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Probability question

2022 · 25 Jun · Shift 1 · Q30
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  5. /2022 · 25 Jun · Shift 1 · Q30

Probability question

2022 · 25 Jun · Shift 1 · Q30

JEE MainMathematicsProbabilityMCQ+4 / −1
Let E1 and E2 be two events such that the conditional probabilities P(E1∣E2)=12P({E_1}|{E_2}) = {1 \over 2}P(E1​∣E2​)=21​, P(E2∣E1)=34P({E_2}|{E_1}) = {3 \over 4}P(E2​∣E1​)=43​ and P(E1∩E2)=18P({E_1} \cap {E_2}) = {1 \over 8}P(E1​∩E2​)=81​. Then :
  1. A
    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​)
  2. B
    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​)
  3. C
    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​)
  4. D
    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​)
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