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Probability question

2008 · Shift 0 · Q34
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Probability question

2008 · Shift 0 · Q34

JEE MainMathematicsProbabilityMCQ+4 / −1
It is given that the events AAA and BBB are such that P(A)=14,P(A∣B)=12P\left( A \right) = {1 \over 4},P\left( {A|B} \right) = {1 \over 2}P(A)=41​,P(A∣B)=21​ and P(B∣A)=23.P\left( {B|A} \right) = {2 \over 3}.P(B∣A)=32​. Then P(B)P(B)P(B) is :
  1. A
    16{1 \over 6}61​
  2. B
    13{1 \over 3}31​
  3. C
    23{2 \over 3}32​
  4. D
    12{1 \over 2}21​
View written solutionFree

Correct answer: B

  1. We are given: P(A)=14,P(A∣B)=12,P(B∣A)=23.P(A)=\frac{1}{4}, \quad P(A\mid B)=\frac{1}{2}, \quad P(B\mid A)=\frac{2}{3}.P(A)=41​,P(A∣B)=21​,P(B∣A)=32​.

  2. Use the definition of conditional probability: P(B∣A)=P(A∩B)P(A).P(B\mid A)=\frac{P(A\cap B)}{P(A)}.P(B∣A)=P(A)P(A∩B)​. Substituting the given values: 23=P(A∩B)1/4.\frac{2}{3}=\frac{P(A\cap B)}{1/4}.32​=1/4P(A∩B)​. Therefore, P(A∩B)=23⋅14=16.P(A\cap B)=\frac{2}{3}\cdot \frac{1}{4}=\frac{1}{6}.P(A∩B)=32​⋅41​=61​.

  3. Now use P(A∣B)=P(A∩B)P(B).P(A\mid B)=\frac{P(A\cap B)}{P(B)}.P(A∣B)=P(B)P(A∩B)​. Substitute the known values: 12=1/6P(B).\frac{1}{2}=\frac{1/6}{P(B)}.21​=P(B)1/6​.

  4. Solve for P(B)P(B)P(B): P(B)=1/61/2=16⋅2=13.P(B)=\frac{1/6}{1/2}=\frac{1}{6}\cdot 2=\frac{1}{3}.P(B)=1/21/6​=61​⋅2=31​.

  5. Hence, P(B)=13.P(B)=\frac{1}{3}.P(B)=31​.

  6. Comparing with the options:

    • A: 16\frac{1}{6}61​
    • B: 13\frac{1}{3}31​
    • C: 23\frac{2}{3}32​
    • D: 12\frac{1}{2}21​

    The correct option is B.

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