Probability Problem Analysis

This problem involves calculating the probability of a specific event occurring when multiple independent events happen simultaneously. We need to determine the likelihood that at least two out of three individuals (A, B, and C) successfully hit the target with their single shots.

Defining Individual Probabilities

First, let's determine the probability of each person hitting the target and the probability of them missing the target. We assume each shot is an independent event.

• Person A: Hits 3 out of 5 shots.

  Probability of A hitting, $P(A) = \frac{3}{5}$

  Probability of A missing, $P(A') = 1 - P(A) = 1 - \frac{3}{5} = \frac{2}{5}$

• Person B: Hits 2 out of 5 shots.

  Probability of B hitting, $P(B) = \frac{2}{5}$

  Probability of B missing, $P(B') = 1 - P(B) = 1 - \frac{2}{5} = \frac{3}{5}$

• Person C: Hits 3 out of 4 shots.

  Probability of C hitting, $P(C) = \frac{3}{4}$

  Probability of C missing, $P(C') = 1 - P(C) = 1 - \frac{3}{4} = \frac{1}{4}$

Calculating Probability of 'At Least Two Shots Hit'

The event 'at least two shots hit' includes the following mutually exclusive scenarios:

• Exactly two shots hit the target.
• Exactly three shots hit the target.

We will calculate the probability of each scenario and then sum them up.

Scenario 1: Exactly Two Shots Hit

This scenario can occur in three distinct ways, based on who misses the target:

• A hits, B hits, C misses: The probability is $P(A \cap B \cap C') = P(A) \times P(B) \times P(C')$
• A hits, B misses, C hits: The probability is $P(A \cap B' \cap C) = P(A) \times P(B') \times P(C)$
• A misses, B hits, C hits: The probability is $P(A' \cap B \cap C) = P(A') \times P(B) \times P(C)$

Let's calculate the probability for each specific combination:

• Probability (A hits, B hits, C misses) = $\frac{3}{5} \times \frac{2}{5} \times \frac{1}{4} = \frac{6}{100}$
• Probability (A hits, B misses, C hits) = $\frac{3}{5} \times \frac{3}{5} \times \frac{3}{4} = \frac{27}{100}$
• Probability (A misses, B hits, C hits) = $\frac{2}{5} \times \frac{2}{5} \times \frac{3}{4} = \frac{12}{100}$

The total probability of exactly two shots hitting is the sum of these probabilities:

P(Exactly 2 hits) = $\frac{6}{100} + \frac{27}{100} + \frac{12}{100} = \frac{45}{100}$

Scenario 2: Exactly Three Shots Hit

This occurs only when all three individuals, A, B, and C, hit the target simultaneously:

$P(A \cap B \cap C) = P(A) \times P(B) \times P(C)$

$P(A \cap B \cap C) = \frac{3}{5} \times \frac{2}{5} \times \frac{3}{4} = \frac{18}{100}$

So, the probability of exactly three shots hitting is P(Exactly 3 hits) = $\frac{18}{100}$

Total Probability Calculation

To find the overall probability that at least two shots hit the target, we add the probabilities calculated for the two scenarios (exactly 2 hits and exactly 3 hits):

P(At least 2 hits) = P(Exactly 2 hits) + P(Exactly 3 hits)

P(At least 2 hits) = $\frac{45}{100} + \frac{18}{100}$

P(At least 2 hits) = $\frac{63}{100}$

Final Answer

The calculated probability that at least two shots hit the target is $\frac{63}{100}$.