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pentagon [3]
3 years ago
5

Problem 1: Given the following probabilities P(Ac) = 0.45, P(B) = 0.34, and P(B|A) = 0.23. Find the following probabilities:

Mathematics
1 answer:
-BARSIC- [3]3 years ago
8 0

Answer:

P(A)=0.55

P(A and B)=P(A∩B)=0.1265

P(A or B)=P(A∪B)=0.7635

P(A|B)=0.3721

Step-by-step explanation:

P(A')=0.45

P(A)=1-0.45=0.55

P(B∩A)=?

P(B|A)=0.23

P(B|A)=(P(A∩B))/P(A)

0.23=(P(A∩B))/0.55

P(A∩B)=0.23×0.55=0.1265

P(A∪B)=P(A)+P(B)-P(A∩B)

=0.55+0.34-0.1265

=0.7635

P(A|B)=[P(A∩B)]/P(B)=0.1265/0.34 ≈0.3721

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Answer:

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Step-by-step explanation:

We need to simplify the expression (6^{-2})(3^{-3})(3*6)^4

Solving:

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Using exponent rule: (a^m)^n=a^{m\times n}

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Using exponent rule: a^m.a^n=a^{m+n}

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