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...

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Find the expected value as the sum of each outcome times its probability.
Updated On: Oct 1, 2026
  • \(\frac{19}{6}\)
  • \(-\frac{19}{6}\)
  • \(\frac{3}{2}\)
  • \(-\frac{3}{2}\)
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The Correct Option is B

Solution and Explanation

Step 1: Setup:
Each face has probability \(\tfrac16\). If the face is 6 the player wins Rs. 36. For \(k=1,2,3,4,5\), he loses Rs. \(k^2\), so his winning is \(-k^2\).

Step 2: Expected Value:
\[ E=\frac16\left[36-(1+4+9+16+25)\right]=\frac16(36-55) \]

Step 3: Compute:
\[ E=\frac{-19}6 \]

Step 4: Check the Options:
A positive value like \(\tfrac{19}6\) comes from reversing the sign of the loss. \(\tfrac32\) and \(-\tfrac32\) do not come from this sum (they would need totals of \(+9\) and \(-9\)). Only \(-\tfrac{19}6\) fits, option (B).

Final Answer:
The expected winning is \(-\dfrac{19}6\) rupees, option (B). \[ \boxed{\text{(B) } -\frac{19}{6}} \]
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