Question

In September, 60% days are rainy & 40% are sunny.Metrology department wrongly predicts 10% of the times in rainy days and 20% on sunny days.Weather forecast indicates a day to be sunny.What is the probability that the forecast will be proved wrong?

Answer 1

Let $R$ be the event that a day is rainy, $S$ be the event that a day is sunny

$$P(R)=0.6, P(S)=0.4, P(R)+P(S)=1$$

Let $W$ be the event that the forecast will be wrong

$$P(W|R)=0.1, P(W|S)=0.2$$

Then by using Bayes’ theorem

$$P(S|W)=\dfrac{P(W|S)P(S)}{P(W|R)P(R)+P(W|S)P(S)}=$$

$$=\dfrac{0.2(0.4)}{0.1(0.6)+0.2(0.4)}=\dfrac{4}{7}\approx0.5714$$

The probability that the forecast will be proved wrong is $\dfrac{4}{7}\approx0.5714$.