If $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are three independent events of a random experiment such that…

If $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are three independent events of a random experiment such that $\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\mathrm{c}} \cap \mathrm{C}^{\mathrm{c}}\right)=\frac{1}{4}, \mathrm{P}\left(\mathrm{A}^{\mathrm{c}} \cap \mathrm{B} \cap \mathrm{C}^{\mathrm{c}}\right)=\frac{1}{8}$ and $\mathrm{P}\left(\mathrm{A}^{\mathrm{c}} \cap \mathrm{B}^{\mathrm{c}} \cap \mathrm{C}^{\mathrm{c}}\right)=\frac{1}{4}$, then $\mathrm{P}(\mathrm{A}), \mathrm{P}(\mathrm{B})$ and $\mathrm{P}(\mathrm{C})$ are respectively
  1. $\frac{1}{2}, \frac{1}{4}, \frac{1}{5}$
  2. $1, \frac{1}{2}, \frac{1}{3}$
  3. $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$
  4. $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}$

Solution

$\because \mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\mathrm{c}} \cap \mathrm{C}^{\mathrm{c}}\right)=\frac{1}{4}$
Let $P(A)=x, P(B)=y, P(C)=z$ Then $x \times(1-y) \times(1-z)=\frac{1}{4}$ ...(i) Similarly, $P\left(A^c \cap B \cap C^c\right)=\frac{1}{8}$ $\Rightarrow(1-x) y(1-z)=\frac{1}{8}$ ...(ii) Also, $\mathrm{P}\left(\mathrm{A}^{\mathrm{c}} \cap \mathrm{B}^{\mathrm{c}} \cap \mathrm{C}^{\mathrm{c}}\right)=\frac{1}{4}$ $\Rightarrow(1-x)(1-y)(1-z)=\frac{1}{4}$ ...(iii) $\mathrm{Eq}^{\mathrm{n}}(\mathrm{i}) /(\mathrm{ii}) \Rightarrow \frac{\mathrm{x}}{1-\mathrm{x}}=1 \Rightarrow \mathrm{x}=\frac{1}{2}$ $\mathrm{Eq}^{\mathrm{n}}(\mathrm{ii}) /(\mathrm{iii}) \Rightarrow \frac{\mathrm{x}}{1-\mathrm{y}}=\frac{1}{2} \Rightarrow \mathrm{y}=\frac{1}{3}$ from $\mathrm{eq}^{\mathrm{n}}$ (i) : $\frac{1}{2} \times \frac{2}{3} \times(1-\mathrm{z})=\frac{1}{4} \Rightarrow 1-\mathrm{z}=\frac{3}{4} \Rightarrow \mathrm{z}=\frac{1}{4}$ $\therefore \mathrm{P}(\mathrm{A})=\frac{1}{2} \mathrm{P}(\mathrm{B})=\frac{1}{3}$ and $\mathrm{P}(\mathrm{C})=\frac{1}{4}$

Asked in: AP EAMCET 2023 (18 May Shift 1)

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