A) \[\frac{18}{595}\]
B) \[\frac{6}{17}\]
C) \[\frac{3}{10}\]
D) \[\frac{2}{7}\]
Correct Answer: A
Solution :
We have ratio of the ships A, B and C for arriving safely are 2 : 5, 3 : 7 and 6 : 11 respectively. \The probability of ship A for arriving safely \[=\frac{2}{2+5}=\frac{2}{7}\] Similarly, for \[B=\frac{3}{3+7}=\frac{3}{10}\] and for C = \[\frac{6}{6+11}=\frac{6}{17}\] \[\therefore\] Probability of all the ships for arriving safely = \[\frac{2}{7}\times \frac{3}{10}\times \frac{6}{17}=\frac{18}{595}\] .
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