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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Mathematics
Chapter
Probability

BITSAT 2009 · Mathematics · Probability

It is given that the events A and B are such that (P(A)= (1)/(4),...

It is given that the events AA and BB are such that (P(A)=14,\frac{1}{4},P(A \mid B)=12\frac{1}{2} \quad) and (\quad P(B \mid A)=23\frac{2}{3}). Then P(B)P(B) is

Options

  1. A(16\frac{1}{6})
  2. B(13\frac{1}{3})
  3. C(23\frac{2}{3})
  4. D(12\frac{1}{2})

Correct answer

B. (13\frac{1}{3})

Step-by-step solution

(P(A)=1 / 4, P(A / B)=12,B)=\frac{1}{2},P(B / A)=2 / 3) By conditional probability,P(A∩B)=P(A)P(B/A)=P(B)P(A/B)⇒14×23=P(B)×12\RightarrowP(B)=13\begin{aligned} & P(A \cap B)=P(A) P(B / A)=P(B) P(A / B) \\ & \Rightarrow \frac{1}{4} \times \frac{2}{3}=P(B) \times \frac{1}{2} \RightarrowP(B)=\frac{1}{3} \end{aligned}
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