Question

A, B, C can hit the target with probability \((1 / 2),(1 / 3),(1 / 4)\) respectively. What is the probability that exactly two of them can hit the target? (a) \((1 / 2)\) (b) \((1 / 3)\) (c) \((1 / 4)\) (d) \((1 / 5)\)

Short answer

The probability that exactly two of them can hit the target is \(\frac{3}{8}\), which doesn't match any of the given options.

Step-by-step solution

Compute the probability of each person missing the target

To compute the probability of exactly two people hitting the target, we need to first find the probability of each person missing the target. For A, the probability of hitting the target is \(\frac{1}{2}\), hence the probability of missing is \(1 - \frac{1}{2} = \frac{1}{2}\). For B, the probability of hitting the target is \(\frac{1}{3}\), hence the probability of missing is \(1 - \frac{1}{3} = \frac{2}{3}\). For C, the probability of hitting the target is \(\frac{1}{4}\), hence the probability of missing is \(1 - \frac{1}{4} = \frac{3}{4}\).

Enumerate the scenarios where exactly two hit the target

There are three possible scenarios where exactly two people can hit the target: 1. A and B hit the target, C misses. 2. A and C hit the target, B misses. 3. B and C hit the target, A misses.

Calculate the probability of each scenario

1. Probability of A and B hitting the target while C misses: \(\frac{1}{2} \times \frac{1}{3} \times \frac{3}{4} = \frac{1}{8}\). 2. Probability of A and C hitting the target while B misses: \(\frac{1}{2} \times \frac{2}{3} \times \frac{1}{4} = \frac{1}{12}\). 3. Probability of B and C hitting the target while A misses: \(\frac{1}{2} \times \frac{1}{3} \times \frac{3}{4} = \frac{1}{8}\).

Add the probabilities of each scenario

To find the total probability that exactly two can hit the target, we just need to add the probabilities of the three possible scenarios calculated above: \(\frac{1}{8} + \frac{1}{12} + \frac{1}{8} = \frac{3}{8}\). So, the probability that exactly two of them can hit the target is \(\frac{3}{8}\). This doesn't match any of the given options (a), (b), (c), or (d). It seems that there might be an error in the options, as the correct answer is not present.

Key concepts

Probability Calculation

When trying to solve a problem that involves finding the likelihood of an event, such as shooting a target, we delve into the concept of probability calculation. Probability quantifies the chance of an event occurring, and it’s always a value between 0 and 1, where 0 means the event is impossible, and 1 means it is certain. To calculate the probability of a compound event, like two people hitting a target, we multiply the chances of each individual event if they're independent.

For instance, if person A has a probability of \(\frac{1}{2}\) to hit the target, then the probability of A missing the target complements this and is also \(\frac{1}{2}\). When wondering about two or more individuals hitting a target, we must consider every potential scenario that satisfies our condition, calculate each of their probabilities, and sum them up. This forms the crux of a basic probability calculation in multiple-event scenarios.

Independent Events in Probability

In probability, events are independent if the occurrence of one event does not influence the probability of another. In our scenario involving A, B, and C attempting to hit a target, we assume that whether A hits the target or not does not impact B's or C's attempts in any way. This is a core concept in probability theory, allowing us to simplify calculations.

To determine the chance of several independent events happening together (like both A and B hitting the target), we simply multiply their individual probabilities. This multiplication rule is a powerful tool as it can extend to any number of independent events, enabling us to easily calculate complex probabilities.

Combinatorial Probability

Combinatorial probability refers to the likelihood of an event occurring when there are several combinations of outcomes. In our target practice problem, there are three people, and we’re interested in the event where 'exactly two' of them hit the target. We have to consider all possible pairs that could accomplish this. It's like picking 2 out of 3 without regard to the order, which, in combinatorics, is called a combination.

The probability of each unique pair hitting the target is calculated, considering they are independent events. Yet, since more than one combination of pairs is possible, we apply the principles of combinatorial probability by adding the probabilities of each distinct pairing. This way, we avoid overcounting and achieve an accurate result, which in this case is a probability that wasn't listed in the supplied multiple-choice answers.