Suppose that we have a sample space S = {E1, E2, E3, E4, E5}, where E1, E2, E3, E4, E5 denote the sample points. The following probability assignments apply P(E1) = 0.1, P(E2) = 0.15, P(E3) = 0.2, P(E4) = 0.25, and P(E5) = 0.3. Let A = {E1,E2}, B = {E2,E3}, and C = {E1,E4,E5}. Find P (Ac ∪ B).

P(AC∪B)=0.9+0.35=1.25

Step-by-step explanation:

I am not sure if A = {E1, E2} and others are combined probability, but try either way and see if it works or not. Are events combined or mutually exclusive?

1. Find P(Ac ∪ B)

Use probability formula (Mutually exclusive): P(A∪B)=P(A)+P(B)

2. Find P(A) and P(B) and P(C)

P(A) = 0.1 * 0.15 or try 0.1 + 0.15

P(A) = 0.0015 or 0.25

P(B) = 0.15 * 0.2 or try 0.15 + 0.2

P(B) = 0.03 or 0.35

P(C) = 0.1 * 0.25 * 0.3 or try 0.1 + 0.25 + 0.35

P(C) = 0.0075 or 0.65

P(AC) = 1.125*10^-5 or 0.9

3. Find answer

P(AC∪B)=P(AC)+P(B)

P(AC∪B)=1.125*10^-5+0.03=0.03

OR

P(AC∪B)=0.9+0.35=1.25

step-by-step explanation: the answer is 27. first multiply .32 by 25 which equals 8. then add 8 to 25 which gives you 33. then you subtract that from 6 since she gave 6 to her brother.

,, add !

its 60

4 x 12 = 48

3 x 4 = 12

48 + 12 = 60

yeeeet