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Probability question

2016 · 9 Apr · Shift 1 · Q37
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Probability question

2016 · 9 Apr · Shift 1 · Q37

JEE MainMathematicsProbabilityMCQ+4 / −1
If A and B are any two events such that P(A) = 25{2 \over 5}52​ and P (A ∩\cap∩ B) =320{3 \over {20}}203​, hen the conditional probability, P(A ∣\left| {} \right.∣(A' ∪\cup∪ B')), where A' denotes the complement of A, is equal to :
  1. A
    14{1 \over 4}41​
  2. B
    517{5 \over 17}175​
  3. C
    817{8 \over 17}178​
  4. D
    1120{11 \over 20}2011​
View written solutionFree

Correct answer: B

  1. We need to find P(A∣(A′∪B′)).P\big(A\mid (A'\cup B')\big).P(A∣(A′∪B′)).

  2. Use the definition of conditional probability: P(A∣E)=P(A∩E)P(E),P(A\mid E)=\frac{P(A\cap E)}{P(E)},P(A∣E)=P(E)P(A∩E)​, where here E=A′∪B′.E=A'\cup B'.E=A′∪B′.

So, P(A∣(A′∪B′))=P(A∩(A′∪B′))P(A′∪B′).P\big(A\mid (A'\cup B')\big)=\frac{P\big(A\cap (A'\cup B')\big)}{P(A'\cup B')}.P(A∣(A′∪B′))=P(A′∪B′)P(A∩(A′∪B′))​.

  1. First simplify the numerator: A∩(A′∪B′)=(A∩A′)∪(A∩B′).A\cap (A'\cup B')=(A\cap A')\cup (A\cap B').A∩(A′∪B′)=(A∩A′)∪(A∩B′). Since A∩A′=∅,A\cap A'=\varnothing,A∩A′=∅, we get A∩(A′∪B′)=A∩B′.A\cap (A'\cup B')=A\cap B'.A∩(A′∪B′)=A∩B′.

Thus, P(A∩(A′∪B′))=P(A∩B′).P\big(A\cap (A'\cup B')\big)=P(A\cap B').P(A∩(A′∪B′))=P(A∩B′). Now, P(A\cap B')=P(A)-P(A\cap B)=\frac25-\frac{3}{20}= rac{8}{20}-\frac{3}{20}=\frac{5}{20}=\frac14.

  1. Now simplify the denominator: P(A′∪B′)=P((A∩B)′).P(A'\cup B')=P((A\cap B)').P(A′∪B′)=P((A∩B)′). By De Morgan's law, A′∪B′=(A∩B)′.A'\cup B'=(A\cap B)'.A′∪B′=(A∩B)′. Hence, P(A′∪B′)=1−P(A∩B)=1−320=1720.P(A'\cup B')=1-P(A\cap B)=1-\frac{3}{20}=\frac{17}{20}.P(A′∪B′)=1−P(A∩B)=1−203​=2017​.

  2. Therefore, P(A∣(A′∪B′))=141720=14⋅2017=517.P\big(A\mid (A'\cup B')\big)=\frac{\frac14}{\frac{17}{20}}=\frac14\cdot\frac{20}{17}=\frac{5}{17}.P(A∣(A′∪B′))=2017​41​​=41​⋅1720​=175​.

  3. Match with options: 517\frac{5}{17}175​ is Option B.

Therefore, the correct answer is B.

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