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(D) Let $p$ denote the probability that a ship departed at $A$ reaches $B$ safely. Then,$p = \frac{9}{10} = 0.9$. Therefore,$q = 1 - p = 1 - 0.9 = 0.1

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$1.4(0.9)^4$

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(D) Let $p$ denote the probability that a ship departed at $A$ reaches $B$ safely. Then,$p = \frac{9}{10} = 0.9$. Therefore,$q = 1 - p = 1 - 0.9 = 0.1$. Let $X$ denote the number of ships that reach $B$ safely out of $n = 5$ ships. This follows a binomial distribution $B(n, p) = B(5, 0.9)$. The required probability is $P(X \geq 4) = P(X = 4) + P(X = 5)$. Using the formula $P(X = k) = {}^nC_k p^k q^{n-k}$: $P(X = 4) = {}^5C_4 (0.9)^4 (0.1)^1 = 5 	imes (0.9)^4 	imes 0.1 = 0.5 	imes (0.9)^4$. $P(X = 5) = {}^5C_5 (0.9)^5 (0.1)^0 = 1 	imes (0.9)^5 	imes 1 = 0.9 	imes (0.9)^4$. Adding these probabilities: $P(X \geq 4) = 0.5(0.9)^4 + 0.9(0.9)^4 = (0.5 + 0.9)(0.9)^4 = 1.4(0.9)^4$.

On an average,nine out of $10$ ships that have departed at $A$ reach $B$ safely. The probability that out of five ships that have departed at $A$,at least four will reach $B$ safely is

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