Problem 55 - Basic Statistics for Business and Economics with Student

Question

During the second round of the 1989 U.S. Open golf tournament, four golfers scored a hole in one on the sixth hole. The odds of a professional golfer making a hole in one are estimated to be 3,708 to \(1,\) so the \(\begin{array}{lllll}\text { probability is } & 1 / 3,709 . & \text { There were } & 155 & \text { golfers }\end{array}\) participating in the second round that day. Estimate the probability that four golfers would score a hole in one on the sixth hole.

Step-by-Step Solution

Answer

Approximately \(1.94 \times 10^{-9}\) is the probability of such an event.

Step 1: Understand the probability of a single event

The probability of a single golfer making a hole in one is given as \(\frac{1}{3709}\). This represents a very small chance of success for each golfer.

Step 2: Use the binomial distribution formula

Since we are looking at a specific number of golfers (4) scoring a hole in one, use the binomial distribution: \[P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\] where \( n = 155 \) (number of golfers), \( k = 4 \) (number of successful outcomes), and \( p = \frac{1}{3709} \) (probability of a hole in one for one golfer).

Step 3: Compute the combinations \(\binom{n}{k}\)

Calculate the binomial coefficient \(\binom{155}{4}\): \[\binom{155}{4} = \frac{155 \times 154 \times 153 \times 152}{4 \times 3 \times 2 \times 1} = 3,829,553\] This represents the number of ways to choose 4 golfers out of 155.

Step 4: Calculate \( p^k \) and \( (1-p)^{n-k} \)

Compute the probability \( p^4 \) and \( (1-p)^{151} \):- \( p^4 = \left(\frac{1}{3709}\right)^4 = \frac{1}{\left(3709^4\right)} \)- \( (1-p)^{151} = \left(1 - \frac{1}{3709}\right)^{151} \approx 0.9597 \) Using calculators, we find:- \( 3709^4 = 1.88257 \times 10^{14} \), hence \( p^4 \approx 5.3 \times 10^{-16} \).

Step 5: Estimate probability using values

Plug these values into the binomial formula:\[P(X=4) = 3,829,553 \times 5.3 \times 10^{-16} \times 0.9597 \approx 1.94 \times 10^{-9}\]This very small number indicates the rarity of 4 golfers scoring a hole in one during the round.

Key Concepts

Probability Calculations

Probability calculations are crucial in understanding the likelihood of events occurring. In this context, the probability of a professional golfer hitting a hole in one is given as \( \frac{1}{3709} \). This fraction indicates a very slim chance, highlighting the rare nature of such an event.
To estimate the probability of multiple successes, like the four golfers achieving holes in one, we use the binomial distribution. It helps model scenarios where there are multiple identical trials, each with the same success probability.
When calculating such probabilities:

• Identify the number of trials (\( n \)), which is 155 golfers here.
• Determine the number of successful outcomes you are interested in (\( k \)), in this case, 4 golfers.
• Compute the probability \( p \) for a single success, given as \( \frac{1}{3709} \).

The rest is plugging values into the binomial formula to get the likelihood of exact outcomes.

Binomial Coefficient

Binomial coefficient, denoted as \( \binom{n}{k} \), plays a key role in solving problems involving multiple events. It essentially captures the number of ways to pick \( k \) successful outcomes from \( n \) trials.
In our golf scenario, it is the number of combinations of selecting 4 golfers out of 155 who achieve a hole in one. The formula for the binomial coefficient is: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] For our calculation:

• \( n=155 \) and \( k = 4 \)
• Compute \( \binom{155}{4} \) as \( \frac{155 \times 154 \times 153 \times 152}{4 \times 3 \times 2 \times 1} \)

This evaluates to 3,829,553, which is a measure of the various ways to achieve our desired outcome.

Rare Events

Rare events, like four golfers getting a hole in one simultaneously, are what make binomial probability so fascinating. These events are characterized by their low probability of occurrence, making predictions about them both challenging and intriguing.
In probability theory, rare events have tiny probabilities, often close to zero. In our exercise, the calculated probability of all four golfers making a hole in one is approximately \(1.94 \times 10^{-9}\).
Key aspects to consider with rare events include:

• Even if individual probabilities are small, with many trials, improbable events may occur.
• When predicting rare events, even smaller changes to parameters can result in significant shifts in probabilities.

These insights help clarify why certain occurrences are so extraordinary and how probability models assist in understanding them.