In london, half of the days have some rain. the weather forecaster is correct 2/3 of thetime, i. e., the probability that it rains, given that she has predicted rain, and the probabilitythat it does not rain, given that she has predicted that it won’t rain, are both equal to 2/3.when rain is forecast, mr. pickwick takes his umbrella. when rain is not forecast, he takesit with probability 1/3. find (a) the probability that pickwick has no umbrella, given that itrains. (b) the probability that it doesn’t rain, given that he brings his umbrella.

(a) The probability that Pickwick has no umbrella, given that it rains is 2/9

(b) The probability that it doesn’t rain, given that he brings his umbrella is 5/9

Step-by-step explanation:

(a) The probability that Pickwick has no umbrella, given that it rains: P1

P1 = P (rain) * P (not umbrella)

This probability have two components, if rain forecasts or not rain forecasts:

P1 = 2/3*0 + 1/3*2/3 = 0 + 2/9 = 2/9

The probability that Pickwick has no umbrella, given that it rains is 2/9

(b) The probability that it doesn’t rain, given that he brings his umbrella: P2

P2 = P (not rain) * P (umbrella)

This probability have two components, if rain forecasts or not rain forecasts:

P1 = 1/3*1 + 2/3*1/3 = 1/3 + 2/9 = 5/9

The probability that it doesn’t rain, given that he brings his umbrella is 5/9

The other person is right

Step-by-step explanation:

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.