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

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

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

answer: x= -2y

step-by-step explanation: