What is the probability of B if they are independent events P(A)= 3/7, P(A then B)=1/3?

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What is the probability of B if they are independent events P(A)= 3/7, P(A then B)=1/3?

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Answer 1 · 2017-04-02T10:04:04

$\frac{7}{9}$ $7/9$

Explanation:

$P (A \to B) = P (A) \cdot P (B)$ $P(A->B)=P(A)*P(B)$

$\frac{1}{3} = \frac{3}{7} \cdot P (B)$ $1/3=3/7*P(B)$

$P (B) = \frac{\frac{1}{3}}{\frac{3}{7}} = \frac{7}{9}$ $P(B)=(1/3)/(3/7)=7/9$

Answer 2 · 2017-04-02T10:12:01

$P (B) = \frac{1}{3} .$ $P(B)=1/3.$

Explanation:

A Clarification : I interpret $P (A then B) a s, P (\frac{B}{A})$ $P(A" then "B) as, P(B/A)$ , i.e., the

Cond. Prob. of an event $B,$ $B,$ knowing that the event $A$ $A$ has

already occurred.

So, if the events $A and B$ $A and B$ are independent,

$P (\frac{B}{A}) = P (B) = \frac{1}{3} .$ $P(B/A)=P(B)=1/3.$

In an otherway round, if we define, Independence of events

$A and B \Leftrightarrow P (A \cap B) = P (A) \cdot P (B),$ $A and B iff P(AnnB)=P(A)*P(B),$ we get the same reult as follows:

$P (A then B) = P (\frac{B}{A}) = \frac{P (B \cap A)}{P (A)} = \frac{P (B) \cdot P (A)}{P (A)} = P (B) .$ $P(A" then "B)=P(B/A)=(P(BnnA))/(P(A))={P(B)*P(A)}/(P(A))=P(B).$

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What is the probability of B if they are independent events P(A)= 3/7, P(A then B)=1/3?

2 Answers

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