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MHT CET · Maths · Probability

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. A \(\frac{15}{64}\)
  2. B \(\frac{21}{64}\)
  3. C \(\frac{39}{64}\)
  4. D \(\frac{9}{64}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{21}{64}\)

Step-by-step Solution

Detailed explanation

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