Understanding the Probability Problem

This question asks us to find the probability of a specific outcome occurring a certain number of times in a fixed number of independent trials. Each trial (firing a shot) has only two possible outcomes: hitting the target (success) or missing the target (failure). The probability of success is constant for each trial. This scenario perfectly fits the definition of a binomial distribution.

Identifying Key Information

Let's break down the information given in the problem:

• Probability of hitting the target in a single shot: \(p = \frac{5}{8}\)
• Probability of not hitting the target in a single shot: \(q = 1 - p = 1 - \frac{5}{8} = \frac{8-5}{8} = \frac{3}{8}\)
• Total number of shots (trials): \(n = 10\)
• Desired number of times hitting the target (successes): \(k = 2\)

Applying the Binomial Probability Formula

The probability of getting exactly \(k\) successes in \(n\) trials in a binomial distribution is given by the formula:

\(P(X=k) = \binom{n}{k} p^k q^{n-k}\)

Where \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\).

In this problem, we need to find the probability of hitting the target exactly 2 times in 10 shots. So, we substitute the values:

\(n = 10\)

\(k = 2\)

\(p = \frac{5}{8}\)

\(q = \frac{3}{8}\)

The formula becomes:

\(P(X=2) = \binom{10}{2} \left(\frac{5}{8}\right)^2 \left(\frac{3}{8}\right)^{10-2}\)

\(P(X=2) = \binom{10}{2} \left(\frac{5}{8}\right)^2 \left(\frac{3}{8}\right)^{8}\)

Calculating the Binomial Coefficient

First, let's calculate the binomial coefficient \(\binom{10}{2}\):

\(\binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10!}{2!8!}\)

\(\binom{10}{2} = \frac{10 \times 9 \times 8!}{2 \times 1 \times 8!}\)

\(\binom{10}{2} = \frac{10 \times 9}{2}\)

\(\binom{10}{2} = \frac{90}{2}\)

\(\binom{10}{2} = 45\)

Completing the Probability Calculation

Now, substitute the value of the binomial coefficient back into the probability formula:

\(P(X=2) = 45 \times \left(\frac{5}{8}\right)^2 \left(\frac{3}{8}\right)^{8}\)

\(P(X=2) = 45 \times \left(\frac{5^2}{8^2}\right) \times \left(\frac{3^8}{8^8}\right)\)

\(P(X=2) = 45 \times \frac{25}{8^2} \times \frac{3^8}{8^8}\)

\(P(X=2) = 45 \times \frac{25 \times 3^8}{8^2 \times 8^8}\)

Using the rule of exponents \(a^m \times a^n = a^{m+n}\), we have \(8^2 \times 8^8 = 8^{2+8} = 8^{10}\).

So, the expression becomes:

\(P(X=2) = 45 \times \frac{25 \times 3^8}{8^{10}}\)

Now, multiply the numbers in the numerator:

\(45 \times 25\)

\(45 \times 25 = (40 + 5) \times 25 = 40 \times 25 + 5 \times 25\)

\(40 \times 25 = 1000\)

\(5 \times 25 = 125\)

\(1000 + 125 = 1125\)

Therefore, the probability of hitting the target exactly twice in 10 shots is:

\(P(X=2) = \frac{1125 \times 3^8}{8^{10}}\)

Comparing with Options

Let's compare our calculated probability with the given options:

• Option 1: \(\frac{1135 \times 3^8}{8^{10}}\)
• Option 2: \(\frac{1165\times 3^8}{8^{10}}\)
• Option 3: \(\frac{1175\times 3^8}{8^{10}}\)
• Option 4: \(\frac{1125\times 3^8}{8^{10}}\)

Our calculated result matches Option 4.

Revision Table: Key Concepts

Additional Information: Binomial Distribution Properties

The binomial distribution is a fundamental concept in probability theory. Here are some additional points about it:

• It models a sequence of independent trials, where each trial has exactly two outcomes (success or failure).
• The probability of success (\(p\)) must remain constant for every trial.
• The number of trials (\(n\)) is fixed.
• The trials must be independent, meaning the outcome of one trial does not affect the outcome of others.
• Examples include flipping a coin a fixed number of times, checking a fixed number of items for defects, or as in this problem, firing a fixed number of shots at a target.
• The sum of probabilities for all possible numbers of successes (from 0 to \(n\)) must equal 1.