Question

Ajit, Ravi and Hari were trying to hit a target. Ajit hits the target 5 times in 8 attempts, Ravi hits it 3 times in 5 attempts, and Hari hits it 2 times in 4 attempts. What is the probability that the target is hit by at least 2 persons?

• A. \(\frac{49}{80}\)
• B. \(\frac{24}{80}\)
• C. \(\frac{45}{80}\)
• D. \(\frac{25}{80}\)

Hint:

Use the complement rule: P(at least 2) = 1 minus P(0 hits) minus P(exactly 1 hit).

Answer 1

The Correct Option is A

The problem requires us to find the probability that at least two out of Ajit, Ravi, and Hari hit a target. Let's break down the process step-by-step:

1. First, we calculate the probability that each individual hits the target:
   • Ajit: Hits 5 times out of 8 attempts. So, probability \( P(A) = \frac{5}{8} \). 
   • Ravi: Hits 3 times out of 5 attempts. So, probability \( P(R) = \frac{3}{5} \).
   • Hari: Hits 2 times out of 4 attempts. So, probability \( P(H) = \frac{1}{2} \).
2. Next, calculate the probability that each individual misses the target:
   • Ajit misses: \( P(A') = 1 - P(A) = 1 - \frac{5}{8} = \frac{3}{8} \).
   • Ravi misses: \( P(R') = 1 - P(R) = 1 - \frac{3}{5} = \frac{2}{5} \).
   • Hari misses: \( P(H') = 1 - P(H) = 1 - \frac{1}{2} = \frac{1}{2} \).
3. Calculate the probability that none of them hit the target:
   • Probability none hit is \( P(A') \times P(R') \times P(H') = \frac{3}{8} \times \frac{2}{5} \times \frac{1}{2} \)
   • Compute: \[ \frac{3}{8} \times \frac{2}{5} \times \frac{1}{2} = \frac{3 \times 2 \times 1}{8 \times 5 \times 2} = \frac{6}{80}. \]
4. Calculate the probability that exactly one person hits the target:
   • Ajit hits, Ravi and Hari miss: \( P(A) \times P(R') \times P(H') = \frac{5}{8} \times \frac{2}{5} \times \frac{1}{2} = \frac{10}{80}. \)
   • Ravi hits, Ajit and Hari miss: \( P(A') \times P(R) \times P(H') = \frac{3}{8} \times \frac{3}{5} \times \frac{1}{2} = \frac{9}{80}. \)
   • Hari hits, Ajit and Ravi miss: \( P(A') \times P(R') \times P(H) = \frac{3}{8} \times \frac{2}{5} \times \frac{1}{2} = \frac{6}{80}. \)
   • Total probability of exactly one hitting: \(\frac{10}{80} + \frac{9}{80} + \frac{6}{80} = \frac{25}{80}. \)
5. The probability that at least two people hit the target is the complement of probabilities of none or exactly one person hitting:
   • \[ P(\text{At least two hit}) = 1 - P(\text{None hits}) - P(\text{Exactly one hits}) \]
   • Compute: \[ 1 - \frac{6}{80} - \frac{25}{80} = \frac{80}{80} - \frac{31}{80} = \frac{49}{80}.\]
6. Conclude that the probability that the target is hit by at least two persons is \(\frac{49}{80}\), which matches the given correct answer option.

Answer 2

Step 1: Write down each person's hit and miss probability.
Ajit hits with probability \(\frac{5}{8}\), so he misses with probability \(\frac{3}{8}\). Ravi hits with probability \(\frac{3}{5}\), so he misses with probability \(\frac{2}{5}\). Hari hits with probability \(\frac{2}{4} = \frac{1}{2}\), so he misses with probability \(\frac{1}{2}\) too.

Step 2: Use the complement rule.
"At least 2 hit" is easier to find by subtracting the cases with 0 hits and exactly 1 hit from 1, since those are the only cases left out.

Step 3: Find the probability that nobody hits the target.
All three miss: \(\frac{3}{8} \times \frac{2}{5} \times \frac{1}{2} = \frac{6}{80}\).

Step 4: Find the probability that exactly one person hits the target.
Only Ajit hits: \(\frac{5}{8} \times \frac{2}{5} \times \frac{1}{2} = \frac{10}{80}\). Only Ravi hits: \(\frac{3}{8} \times \frac{3}{5} \times \frac{1}{2} = \frac{9}{80}\). Only Hari hits: \(\frac{3}{8} \times \frac{2}{5} \times \frac{1}{2} = \frac{6}{80}\). Adding these gives \(\frac{10+9+6}{80} = \frac{25}{80}\).

Step 5: Combine and subtract from 1.
Probability of 0 or 1 hit = \(\frac{6}{80} + \frac{25}{80} = \frac{31}{80}\). So probability of at least 2 hits = \(1 - \frac{31}{80} = \frac{49}{80}\).

Final Answer:
The probability that the target is hit by at least 2 persons is \(\frac{49}{80}\). \[ \boxed{\frac{49}{80}} \]