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(A) Let $p$ be the probability that a person is left-handed,so $p = 0.1$. Let $q$ be the probability that a person is not left-handed,so $q = 1 - p =

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$(0.9)^9$

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(A) Let $p$ be the probability that a person is left-handed,so $p = 0.1$. Let $q$ be the probability that a person is not left-handed,so $q = 1 - p = 0.9$. We use the binomial distribution formula for $n = 10$ trials and $k = 1$ success: $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$ Substituting the values: $P(X = 1) = {}^{10}C_{1} (0.1)^{1} (0.9)^{10-1}$ $P(X = 1) = 10 	imes 0.1 	imes (0.9)^9$ $P(X = 1) = 1 	imes (0.9)^9 = (0.9)^9$

The probability that a person chosen at random is left-handed (in handwriting) is $0.1$. Then the probability that in a group of $10$ people there is exactly one left-handed person is:

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