Questions: Assume that a procedure yields a binomial distribution with a trial repeated n=18 times. Use either the binomial probability formula (or a technology like Excel or StatDisk) to find the probability of k=7 successes given the probability p=37 / 60 of success on a single trial. (Report answer accurate to 4 decimal places.)

Solution Steps

We are tasked with finding the probability of obtaining exactly \( k = 7 \) successes in a binomial distribution where the number of trials is \( n = 18 \) and the probability of success on a single trial is \( p = \frac{37}{60} \).

The probability of obtaining exactly \( k \) successes in \( n \) trials for a binomial distribution is given by the formula: \[ P(X = k) = \binom{n}{k} \cdot p^k \cdot q^{n-k} \] where:

• \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\),
• \(p\) is the probability of success on a single trial,
• \(q = 1 - p\) is the probability of failure on a single trial.

Substitute the given values into the formula:

• \(n = 18\),
• \(k = 7\),
• \(p = \frac{37}{60}\),
• \(q = 1 - \frac{37}{60} = \frac{23}{60}\).

The probability is calculated as: \[ P(X = 7) = \binom{18}{7} \cdot \left(\frac{37}{60}\right)^7 \cdot \left(\frac{23}{60}\right)^{11} \]

The calculated probability of exactly 7 successes is: \[ P(X = 7) = 0.0283 \]

Final Answer