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Probability question

2020 · 8 Jan · Shift 2 · Q23
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Probability question

2020 · 8 Jan · Shift 2 · Q23

JEE MainMathematicsProbabilityMCQ+4 / −1
Let A and B be two events such that the probability that exactly one of them occurs is 25{2 \over 5}52​ and the probability that A or B occurs is 12{1 \over 2}21​ , then the probability of both of them occur together is :
  1. A
    0.20
  2. B
    0.02
  3. C
    0.01
  4. D
    0.10
View written solutionFree

Correct answer: D

  1. Let P(A∩B)=x.P(A\cap B)=x.P(A∩B)=x.

  2. The probability that exactly one of AAA and BBB occurs is P(A⊕B)=P(A∪B)−P(A∩B).P(A\oplus B)=P(A\cup B)-P(A\cap B).P(A⊕B)=P(A∪B)−P(A∩B). This is because in P(A∪B)P(A\cup B)P(A∪B), the region A∩BA\cap BA∩B is included once, but for “exactly one” we must exclude it.

    More directly, P(exactly one)=P(A)+P(B)−2P(A∩B).P(\text{exactly one})=P(A)+P(B)-2P(A\cap B).P(exactly one)=P(A)+P(B)−2P(A∩B). Also, P(A∪B)=P(A)+P(B)−P(A∩B).P(A\cup B)=P(A)+P(B)-P(A\cap B).P(A∪B)=P(A)+P(B)−P(A∩B). Hence, P(exactly one)=P(A∪B)−P(A∩B).P(\text{exactly one})=P(A\cup B)-P(A\cap B).P(exactly one)=P(A∪B)−P(A∩B).

  3. Given: P(exactly one)=25,P(A∪B)=12.P(\text{exactly one})=\frac{2}{5}, \qquad P(A\cup B)=\frac{1}{2}.P(exactly one)=52​,P(A∪B)=21​.

    So, 25=12−x.\frac{2}{5}=\frac{1}{2}-x.52​=21​−x.

  4. Solve for xxx: x=12−25x=\frac{1}{2}-\frac{2}{5}x=21​−52​ x=5−410=110=0.10.x=\frac{5-4}{10}=\frac{1}{10}=0.10.x=105−4​=101​=0.10.

  5. Therefore, the probability that both events occur together is P(A∩B)=0.10.P(A\cap B)=0.10.P(A∩B)=0.10.

  6. Checking options:

    • A: 0.200.200.20 ❌
    • B: 0.020.020.02 ❌
    • C: 0.010.010.01 ❌
    • D: 0.100.100.10 ✅

Hence, the correct answer is D.

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