Question

Let $A$ and $B$ be two independent events such that the probability is $\frac{1}{8}$ that they will occour simultaneously and $\frac{3}{8}$ that neither of them will occur. Find $P\left(A\right)$ and $P\left(B\right)$.

• A. $P\left(A\right)=\frac{3}{4}$ and $P\left(B\right)=\frac{1}{4}$
• B. $P\left(A\right)=\frac{1}{4}$ and $P\left(B\right)=\frac{3}{4}$
• C. $P\left(A\right)=\frac{1}{2}$ and $P\left(B\right)=\frac{1}{2}$
• D. $P\left(A\right)=\frac{1}{3}$ and $P\left(B\right)=\frac{2}{3}$

Answer 1

Answer: (C)

$P\left(A\right)=\frac{1}{2}$ and $P\left(B\right)=\frac{1}{2}$

We are given, $P(A\cap B)=\frac{1}{8}$ ...(1)
and $P(\overline{A}\cap \overline{B})=\frac{3}{8}$
Let $P\left(A\right)=x$ and $P\left(B\right)=y$.
Since the events are independent, we have from (1),
$P\left(A\right)P\left(B\right)=\frac{1}{8}$ i.e. $xy=\frac{1}{8}$ ...(3)
And from (2), we have $P(\overline{A}\cap \overline{B})=P(A\cup B)\frac{3}{8}+1-P(A\cup B)=\frac{3}{8}\Rightarrow 1-P\left(A\right)-P\left(B\right)+P(A\cap B)=\frac{3}{8}$
$\Rightarrow 1-P\left(A\right)-P\left(B\right)+P\left(A\right)P\left(B\right)=\frac{3}{8}\Rightarrow 1-x-y+xy=\frac{3}{8}$ ...(4)
Substracting (3)from (4), we get $1-x-y=\frac{1}{4}orx+y=\frac{3}{4}$ ...(5)
Now $(x-y{)}^2=(x+y{)}^2-4xy=\frac{9}{16}-4\times \frac{1}{8}=\frac{1}{16}\Rightarrow x-y=\frac{1}{4}$ ..(6)
Solving (5) and (6), we get $x=\frac{1}{2}$ and $y=\frac{1}{4}\Rightarrow x=\frac{1}{4}$ and $y=\frac{1}{2}$
Hence $P\left(A\right)=\frac{1}{2}$ and $P\left(B\right)=\frac{1}{2}$