Question:medium

In a game, 3 coins are tossed. A person is paid Rs \( 150 \) if he gets all heads or all tails and he is supposed to pay ₹\( 50 \) if he gets one head or two heads. The amount he can expect to win lose on an average per game in ₹ is

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A fair game is one where the expected value of the winnings is zero.
Updated On: May 14, 2026
  • 100
  • 0
  • 200
  • -100
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
We need to calculate the expected value of a game.
The expected value is the sum of all possible outcomes multiplied by their respective probabilities.
Step 2: Key Formula or Approach:
Expected Value \( E(X) = \sum x_i \cdot P(x_i) \), where \( x_i \) is the payoff and \( P(x_i) \) is its probability.
For 3 coins, the total number of outcomes is \( 2^3 = 8 \).
Step 3: Detailed Explanation:
The sample space for tossing 3 coins is:
\( \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\} \)
Total outcomes = 8.
Case 1: He gets all heads or all tails.
Outcomes = \( \{HHH, TTT\} \) (2 outcomes).
Probability \( P_1 = \frac{2}{8} = \frac{1}{4} \).
Payoff \( x_1 = +150 \) Rs.
Case 2: He gets one head or two heads.
Outcomes = \( \{HHT, HTH, THH, HTT, THT, TTH\} \) (6 outcomes).
Probability \( P_2 = \frac{6}{8} = \frac{3}{4} \).
Payoff \( x_2 = -50 \) Rs (since he has to pay, it's a loss).
Calculate Expected Value:
\[ E(X) = (x_1 \cdot P_1) + (x_2 \cdot P_2) \] \[ E(X) = \left( 150 \times \frac{1}{4} \right) + \left( -50 \times \frac{3}{4} \right) \] \[ E(X) = \frac{150}{4} - \frac{150}{4} \] \[ E(X) = 0 \] The expected amount is 0 Rs.
Step 4: Final Answer:
The expected win/lose amount is 0.
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