Sure, I'd be happy to walk you through the process of determining each of these probabilities using the binomial distribution.
1) The probability that she will make exactly 5 of the next 15 putts
The binomial distribution can be used to find the probability of a given number of successes (making a putt) in a fixed number of trials (15 putts). The formula for the probability mass function (PMF) of a binomial distribution is:
\(\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]\)
where \(\( \binom{n}{k} \)\) is the binomial coefficient, \(\( n \)\) is the number of trials, \(\( k \)\) is the number of successes, and \(\( p \)\) is the probability of a success.
In this case:
- \(\( n = 15 \)\)
- \(\( k = 5 \)\)
- \(\( p = 0.17 \)\)
This tells us:
\(\[ P(X = 5) = \binom{15}{5} (0.17)^5 (0.83)^{10} \]\)
Upon calculating this, the probability that she will make exactly 5 of the next 15 putts is:
\(\[ P(X = 5) \approx 0.0662 \]\)
2) The probability that she will make less than 5 of the next 15 putts
To find the probability that she will make less than 5 putts, we need the cumulative probability for 0 through 4 successes. This can be found using the cumulative distribution function (CDF) of the binomial distribution:
\(\[ P(X < 5) = P(X \leq 4) = \sum_{k=0}^{4} \binom{15}{k} p^k (1-p)^{15-k} \]\)
Upon calculating this, the cumulative probability that she will make less than 5 of the next 15 putts is:
\(\[ P(X < 5) \approx 0.9039 \]\)
3) The probability that she will make at least 5 of the putts
The probability that she will make at least 5 putts is the complement of the probability that she will make less than 5 putts:
\(\[ P(X \geq 5) = 1 - P(X < 5) = 1 - P(X \leq 4) \]\)
From the previous result:
\(\[ P(X \geq 5) = 1 - 0.9039 \]\)
Upon calculating this, the probability that she will make at least 5 of the next 15 putts is:
\(\[ P(X \geq 5) \approx 0.0961 \]\)
So, to summarize the results:
1. The probability that she will make exactly 5 of the next 15 putts is approximately 0.0662.
2. The probability that she will make less than 5 of the next 15 putts is approximately 0.9039.
3. The probability that she will make at least 5 of the putts is approximately 0.0961.
