Problem Solving

Probability of rain on the first day = 2/5
Probability of rain on the last day = 4/5
Probability of rain on other days = 1/6
$$\Pr obability\ of\ not\ raining\ on\ first\ day=1-\frac{2}{5}=\frac{3}{5}$$
$$\Pr obability\ of\ not\ raining\ on\ last\ day=1-\frac{4}{5}=\frac{1}{5}$$
$$\Pr obability\ of\ not\ raining\ on\ other\ days=1-\frac{1}{6}=\frac{5}{6}$$
What is the probability that rain will occur on at least one of the 5 days?
P(of raining on at least 1 day) = 1 - P(of not raining at all)
$$=1-\left(\frac{3}{5}\cdot\frac{1}{5}\cdot\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}\right)$$
$$=1-\frac{5}{72}$$
$$=\frac{72-5}{72}=\frac{67}{72}\ \ \ \ \ \ Answer\ =option\ E$$