Probability 3 Question 2
2. Four persons can hit a target correctly with probabilities $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ and $\frac{1}{8}$ respectively. If all hit at the target independently, then the probability that the target would be hit, is
(2019 Main, 9 April I)
(a) $\frac{1}{192}$
(b) $\frac{25}{32}$
(c) $\frac{7}{32}$
(d) $\frac{25}{192}$
Show Answer
Answer:
Correct Answer: 2. (b)
| 1. (d) | 2. (b) | 3. (b) | 4. (a) |
|---|---|---|---|
| 5. (c) | 6. (d) | 7. (a) | 8. (a) |
| 9. (d) | 10. (c) | 11. (c) | 12. (b) |
| 13. (b) | 14. (a) | 15. (a) | 16. (b) |
| 17. (a) | 18. (c) | 19. (b) | 20. (b) |
| 21. (a, b) | 22. (a,b) | 23. (a, d) | 24. (b, c) |
| 25. (a, d) | 26. (a, d) | 27. (a, c) | 28. (b, c, d) |
| 29. $\frac{1}{4}$ | 30. $\frac{5}{7}$ | 31. $\frac{2}{5}$ | 32. $\frac{32}{55}$ |
| 33. $\frac{1}{9}$ | 35. $\frac{1}{2}$ | 36. $2 p^{2}-p^{3}$ | 38. $\frac{193}{792}$ |
Solution:
- Key Idea Use $P(\bar{A})=1-P(A)$ and condition of independent events i.e $P(A \cap B)=P(A) \cdot P(B)$
Given that probability of hitting a target independently by four persons are respectively
$$ P _1=\frac{1}{2}, P _2=\frac{1}{3}, P _3=\frac{1}{4} \text { and } P _4=\frac{1}{8} $$
Then, the probability of not hitting the target is
$$ =1-\frac{1}{2} \quad 1-\frac{1}{3} \quad 1-\frac{1}{4} \quad 1-\frac{1}{8} $$
$$ =\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{7}{8}=\frac{7}{32} $$
$[\because$ events are independent]
Therefore, the required probability of hitting the target $=1-($ Probability of not hitting the target)
$=1-\frac{7}{32}=\frac{25}{32}$
