Four persons can hit a target correctly with probabilities $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ and…

Four persons can hit a target correctly with probabilities $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ and $\frac{1}{5}$ respectively. If all hit at the target independently, then the probability that the target would be hit, is
  1. $\frac{1}{5}$
  2. $\frac{3}{5}$
  3. $\frac{2}{5}$
  4. $\frac{4}{5}$

Solution

Let event A : A can hit the target event B : B can hit the target event $C$ : $C$ can hit the target event D : D can hit the target $\mathrm{P}(\mathrm{~A})=\frac{1}{2}, \mathrm{P}(\mathrm{~B})=\frac{1}{3}, \mathrm{P}(\mathrm{C})=\frac{1}{4}, \mathrm{P}(\mathrm{D})=\frac{1}{5}$
P (Target is not hit by any one of them) $=\mathrm{P}\left(\mathrm{~A}^{\prime} \cap \mathrm{B}^{\prime} \cap \mathrm{C}^{\prime} \cap \mathrm{D}^{\prime}\right)$ $=\mathrm{P}\left(\mathrm{A}^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{B}^{\prime}\right) \cdot \mathrm{P}^{\prime}\left(\mathrm{C}^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{D}^{\prime}\right)$ $=\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)\left(\frac{3}{4}\right)\left(\frac{4}{5}\right)$ $=\frac{1}{5}$ $\begin{aligned} & \mathrm{P}(\text { Target is hit }) \\ & =1-\mathrm{P}(\text { Target is not hit by any one of them }) \\ & =1-\frac{1}{5} \\ & =\frac{4}{5}\end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 2)

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