Three ships A, B, and C sail from England to | vedclass

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Question

(A) Let $P(A)$, $P(B)$, and $P(C)$ be the probabilities of safe arrival for ships $A$, $B$, and $C$ respectively.Given odds in favour are $2:5$, $3:7$

Vedclass · Accepted Answer

$\frac{24}{119}$

Vedclass · Answer

(A) Let $P(A)$, $P(B)$, and $P(C)$ be the probabilities of safe arrival for ships $A$, $B$, and $C$ respectively.Given odds in favour are $2:5$, $3:7$, and $6:11$.$P(A) = \frac{2}{2+5} = \frac{2}{7}$, $P(A') = 1 - \frac{2}{7} = \frac{5}{7}$.$P(B) = \frac{3}{3+7} = \frac{3}{10}$, $P(B') = 1 - \frac{3}{10} = \frac{7}{10}$.$P(C) = \frac{6}{6+11} = \frac{6}{17}$, $P(C') = 1 - \frac{6}{17} = \frac{11}{17}$.Probability that exactly two ships arrive safely is $P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A' \cap B \cap C)$.$= (\frac{2}{7} 	imes \frac{3}{10} 	imes \frac{11}{17}) + (\frac{2}{7} 	imes \frac{7}{10} 	imes \frac{6}{17}) + (\frac{5}{7} 	imes \frac{3}{10} 	imes \frac{6}{17})$.$= \frac{66}{1190} + \frac{84}{1190} + \frac{90}{1190} = \frac{240}{1190} = \frac{24}{119}$.

Three ships $A$, $B$, and $C$ sail from England to India. If the odds in favour of their safe arrival are $2:5$, $3:7$, and $6:11$ respectively, then the probability that exactly two ships arrive safely is

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