Question

in a certain city where it rains frequently, records have been kept and relative frequencies have been used to estimate these probabilities. the probability that rain is predicted on a day is 0.2. the probability that it actually rains on a day that rain is predicted is 0.9. the probability that it actually rains on a day that rain is not predicted is 0.3. for a randomly selected day, what is the probability that the prediction is correct? group of answer choices 0.74 0.18 0.56

Answer 1

0.74

Explanation:

Let X be the event that it actually rains

Let Y be the event that rain is predicted

We are given

P(Y) = 0.2 probability that rain is predicted

P(X|Y) = 0.9 probability that it rains given rain is predicted

P(X|Y') =0.3 probability that it rains given rain is not predicted
Y' stands for the complement event for Y: not predicted

We are asked to find the probability that the prediction is correct.

This can occur in two ways
- it rains after prediction of rain P(X and Y)
- it does not rain after a prediction of no rain(X' and Y')

The sum of the above two probabilities is the required probability

P(correct prediction) = P(X and Y) + P(X' and Y')

Computing the individual probabilities:

P(X and Y) = P(Y) P(X|Y)= 0.2 x 0.9 = 0.18

P(x' and Y') = P(Y') P(X'|Y')

P(Y') = 1 - P(Y) = 1 - 0.2 = 0.8

P(X'|Y') = 1 - P(X|y') = 1 - 0.3 = 0.7
P(x' and Y') = P(Y') P(X'|Y') = 0.8 x 0.7 = 0.56

P(correct prediction) = P(X and Y) + P(X' and Y')

= 0.18 + 0.56 = 0.74

It is much easier if you draw a tree diagram and list the probabilities along the branch.