Assume that A and B are events in a sample space and that P(A)=.40 and P(BA)=.25. What is the probability that both A and B occur?

Solution: According to a definition of P(BA) we have: P(BA)=P(B∩A)P(A), so we can substitute the numbers: 0.25=P(B∩A)0.4 P(B∩A)=0.25⋅0.4 P(B∩A)=0.1

Step 1 We use the definition of conditional probability: P(A∣B)=P(A∩B)P(B) So swapping the roles of A and B we get: P(B∣A)=P(B∩A)P(A) We are given that P(B∣A)=0.25, and P(A)=0.40, so: 0.25=P(A∩B)0.40⇒P(A∩B)}{0.40}⇒P(A∩B)=0.25×0.40=0.1

Solution: According to a definition of P(BA) we have: P(BA)=P(B∩A)P(A), so we can substitute the numbers: 0.25=P(B∩A)0.4 P(B∩A)=0.25⋅0.4 P(B∩A)=0.1

Step 1 We use the definition of conditional probability: P(A∣B)=P(A∩B)P(B) So swapping the roles of A and B we get: P(B∣A)=P(B∩A)P(A) We are given that P(B∣A)=0.25, and P(A)=0.40, so: 0.25=P(A∩B)0.40⇒P(A∩B)}{0.40}⇒P(A∩B)=0.25×0.40=0.1

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MMDCCVIII446

Answered question

2022-02-15

Assume that A and B are events in a sample space and that

P (A) = .40 and P (\frac{B}{A}) = .25

. What is the probability that both A and B occur?

Answer & Explanation

laiedkn

Beginner2022-02-16Added 15 answers

Solution:
According to a definition of

P (\frac{B}{A})

we have:

P (\frac{B}{A}) = \frac{P (B \cap A)}{P (A)}

, so we can substitute the numbers:

0.25 = \frac{P (B \cap A)}{0.4}

P (B \cap A) = 0.25 \cdot 0.4

P (B \cap A) = 0.1

Deacon Osborn

Beginner2022-02-17Added 18 answers

Step 1
We use the definition of conditional probability:

P (A ∣ B) = \frac{P (A \cap B)}{P (B)}

So swapping the roles of A and B we get:

P (B ∣ A) = \frac{P (B \cap A)}{P (A)}

We are given that

P (B ∣ A) = 0.25

, and

P (A) = 0.40

, so:

0.25 = \frac{P (A \cap B)}{0.40} \Rightarrow P (A \cap B)} {0.40} \Rightarrow P (A \cap B) = 0.25 \times 0.40 = 0.1

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