Question

Let E and F be two independent events. If the probability that both E and F happen is 1/12 and the probability that neither E nor F happen 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) = 1/6, P(F) = 1/2
• D. P(E) = 1/4, P(F) = 1/3

Answer 1

Answer: (D)

P(E) = 1/4, P(F) = 1/3

Answer 2

Both E and F happen $\Rightarrow \, \, \, P(E \cap F)=\frac{1}{12}$ and neither E nor F happens $\rightarrow \, \, \, P(\overline{E}\cap \overline{F})=\frac{1}{2}$
But for independent events, we have
\hspace20mm P(E \cap F)=P(E)P(F)=\frac{1}{12} \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(i)
and $\, \, \, \, \, \, \, \, \, \, \, \, \, P(\overline{E} \cap \overline{F})=P(\overline{E})P(\overline{F})$
\hspace20mm \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\\{ 1-P(E)\\} \\{1-P(F)\\}
\hspace20mm \, \, \, \, \, \, \, \, \, \, \, =1-P(E)-P(F)+P(E)P(F)
$\Rightarrow \, \, \, \, \, \, \, \, P(E)+P(F)=1-\frac{1}{2}+\frac{1}{12}=\frac{7}{12} \, \, \, \, \, \, \, \, \, \, \,$...(ii)
On solving Eqs. (i) and (ii), we get
either $\, \, \, \, P(E)=\frac{1}{3}$ and $P(F)=\frac{1}{4}$
or $\, \, \, \, \, \, \, \, \, \, \, \, \, \, P(E)=\frac{1}{4}$ and $P(F)=\frac{1}{3}$