Question

Events $A$ and $C$ are independent. If the probabilities relating $A,B$ and $C$ are $P\left(A\right)=\frac{1}{5},P\left(B\right)=\frac{1}{6},$ $P\left(A\cap C\right)=\frac{1}{20},P\left(B\cup C\right)=\frac{3}{8}$ then

• A. events $B$ and $C$ are independent
• B. events $B$ and $C$ are mutually exclusive
• C. events $B$ and $C$ are neither independent nor mutually exclusive
• D. events $B$ and $C$ are equiprobable

Answer 1

The correct option is C events $B$ and $C$ are neither independent nor mutually exclusive

Solution -

$P\left(A\right)=\frac{1}{5}$ $P\left(B\right)=\frac{1}{6}$ $P(A\cap C)=\frac{1}{20}$ $P(B\cup C)=\frac{3}{8}$

Events A and C are independent

then,

$P\left(\frac{A}{C}\right)=\frac{P(A\cap C)}{P\left(C\right)}=P\left(A\right)\left(P\right(C\left)\right)$

$\frac{1/20}{P\left(C\right)}=\frac{1}{5}P\left(C\right)$

$P\left(C\right)=\frac{1}{2}$

$P(B\cap C)=P\left(B\right)+P\left(C\right)-P(B\cup C)$

$=\frac{1}{6}+\frac{1}{2}-\frac{3}{8}=\frac{7}{24}$

$P\left(\frac{B}{C}\right)=\frac{P(B\cap C)}{P\left(C\right)}=\frac{7/24}{1/2}=\frac{7}{12}$

$P\left(B\right).P\left(C\right)=\frac{1}{6}\times \frac{1}{2}=\frac{1}{2}$

Events B & C are neither independent nor mutually exclusive

C is correct