P(A)=0.56,P(B)=0.52,P(B∩A¯¯¯¯)=0.32P(A)=0.56,P(B)=0.52,P(B∩A¯)=0.32 Find the following probabilities. P(A)P(A) =  P(A∪B)P(A∪B) =  P(A∪B¯¯¯¯¯¯¯¯¯¯¯¯¯)P(A∪B¯) =  P(A∩B)P(A∩B) =  P(A¯¯¯¯)P(A¯) =  P(A∩B¯¯¯¯¯¯¯¯¯¯¯¯¯)P(A∩B¯) =  P(A¯¯¯¯∪B)P(A¯∪B) =

Can you do a, b, c?

Suppose you are deciding whether or not to bring an umbrella to school and the weathercan be either sunny or rainy. Initially, you think the probability that it will be sunny isP (S) = 0.6, and the probability that it will be rainy is P (R) = 0.4. If you bring an umbrellaand it rains, your payoff is 6. If you bring an umbrella and it is sunny, your payoff is 9. Ifyou don’t bring an umbrella and it rains, your payoff is 0. If you don’t bring an umbrellaand it is sunny, your payoff is 10. Suppose that you check two websites for weather forecasts.You know that the two forecasts are independent of each other, and each forecast is correct70% of the time. That is, the probability that the website says it will be sunny given that itactually will be sunny is 0.7, and the probability that the website says it will rain given thatit actually will rain is 0.7. Suppose that you suffer from confirmation bias. You initially thinkit is more likely to be sunny, so you misinterpret rain forecasts as sun…

first 2 parts please

For each set of probabilities, determine whether the events A and B are independent or dependent. (If necessary, consult a list of formulas.) Probabilities Independent Dependent = P(A 18) - 1 (a) P(A)=-;P(B) =;P(A\B) = 5 1 1 P(A) =;P (B) = P(A and B) = %3D 4 6. 1 1 P (B|A) = 1 (c) P(4)-극: P(B)-P(Bl4): %3D %3D 1 1 (d) P(A) = : P(B) = P(4 |B) = | %3D

Possible values of X, the number of components in a system submitted for repair that must be replaced, are 1, 2, 3, and 4 with corresponding probabilities 0.35, 0.15, 0.35, and 0.15, respectively. (a) Calculate E(X) and then E(5 - X). E(X) E(5-X) = (b) Would the repair facility be better off charging a flat fee of $70 or else the amount $ 150 (5-X) The repair facility ---Select--- be better off charging a flat fee of $70 because E ]= 150 (5-X)] ? Note: It is not generally true that : E ( =) = E(Y)

Possible values of X, the number of components in a system submitted for repair that must be replaced, are 1, 2, 3, and 4 with corresponding probabilities 0.45, 0.05, 0.45, and 0.05, respectively. (0) Calculate E(X) and then E(5 - X). E(X) - E(5 - X) - 150 (b) Would the repair facility be better off charging a flat fee of $65 or else the amounts (5- X). (Note: It is not generally true that E()- The repair facility Select--- be better off charging a flat fee of s65 because E150 [(5 - x)] -

P(A)=0.55,P(B)=0.58,P(B∪A)=0.86 Find the following probabilities. P(A∩B) =  P(B|A) =  P(A|B) =