Let E and F be two independent events such that P(E)P(F) . The probability that both E and F happened is (1/12) and the probability that neither E nor F happens is (1/2), then

Question

Let E and F be two independent events such that P(E)>P(F) . The probability that both E and F happened is (1/12) and the probability that neither E nor F happens is (1/2), then (A) P(E)=(1/3),P(F)=(1/4) (B) P(E)=(1/2),P(F)=(1/6) (C) P(E)=(1/2),P(F)=(1/8) (D) none of these

Tardigrade Admin · Accepted Answer

P E ∩ F=P E ⋅ P F=(1/12), P E prime ∩ F prime=(1/2) 1-P E 1-P F=(1/2)

Q. Let $E$ and $F$ be two independent events such that $P (E) > P (F)$ . The probability that both $E$ and $F$ happened is $\frac{1}{12}$ and the probability that neither $E$ nor $F$ happens is $\frac{1}{2},$ then

$P (E) = \frac{1}{3}, P (F) = \frac{1}{4}$

$P (E) = \frac{1}{2}, P (F) = \frac{1}{6}$

$P (E) = \frac{1}{2}, P (F) = \frac{1}{8}$

none of these

$PE \cap F = PE \cdot PF = \frac{1}{12}, P E^{'} \cap F^{'} = \frac{1}{2}$
$1 - PE 1 - PF = \frac{1}{2}$

Q. Let E and F be two independent events such that P(E)>P(F) . The probability that both E and F happened is 121​ and the probability that neither E nor F happens is 21​, then

P(E)=31​,P(F)=41​

P(E)=21​,P(F)=61​

P(E)=21​,P(F)=81​