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laila [671]
3 years ago
7

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 p

robability 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).
Mathematics
1 answer:
Zepler [3.9K]3 years ago
5 0

Answer:

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

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4 years ago
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The answer is the number that 'Y' must be in order for
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That number is called the 'solution' to the equation.
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4 years ago
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