, 3, or 4) = $\frac{4}{6} = \frac{2}{3}$

---

Scenario 1: Wins on 1st throw

Gets a number > 4 right away

Probability: $\frac{1}{3}$

Money: Rs. 5

---

Scenario 2: Loses 1st, Wins on 2nd throw

Gets ≤ 4, then gets > 4

Probability: $\frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$

Money: $-1 + 5 =$ Rs. 4

---

Scenario 3: Loses 1st, Loses 2nd, Wins on 3rd throw

Gets ≤ 4, then ≤ 4, then > 4

Probability: $\frac{2}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{27}$

Money: $-1 - 1 + 5 =$ Rs. 3

---

Scenario 4: Loses all 3 throws

Gets ≤ 4 three times in a row

Probability: $\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \frac{8}{27}$

Money: $-1 - 1 - 1 =$ Rs. $-3$

---

Expected Value:

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

Converting to common denominator 27:

$E(X) = \frac{9}{27} \times 5 + \frac{6}{27} \times 4 + \frac{4}{27} \times 3 + \frac{8}{27} \times (-3)$

$E(X) = \frac{45}{27} + \frac{24}{27} + \frac{12}{27} - \frac{24}{27}$

$E(X) = \frac{45 + 24 + 12 - 24}{27}$

$E(X) = \frac{57}{27}$

$E(X) = \frac{19}{9}$

Therefore, the expected value is $\frac{19}{9}$ Rs.