A boy tries to message his friend.
Each time, the chance the message is delivered is \( \frac{1}{6} \), and the chance it fails is \( \frac{5}{6} \).
He sends 6 messages.
Find the probability that exactly 5 messages are delivered.
Show Hint
When solving binomial probability problems, remember to use the binomial distribution formula:
\[
P(X = k) = \binom{n}{k} p^k q^{n-k}
\]
where \( p \) is the probability of success and \( q \) is the probability of failure.
This scenario adheres to a binomial distribution due to independent trials (message delivery) with two outcomes: success (delivered) or failure (undelivered).Parameters:- Probability of success (delivery): \( p = \frac{1}{6} \)- Probability of failure (non-delivery): \( q = 1 - p = \frac{5}{6} \)- Number of trials (messages sent): \( n = 6 \)- Target: Probability of exactly 5 successes (5 messages delivered). Step 1: Binomial Distribution FormulaThe probability of exactly \( k \) successes in \( n \) trials is given by:\[P(X = k) = \binom{n}{k} p^k q^{n-k}\]where \( \binom{n}{k} \) is the binomial coefficient. Step 2: Value SubstitutionFor this problem, \( k = 5 \), \( n = 6 \), \( p = \frac{1}{6} \), and \( q = \frac{5}{6} \). Substituting these values:\[P(X = 5) = \binom{6}{5} \left( \frac{1}{6} \right)^5 \left( \frac{5}{6} \right)^{6-5}\]\[P(X = 5) = \binom{6}{5} \left( \frac{1}{6} \right)^5 \left( \frac{5}{6} \right)\] Step 3: Simplification\( \binom{6}{5} = 6 \).\[P(X = 5) = 6 \left( \frac{1}{6} \right)^5 \left( \frac{5}{6} \right)\]\[P(X = 5) = 6 \times \frac{1}{7776} \times \frac{5}{6}\]\[P(X = 5) = \frac{30}{7776} = \frac{5}{1296}\]The probability of exactly 5 messages being delivered is \( \boxed{\frac{5}{1296}} \).