MHTCET 10 May 2024 Shift 1 Paper

Concept:Expected value is the weighted average of all possible outcomes: $E(X) = \sum x_i P(X = x_i)$.Explanation:When 3 coins are tossed, total outcomes are $2^3 = 8$.All heads or all tails means only HHH and TTT, i.e., 2 favourable outcomes.$P(\text{all heads or all tails}) = \frac{2}{8} = \frac{1}{4}$One head or two heads covers the remaining 6 outcomes.$P(\text{one head or two heads}) = \frac{6}{8} = \frac{3}{4}$Let $X$ be the amount received per game.$P(X = 100) = \frac{1}{4}$, since he wins ₹100 in this case.$P(X = -40) = \frac{3}{4}$, since he pays ₹40 in this case.Now, expected value is:$$E(X) = 100 \times \frac{1}{4} + (-40) \times \frac{3}{4}$$$$E(X) = 25 - 30 = -5$$Thus, on average, he loses ₹5 per game.Answer:₹5 loss (Option B).