Three ships A, B, 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 the exactly two ships arrive safely is

Let $P(A)$, $P(B)$, and $P(C)$ be the probabilities of safe arrival for ships A, B, and C respectively.

Given the odds in favour of their safe arrival, we can find their probabilities:
$P(A) = \dfrac{2}{2+5} = \dfrac{2}{7} \Rightarrow P(\bar{A}) = 1 - \dfrac{2}{7} = \dfrac{5}{7}$

$P(B) = \dfrac{3}{3+7} = \dfrac{3}{10} \Rightarrow P(\bar{B}) = 1 - \dfrac{3}{10} = \dfrac{7}{10}$

$P(C) = \dfrac{6}{6+11} = \dfrac{6}{17} \Rightarrow P(\bar{C}) = 1 - \dfrac{6}{17} = \dfrac{11}{17}$

The probability that exactly two ships arrive safely is the sum of the probabilities of three mutually exclusive cases: A and B arrive but C does not, A and C arrive but B does not, and B and C arrive but A does not.

Required probability $= P(A)P(B)P(\bar{C}) + P(A)P(\bar{B})P(C) + P(\bar{A})P(B)P(C)$

Substituting the values:
$= \left(\dfrac{2}{7} \times \dfrac{3}{10} \times \dfrac{11}{17}\right) + \left(\dfrac{2}{7} \times \dfrac{7}{10} \times \dfrac{6}{17}\right) + \left(\dfrac{5}{7} \times \dfrac{3}{10} \times \dfrac{6}{17}\right)$

$= \dfrac{66}{1190} + \dfrac{84}{1190} + \dfrac{90}{1190}$

$= \dfrac{66 + 84 + 90}{1190}$

$= \dfrac{240}{1190} = \dfrac{24}{119}$

Answer: $\dfrac{24}{119}$