A fair coin is tossed 9 times. On each toss, | vedclass

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(C) Let $X$ be the number of successful predictions (heads). Since the coin is fair, the probability of heads is $p = 1/2$ and tails is $q = 1/2$. The

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$1/2$

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(C) Let $X$ be the number of successful predictions (heads). Since the coin is fair, the probability of heads is $p = 1/2$ and tails is $q = 1/2$. The number of tosses is $n = 9$.$X$ follows a binomial distribution $B(9, 1/2)$.We want the probability that the number of successful predictions is strictly greater than the number of unsuccessful predictions.Let $S$ be the number of successes and $F$ be the number of failures. We have $S + F = 9$ and we want $S > F$.Since $F = 9 - S$, the condition $S > 9 - S$ implies $2S > 9$, or $S > 4.5$.Since $S$ must be an integer, we need $S \in \{5, 6, 7, 8, 9\}$.Because the distribution is symmetric for $p = 1/2$, $P(S \ge 5) = P(S \le 4)$.Since $P(S \le 9) = 1$, we have $P(S \le 4) + P(S \ge 5) = 1$.Therefore, $2 \cdot P(S \ge 5) = 1$, which gives $P(S \ge 5) = 1/2$.

$A$ fair coin is tossed $9$ times. On each toss, a man predicts that the outcome will be heads. The probability that the number of successful predictions is strictly greater than the number of unsuccessful predictions is...

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