Question

Consider a game of tossing a six sided fair die. If the face that comes up is \(6\), the player wins Rs. \(36\) and he loses Rs. \(k^2\), where \(k\) is the face that comes up \(k = \{1,2,3,4,5\}\), then the expected winning amount in this game in Rs. is...

• A. \(\frac{19}{6}\)
• B. \(-\frac{19}{6}\)
• C. \(\frac{3}{2}\)
• D. \(-\frac{3}{2}\)

Hint:

Find the expected value as the sum of each outcome times its probability.

Answer 1

The Correct Option is B

Step 1: Probability Table:
Winning $X$: $-1,-4,-9,-16,-25$ each with probability $\tfrac16$, and $+36$ with probability $\tfrac16$.

Step 2: Sum of Losses:
$1+4+9+16+25=55$, which is $\dfrac{5\cdot6\cdot11}6$, the sum of squares formula.

Step 3: Expectation:
$E=\dfrac{36-55}{6}=-\dfrac{19}{6}$. Option (B).

Final Answer:
Option (B). \[ \boxed{\text{(B) } -\frac{19}{6}} \]