Aspire Faculty ID #10319 · Topic: NIMCET 2013 · Just now
NIMCET 2013

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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