$\mathrm{A}, \mathrm{B}, \mathrm{C}$ are mutually exclusive and exhaustive events of a random experiment and…

$\mathrm{A}, \mathrm{B}, \mathrm{C}$ are mutually exclusive and exhaustive events of a random experiment and E is an event that occurs in conjunction with one of the events $\mathrm{A}, \mathrm{B}, \mathrm{C}$. The conditional Probabilities of E given the happening of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are respectively $0.6,0.3$ and 0.1 . If $\mathrm{P}(\mathrm{A})=0.30$ and $\mathrm{P}(\mathrm{B})=$ 0.50 , then $P(C \mid E)=$
  1. $\frac{2}{35}$
  2. $\frac{15}{35}$
  3. $\frac{18}{35}$
  4. $\frac{17}{35}$

Solution

$P(\mathrm{~A})=0.30, P(\mathrm{~B})=0.50$ Since $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are mutually exclusive and exhaustive $\therefore P(A)+P(B)+P(C)=1$ $\Rightarrow \mathrm{P}(\mathrm{C})=0.20, P(E / A)=0.6, P(E / B)=0.3, P(E / C)=0.1$ Now, $P(C / E)=\frac{P(C) P(E / C)}{P(A) P(E / A)+P(B) P(E / B)+P(C) P(E / C)}$ $=\frac{0.2 \times 0.1}{(0.3)(0.6)+(0.5)(0.3)+(0.2)(0.1)}=\frac{0.02}{0.35}=\frac{2}{35}$

Asked in: AP EAMCET 2024 (20 May Shift 1)

Practice more Probability questions on Aicharya