Question

a certain city has one chance in two of receiving rain on june 1, one chance in five of receiving rain on july 1, and two chances in three of receiving rain on august 1. what is the probability that the city will receive rain on none of these days?

Answer 1

Probability of no rain on June 1  = 1 - 1/2 = 1/2.
Probability of no rain on July 1 = 1 - 1/5 = 4/5.
Probability of no rain on August 1 = 1 - 2/3 = 1/3.
The probability that the city will receive rain on none of these days is:
$\frac{1}{2}\times\frac{4}{5}\times\frac{1}{3}=\frac{2}{15}$
The required value of probability is 2/15.