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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Person A can hit a target 4 times in 5 attempts. Person B - 3 times in four attempts. Person C – 2
times in 3 attempts. They fire a volley. The probability that the target is hit at least two times. Is
Solution
Person A can hit the target with probability $\frac{4}{5}$ Person B with probability $\frac{3}{4}$
Person C with probability $\frac{2}{3}$
They fire together.
We need:
Probability(target is hit at least two times)
This means: exactly 2 hits OR exactly 3 hits.
Let $p_A=\frac{4}{5}$, $p_B=\frac{3}{4}$, $p_C=\frac{2}{3}$
and miss probabilities:
$q_A=\frac{1}{5}$, $q_B=\frac{1}{4}$, $q_C=\frac{1}{3}$.
1.Exactly 2 hits:
-A and B hit, C misses: $p_A p_B q_C$
- A and C hit, B misses: $p_A q_B p_C$
- B and C hit, A misses: $q_A p_B p_C$
So,
$P(\text{exactly 2}) = p_A p_B q_C + p_A q_B p_C + q_A p_B p_C$
$= \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{1}{3} \;+\; \frac{4}{5} \cdot \frac{1}{4} \cdot \frac{2}{3} \;+\; \frac{1}{5} \cdot \frac{3}{4} \cdot \frac{2}{3}$
$= \frac{1}{5} + \frac{2}{15} + \frac{2}{20}$
$= \frac{4}{20} + \frac{8}{60} + \frac{6}{60}$
$= \frac{12}{60} + \frac{14}{60}$
$= \frac{26}{60} = \frac{13}{30}$
2. Exactly 3 hits:
$P(\text{all hit}) = p_A p_B p_C$
$= \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3}$
$= \frac{2}{5}$
Therefore,
$P(\text{at least 2 hits}) = P(\text{exactly 2}) + P(\text{exactly 3})$
$= \frac{13}{30} + \frac{2}{5}$
$= \frac{13}{30} + \frac{12}{30}$
$= \frac{25}{30}$
$= \frac{5}{6}$
Final Answer: $\frac{5}{6}$
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