Q. Past records reveal that during a particular was, out of 9 vessels expected to arrive at the Mumbai harbour exactly 7 reached the harbour safely. If 3 vessels were expected to arrive there on a particular data, the probability that exactly two would arrive at the harbour safely, is
A) $\frac{91}{243}$
B) $\frac{92}{243}$
C) $\frac{95}{243}$
D) None of these
Explanation
(d)
On the basis of past records, the probability of safe arrival of a vessel on Mumbai harbor is 7/9. Since, arrival of all the vessels is independent and there are only two possibilities on every namely safe arrival and unsafe arrival. We can use the binomial distribution. Here
$n = 3,\ p = \frac{7}{9},\ q = \frac{2}{9}\ and\ r = 2$
$\therefore P(2) = {\ ^{3}C}_{2}\left( \frac{7}{9} \right)^{2}\left( \frac{2}{9} \right) = \frac{98}{243}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.