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Probability question

2022 · 28 Jul · Shift 2 · Q34
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  5. /2022 · 28 Jul · Shift 2 · Q34

Probability question

2022 · 28 Jul · Shift 2 · Q34

JEE MainMathematicsProbabilityMCQ+4 / −1
Let A\mathrm{A}A and B\mathrm{B}B be two events such that P(B∣A)=25,P(A∣B)=17P(B \mid A)=\frac{2}{5}, P(A \mid B)=\frac{1}{7}P(B∣A)=52​,P(A∣B)=71​ and P(A∩B)=19⋅P(A \cap B)=\frac{1}{9} \cdotP(A∩B)=91​⋅ Consider (S1) P(A′∪B)=56P\left(A^{\prime} \cup B\right)=\frac{5}{6}P(A′∪B)=65​, (S2) P(A′∩B′)=118P\left(A^{\prime} \cap B^{\prime}\right)=\frac{1}{18}P(A′∩B′)=181​ Then :
  1. A
    Both (S1) and (S2) are true
  2. B
    Both (S1) and (S2) are false
  3. C
    Only (S1) is true
  4. D
    Only (S2) is true
View written solutionFree

Correct answer: A

  1. Given data

We have: P(B∣A)=25,P(A∣B)=17,P(A∩B)=19.P(B\mid A)=\frac{2}{5}, \qquad P(A\mid B)=\frac{1}{7}, \qquad P(A\cap B)=\frac{1}{9}.P(B∣A)=52​,P(A∣B)=71​,P(A∩B)=91​.

We use the conditional probability formulas: P(B∣A)=P(A∩B)P(A),P(A∣B)=P(A∩B)P(B).P(B\mid A)=\frac{P(A\cap B)}{P(A)}, \qquad P(A\mid B)=\frac{P(A\cap B)}{P(B)}.P(B∣A)=P(A)P(A∩B)​,P(A∣B)=P(B)P(A∩B)​.

  1. Find P(A)P(A)P(A)

From P(A∩B)P(A)=25,\frac{P(A\cap B)}{P(A)}=\frac{2}{5},P(A)P(A∩B)​=52​, we get 1/9P(A)=25.\frac{1/9}{P(A)}=\frac{2}{5}.P(A)1/9​=52​.

So, P(A)=19⋅52=518.P(A)=\frac{1}{9}\cdot\frac{5}{2}=\frac{5}{18}.P(A)=91​⋅25​=185​.

  1. Find P(B)P(B)P(B)

From P(A∩B)P(B)=17,\frac{P(A\cap B)}{P(B)}=\frac{1}{7},P(B)P(A∩B)​=71​, we get 1/9P(B)=17.\frac{1/9}{P(B)}=\frac{1}{7}.P(B)1/9​=71​.

So, P(B)=19⋅7=79.P(B)=\frac{1}{9}\cdot 7=\frac{7}{9}.P(B)=91​⋅7=97​.

  1. Check statement (S1): P(A′∪B)=56P(A'\cup B)=\frac{5}{6}P(A′∪B)=65​

Using De Morgan's law, A′∪B=(A∩B′)′.A'\cup B=(A\cap B')'.A′∪B=(A∩B′)′. So, P(A′∪B)=1−P(A∩B′).P(A'\cup B)=1-P(A\cap B').P(A′∪B)=1−P(A∩B′).

Now, P(A∩B′)=P(A)−P(A∩B)=518−19=518−218=318=16.P(A\cap B')=P(A)-P(A\cap B)=\frac{5}{18}-\frac{1}{9}=\frac{5}{18}-\frac{2}{18}=\frac{3}{18}=\frac{1}{6}.P(A∩B′)=P(A)−P(A∩B)=185​−91​=185​−182​=183​=61​.

Therefore, P(A′∪B)=1−16=56.P(A'\cup B)=1-\frac{1}{6}=\frac{5}{6}.P(A′∪B)=1−61​=65​.

So, (S1) is true.

  1. Check statement (S2): P(A′∩B′)=118P(A'\cap B')=\frac{1}{18}P(A′∩B′)=181​

We know P(A′∩B′)=1−P(A∪B).P(A'\cap B')=1-P(A\cup B).P(A′∩B′)=1−P(A∪B).

Now, 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). Substituting values: P(A∪B)=518+79−19.P(A\cup B)=\frac{5}{18}+\frac{7}{9}-\frac{1}{9}.P(A∪B)=185​+97​−91​.

Convert to denominator 181818: 79=1418,19=218.\frac{7}{9}=\frac{14}{18}, \qquad \frac{1}{9}=\frac{2}{18}.97​=1814​,91​=182​. Hence, P(A∪B)=518+1418−218=1718.P(A\cup B)=\frac{5}{18}+\frac{14}{18}-\frac{2}{18}=\frac{17}{18}.P(A∪B)=185​+1814​−182​=1817​.

Therefore, P(A′∩B′)=1−1718=118.P(A'\cap B')=1-\frac{17}{18}=\frac{1}{18}.P(A′∩B′)=1−1817​=181​.

So, (S2) is true.

  1. Conclusion

Both (S1) and (S2) are true.

Hence, the correct option is: A\boxed{\text{A}}A​

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