$\mathrm{P}, \mathrm{Q}$ and R try to hit the same target one after the other. If their probabilities of…

$\mathrm{P}, \mathrm{Q}$ and R try to hit the same target one after the other. If their probabilities of hitting the target are $\frac{2}{3}, \frac{3}{5}, \frac{5}{7}$ respectively, then the probability that the target is hit by P or Q but not by R is
  1. $\frac{26}{105}$
  2. $\frac{79}{105}$
  3. $0$
  4. $\frac{75}{105}$

Solution

Let $P=$ The event that $P$ hit target $\mathrm{Q}=$ The event that Q hit target $\mathrm{R}=$ The event that R hit target $\because \mathrm{P}(\mathrm{P})=\frac{2}{3}, \mathrm{P}(\mathrm{Q})=\frac{3}{5}$ and $\mathrm{P}(\mathrm{R})=\frac{5}{7}$ Now, required probability $\begin{aligned} & =\mathrm{P}(\mathrm{P}) \mathrm{P}\left(\mathrm{Q}^{\prime}\right) \mathrm{P}\left(\mathrm{R}^{\prime}\right)+\mathrm{P}\left(\mathrm{P}^{\prime}\right) \mathrm{P}(\mathrm{Q}) \mathrm{P}\left(\mathrm{R}^{\prime}\right)+\mathrm{P}(\mathrm{P}) \mathrm{P}(\mathrm{Q}) \mathrm{P}\left(\mathrm{R}^{\prime}\right) \\ & =\frac{2}{3} \times \frac{2}{5} \times \frac{2}{7}+\frac{1}{3} \times \frac{3}{5} \times \frac{2}{7}+\frac{2}{3} \times \frac{3}{5} \times \frac{2}{7}=\frac{8+6+12}{105}=\frac{26}{105}\end{aligned}$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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