Each day, a weather forecaster predicts whether or not it will rain. For 80% of rainy days, she correctly predicts that it will rain. For 90% of non-rainy days, she correctly predicts that it will not rain. Suppose that 10% of all days are rainy and 90% of all days are not rainy. (a) What proportion of the forecasts are correct? (b) Another forecaster always (100%) predicts that there will be no rain. What proportion of these forecasts are correct?

Proportion of correct forecast for first forecaster = 0.89 i. e. 89/100

For second forecaster proportion of correct forecast = 0.9 i. e. 90/100

Step-by-step explanation:

Consider,

Events of rainy days = R₁

Events of non-rainy days = R₂

Events of correct forecast = C

A) for first forecaster:

correct forecast for rainy days = 80%

P (C|R₁) = 0.8

correct forecast for non-rainy days = 90%

P (C|R₂) = 0.9

%age rainy days = 10%

P (R₁) = 0.1

%age of non-rainy days = 90%

P (R₂) = 0.9

Using Baye's formula of conditional probability,

proportion of correct forecast = P (C) = P (C|R₁) * P (R₁) + P (C|R₂) * P (R₂)

= (0.8) (0.1) + (.9) (0.9)

= 0.89

i. e. proportion of correct forecast for first forecaster = 89/100

B) for second forecaster:

forecast for non-rainy days = 100%

P (C|R₂) = 0.9

forecast for non-rainy days = 0%

P (C|R₁) = 0

Using Baye's formula of conditional probability,

proportion of correct forecast = P (C) = P (C|R₁) * P (R₁) + P (C|R₂) * P (R₂)

= (0) (0.1) + (1) (0.9)

= 0.90

i. e. proportion of correct forecast for first forecaster = 90/100