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Probability question

2020 · 2 Sep · Shift 2 · Q38
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Probability question

2020 · 2 Sep · Shift 2 · Q38

JEE MainMathematicsProbabilityMCQ+4 / −1
Let EC denote the complement of an event E. Let E1 , E2 and E3 be any pairwise independent events with P(E1) > 0 and P(E1 ∩\cap∩ E2 ∩\cap∩ E3) = 0. Then P(E2C∩E3C/E1E_2^C \cap E_3^C/{E_1}E2C​∩E3C​/E1​) is equal to :
  1. A
    P(E3C)P\left( {E_3^C} \right)P(E3C​) - P(E2)
  2. B
    P(E2C)P\left( {E_2^C} \right)P(E2C​) + P(E3)
  3. C
    P(E3C)P\left( {E_3^C} \right)P(E3C​)-P(E2C)P\left( {E_2^C} \right)P(E2C​)
  4. D
    P(E3) - P(E2C)P\left( {E_2^C} \right)P(E2C​)
View written solutionFree

Correct answer: A

  1. We need to find P(E2C∩E3C∣E1).P(E_2^C\cap E_3^C\mid E_1).P(E2C​∩E3C​∣E1​).

Using conditional probability, P(E2C∩E3C∣E1)=P(E1∩E2C∩E3C)P(E1).P(E_2^C\cap E_3^C\mid E_1)=\frac{P(E_1\cap E_2^C\cap E_3^C)}{P(E_1)}.P(E2C​∩E3C​∣E1​)=P(E1​)P(E1​∩E2C​∩E3C​)​.

  1. Now simplify the numerator: E1∩E2C∩E3C=E1∖(E2∪E3).E_1\cap E_2^C\cap E_3^C = E_1\setminus (E_2\cup E_3).E1​∩E2C​∩E3C​=E1​∖(E2​∪E3​). So, P(E1∩E2C∩E3C)=P(E1)−P(E1∩(E2∪E3)).P(E_1\cap E_2^C\cap E_3^C)=P(E_1)-P(E_1\cap(E_2\cup E_3)).P(E1​∩E2C​∩E3C​)=P(E1​)−P(E1​∩(E2​∪E3​)).

Also, P(E1∩(E2∪E3))=P(E1∩E2)+P(E1∩E3)−P(E1∩E2∩E3).P(E_1\cap(E_2\cup E_3))=P(E_1\cap E_2)+P(E_1\cap E_3)-P(E_1\cap E_2\cap E_3).P(E1​∩(E2​∪E3​))=P(E1​∩E2​)+P(E1​∩E3​)−P(E1​∩E2​∩E3​). Given P(E1∩E2∩E3)=0,P(E_1\cap E_2\cap E_3)=0,P(E1​∩E2​∩E3​)=0, we get P(E1∩E2C∩E3C)=P(E1)−P(E1∩E2)−P(E1∩E3).P(E_1\cap E_2^C\cap E_3^C)=P(E_1)-P(E_1\cap E_2)-P(E_1\cap E_3).P(E1​∩E2C​∩E3C​)=P(E1​)−P(E1​∩E2​)−P(E1​∩E3​).

  1. Since E1,E2,E3E_1,E_2,E_3E1​,E2​,E3​ are pairwise independent, P(E1∩E2)=P(E1)P(E2),P(E_1\cap E_2)=P(E_1)P(E_2),P(E1​∩E2​)=P(E1​)P(E2​), P(E1∩E3)=P(E1)P(E3).P(E_1\cap E_3)=P(E_1)P(E_3).P(E1​∩E3​)=P(E1​)P(E3​). Hence, P(E1∩E2C∩E3C)=P(E1)−P(E1)P(E2)−P(E1)P(E3).P(E_1\cap E_2^C\cap E_3^C)=P(E_1)-P(E_1)P(E_2)-P(E_1)P(E_3).P(E1​∩E2C​∩E3C​)=P(E1​)−P(E1​)P(E2​)−P(E1​)P(E3​). Factor out P(E1)P(E_1)P(E1​): P(E1∩E2C∩E3C)=P(E1)(1−P(E2)−P(E3)).P(E_1\cap E_2^C\cap E_3^C)=P(E_1)\big(1-P(E_2)-P(E_3)\big).P(E1​∩E2C​∩E3C​)=P(E1​)(1−P(E2​)−P(E3​)).

  2. Therefore, P(E2C∩E3C∣E1)=P(E1)(1−P(E2)−P(E3))P(E1)=1−P(E2)−P(E3).P(E_2^C\cap E_3^C\mid E_1)=\frac{P(E_1)(1-P(E_2)-P(E_3))}{P(E_1)}=1-P(E_2)-P(E_3).P(E2C​∩E3C​∣E1​)=P(E1​)P(E1​)(1−P(E2​)−P(E3​))​=1−P(E2​)−P(E3​).

  3. Rewrite in terms of the options: Since P(E3C)=1−P(E3),P(E_3^C)=1-P(E_3),P(E3C​)=1−P(E3​), we have 1−P(E2)−P(E3)=P(E3C)−P(E2).1-P(E_2)-P(E_3)=P(E_3^C)-P(E_2).1−P(E2​)−P(E3​)=P(E3C​)−P(E2​). This matches option A.

  4. Check other options briefly:

  • B: P(E2C)+P(E3)=1−P(E2)+P(E3)P(E_2^C)+P(E_3)=1-P(E_2)+P(E_3)P(E2C​)+P(E3​)=1−P(E2​)+P(E3​), not equal in general.
  • C: P(E3C)−P(E2C)=(1−P(E3))−(1−P(E2))=P(E2)−P(E3)P(E_3^C)-P(E_2^C)=(1-P(E_3))-(1-P(E_2))=P(E_2)-P(E_3)P(E3C​)−P(E2C​)=(1−P(E3​))−(1−P(E2​))=P(E2​)−P(E3​), not equal.
  • D: P(E3)−P(E2C)=P(E3)−(1−P(E2))=P(E2)+P(E3)−1P(E_3)-P(E_2^C)=P(E_3)-(1-P(E_2))=P(E_2)+P(E_3)-1P(E3​)−P(E2C​)=P(E3​)−(1−P(E2​))=P(E2​)+P(E3​)−1, not equal.

Hence the correct answer is A.

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