Assume that each of the n trials is independent and that pis given trial. Use the binomial probability formula to find P(x). n=15, x = 3, p=0.5 P(x) = (Round to three decimal places as needed.)

To find the probability of exactly x successes in n independent trials with probability p of success on a given trial, you can use the binomial probability formula:

\[ P(x) = \binom{n}{x} p^x (1-p)^{n-x} \]

Where: - \( P(x) \) is the probability of getting exactly x successes in n trials. - \( \binom{n}{x} \) is the binomial coefficient, which calculates the number of ways to choose x successes from n trials, and is calculated as \(\frac{n!}{x!(n-x)!}\). - \( p \) is the probability of success on a single trial. - \( 1-p \) is the probability of failure on a single trial.

Given n=15, x=3, and p=0.5, let's plug the values into the formula:

\[ P(3) = \binom{15}{3} (0.5)^3 (1 - 0.5)^{15-3} \]

First, calculate the binomial coefficient \(\binom{15}{3}\):

\[ \binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15!}{3! \times 12!}\]

Since \(15! = 15 \times 14 \times 13 \times 12! \), many terms cancel out when divided by \(12!\), resulting in:

\[ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 5 \times 7 \times 13\]

Now calculate the value:

\[ \binom{15}{3} = 5 \times 7 \times 13 = 455 \]

Now use this coefficient to calculate \( P(3) \):

\[ P(3) = 455 \times (0.5)^3 \times (0.5)^{12} \]

Remember that \( (0.5)^3 = 0.125 \) and \( (0.5)^{12} = (0.5)^{12} \). When you multiply a number by itself, the powers add up, so \( (0.5)^3 \times (0.5)^{12} = (0.5)^{15} \).

Because \( p = 0.5 \), which is \( \frac{1}{2} \), \( (0.5)^{15} = ( \frac{1}{2} )^{15} \).

Perform the final calculation:

\[ P(3) = 455 \times (0.5)^{15} \]

\[ P(3) = 455 \times \frac{1}{32768} \]

\[ P(3) = \frac{455}{32768} \]

\[ P(3) \approx 0.0139 \]

Rounded to three decimal places, \( P(3) \) is approximately 0.014.

Extra: The binomial probability formula is a fundamental concept in statistics, used when you have a series of independent trials, each with the same probability of success. Independent trials mean the outcome of one trial does not affect the outcome of another.

In this example, we looked at a situation with a probability of success (p) of 0.5 for each trial, like flipping a fair coin. With a fair coin, there are only two outcomes, head or tail, which is why the probability of success (landing a head) or failure (landing a tail) is 0.5.

The binomial coefficient (\(\binom{n}{x}\)) effectively counts how many different ways x successes can occur among n trials, without considering the order of occurrence. This is important because, in a binomial setting, we only care about the number of successes, not the sequence in which they occurred.

The binomial distribution has many practical applications in various fields, such as quality control, survey analysis, and many disciplines that require probability assessments for discrete outcomes. It provides a straightforward way to calculate probabilities for scenarios where there are clear definitions of 'success' and 'failure.'