Calculating Probability in a Shooting Test

The question asks for the probability that exactly one person hits the target when three people (A, B, and C) fire at it independently. We are given the individual probabilities of each person hitting the target.

Understanding Individual Probabilities

Let $P(A)$ be the probability that A hits the target, $P(B)$ be the probability that B hits the target, and $P(C)$ be the probability that C hits the target.

• Probability of A hitting the target, $P(A) = \frac{1}{2}$
• Probability of B hitting the target, $P(B) = \frac{2}{3}$
• Probability of C hitting the target, $P(C) = \frac{3}{4}$

Calculating Probabilities of Missing the Target

If the probability of hitting the target is $P(Hit)$, the probability of missing the target is $P(Miss) = 1 - P(Hit)$.

• Probability of A missing the target, $P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}$
• Probability of B missing the target, $P(B') = 1 - P(B) = 1 - \frac{2}{3} = \frac{3-2}{3} = \frac{1}{3}$
• Probability of C missing the target, $P(C') = 1 - P(C) = 1 - \frac{3}{4} = \frac{4-3}{4} = \frac{1}{4}$

Identifying Scenarios Where Only One Person Hits

We are interested in the event where exactly one of the three shooters hits the target. Since the events are independent, we can multiply their individual probabilities for each scenario. There are three possible scenarios:

1. Only A hits the target (A hits, B misses, C misses).
2. Only B hits the target (A misses, B hits, C misses).
3. Only C hits the target (A misses, B misses, C hits).

Calculating the Probability for Each Scenario

• Scenario 1: Only A hits.

  This occurs when A hits, B misses, and C misses. Since the events are independent, the probability is:

  $\quad P(\text{Only A hits}) = P(A) \times P(B') \times P(C') = \frac{1}{2} \times \frac{1}{3} \times \frac{1}{4} = \frac{1 \times 1 \times 1}{2 \times 3 \times 4} = \frac{1}{24}$

• Scenario 2: Only B hits.

  This occurs when A misses, B hits, and C misses. The probability is:

  $\quad P(\text{Only B hits}) = P(A') \times P(B) \times P(C') = \frac{1}{2} \times \frac{2}{3} \times \frac{1}{4} = \frac{1 \times 2 \times 1}{2 \times 3 \times 4} = \frac{2}{24} = \frac{1}{12}$

• Scenario 3: Only C hits.

  This occurs when A misses, B misses, and C hits. The probability is:

  $\quad P(\text{Only C hits}) = P(A') \times P(B') \times P(C) = \frac{1}{2} \times \frac{1}{3} \times \frac{3}{4} = \frac{1 \times 1 \times 3}{2 \times 3 \times 4} = \frac{3}{24} = \frac{1}{8}$

Calculating the Total Probability

The event that exactly one person hits the target is the union of the three mutually exclusive scenarios listed above. Therefore, the total probability is the sum of the probabilities of these three scenarios.

$\quad P(\text{Only one hits}) = P(\text{Only A hits}) + P(\text{Only B hits}) + P(\text{Only C hits})$

$\quad P(\text{Only one hits}) = \frac{1}{24} + \frac{2}{24} + \frac{3}{24}$ (using common denominator 24 for 1/12 and 1/8)

$\quad P(\text{Only one hits}) = \frac{1+2+3}{24} = \frac{6}{24}$

Simplifying the fraction:

$\quad P(\text{Only one hits}) = \frac{6 \div 6}{24 \div 6} = \frac{1}{4}$

Thus, the probability that only one of them hits the target is $\frac{1}{4}$.

Probability Revision Table

Additional Information on Probability Concepts

This problem involves a few key probability concepts:

• Independent Events: Events are independent if the outcome of one event does not affect the outcome of another. In this problem, the shooters firing at the target are considered independent events. The probability of multiple independent events occurring together is the product of their individual probabilities.
• Complementary Events: For any event E, the event that E does not occur is called its complement, denoted by E'. The sum of the probabilities of an event and its complement is always 1, i.e., $P(E) + P(E') = 1$. We used this to find the probability of each shooter missing the target.
• Mutually Exclusive Events: Events are mutually exclusive if they cannot happen at the same time. In this problem, the three scenarios (only A hits, only B hits, only C hits) are mutually exclusive because in each case, exactly one specific person hits, and the others miss. The probability of any one of several mutually exclusive events occurring is the sum of their individual probabilities.

Understanding these concepts is crucial for solving probability problems involving multiple events.