Three persons $\mathrm{P}, \mathrm{Q}$ and R independently try to hit a target. If the probabilities of…

Three persons $\mathrm{P}, \mathrm{Q}$ and R independently try to hit a target. If the probabilities of their hitting the target are $\frac{3}{4}, \frac{1}{2}$ and $\frac{5}{8}$ respectively, then the probability that the target is hit by P or Q but not by $R$, is
  1. $\frac{15}{64}$
  2. $\frac{21}{64}$
  3. $\frac{39}{64}$
  4. $\frac{9}{64}$

Solution

$\mathrm{P}(\mathrm{P})=\frac{3}{4}, \mathrm{P}(\mathrm{Q})=\frac{1}{2}, \mathrm{P}(\mathrm{R})=\frac{5}{8}$
Since $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ are independent events, $\mathrm{P}^{\prime}, \mathrm{Q}^{\prime}, \mathrm{R}^{\prime}$ are also independent events. $\begin{aligned} \therefore \quad & P(\text { Target is hit by } P \text { or } Q \text { but not } R) \\ = & P\left(P \cap \mathrm{Q}^{\prime} \cap \mathrm{R}^{\prime}\right)+P\left(\mathrm{P}^{\prime} \cap \mathrm{Q} \cap \mathrm{R}^{\prime}\right) \\ \quad & \quad \mathrm{P}\left(\mathrm{P} \cap \mathrm{Q} \cap \mathrm{R}^{\prime}\right) \\ = & \mathrm{P}(\mathrm{P}) \cdot \mathrm{P}\left(\mathrm{Q}^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{R}^{\prime}\right)+\mathrm{P}\left(\mathrm{P}^{\prime}\right) \cdot \mathrm{P}(\mathrm{Q}) \cdot \mathrm{P}\left(\mathrm{R}^{\prime}\right) \\ & \quad+\mathrm{P}(\mathrm{P}) \cdot \mathrm{P}(\mathrm{Q}) \cdot \mathrm{P}\left(\mathrm{R}^{\prime}\right) \\ = & \left(\frac{3}{4}\right)\left(\frac{1}{2}\right)\left(\frac{3}{8}\right)+\left(\frac{1}{4}\right)\left(\frac{1}{2}\right)\left(\frac{3}{8}\right)+\left(\frac{3}{4}\right)\left(\frac{1}{2}\right)\left(\frac{3}{8}\right) \\ = & \frac{9+3+9}{64} \\ = & \frac{21}{64} \end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 1)

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