Understanding the Probability Problem

This question asks for the probability that exactly one out of two men hits a target, given their individual probabilities of hitting the target. We are given the probability that the first man hits the target and the probability that the second man hits the target.

Defining the Events and Probabilities

Let's define the events:

• \(H_1\): The first man hits the target.
• \(H_2\): The second man hits the target.

We are given the following probabilities:

• Probability of the first man hitting the target, \(P(H_1) = \frac{1}{2}\).
• Probability of the second man hitting the target, \(P(H_2) = \frac{1}{3}\).

To find the probability that exactly one man hits the target, we also need the probabilities that each man *misses* the target. The probability of an event not happening is 1 minus the probability of the event happening.

• Probability of the first man missing the target, \(P(H_1') = 1 - P(H_1) = 1 - \frac{1}{2} = \frac{1}{2}\).
• Probability of the second man missing the target, \(P(H_2') = 1 - P(H_2) = 1 - \frac{1}{3} = \frac{2}{3}\).

Finding the Probability of Exactly One Hit

The event "exactly one of them hits the target" can occur in two mutually exclusive ways:

1. The first man hits the target AND the second man misses the target.
2. The first man misses the target AND the second man hits the target.

Assuming the events are independent (one man's success doesn't affect the other's), we can calculate the probability of each case:

Case 1: Man 1 hits and Man 2 misses

The probability of this case is \(P(H_1 \text{ and } H_2')\). Since the events are independent, this is \(P(H_1) \times P(H_2')\).

Case 2: Man 1 misses and Man 2 hits

The probability of this case is \(P(H_1' \text{ and } H_2)\). Since the events are independent, this is \(P(H_1') \times P(H_2)\).

Since these two cases are mutually exclusive (they cannot both happen at the same time), the total probability that exactly one of them hits the target is the sum of the probabilities of these two cases.

To add these fractions, we find a common denominator, which is 6.

Simplifying the fraction \(\frac{3}{6}\), we get:

Therefore, the probability that exactly one of them hits the target is \(\frac{1}{2}\).

Conclusion

The probability that exactly one of the two men hits the target is \(\frac{1}{2}\). This is calculated by considering the two possible scenarios where exactly one hit occurs (Man 1 hits and Man 2 misses, OR Man 1 misses and Man 2 hits) and summing their individual probabilities, assuming independence of their attempts.

Revision Table: Probability Concepts

Additional Information: Types of Probability Problems

Probability questions often involve different scenarios. Understanding key terms helps solve them:

• "And" in Probability: Often implies multiplication of probabilities, especially for independent events. \(P(A \text{ and } B)\).
• "Or" in Probability: Often implies addition of probabilities, especially for mutually exclusive events. \(P(A \text{ or } B)\).
• "Exactly one" / "Exactly two" etc.: Requires identifying specific combinations of successes and failures and summing their probabilities.
• Conditional Probability: When the probability of one event depends on another event having already occurred. \(P(A | B)\). This problem assumed independence, so conditional probability wasn't directly needed in that form, but understanding independence is key to multiplying probabilities correctly.

Practice with different types of probability questions helps build confidence in applying the correct formulas and concepts like independence and mutual exclusivity.