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Chapter 6: Problem 55

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.

Short Answer

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

Step by step solution

01

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.
02

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).
03

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.
04

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 \( p^4 \approx 5.3 \times 10^{-16} \).
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A recent CBS News survey reported that 67 percent of adults felt the U.S. Treasury should continue making pennies. Suppose we select a sample of 15 adults. a. How many of the 15 would we expect to indicate that the Treasury should continue making pennies? What is the standard deviation? b. What is the likelihood that exactly 8 adults would indicate the Treasury should continue making pennies? c. What is the likelihood at least 8 adults would indicate the Treasury should continue making pennies?

Tire and Auto Supply is considering a 2 -for- 1 stock split. Before the transaction is finalized, at least two thirds of the 1,200 company stockholders must approve the proposal. To evaluate the likelihood the proposal will be approved, the CFO selected a sample of 18 stockholders. He contacted each and found 14 approved of the proposed split. What is the likelihood of this event, assuming two-thirds of the stockholders approve?

In a Poisson distribution \(\mu=0.4\). a. What is the probability that \(x=0 ?\) b. What is the probability that \(x>0 ?\)

Thirty percent of the population in a southwestern community are Spanish- speaking Americans. A Spanish speaking person is accused of killing a non- Spanish speaking American and goes to trial. Of the first 12 potential jurors, only 2 are Spanish-speaking Americans, and 10 are not. The defendant's lawyer challenges the jury selection, claiming bias against her client. The government lawyer disagrees, saying that the probability of this particular jury composition is common. Compute the probability and discuss the assumptions.

A telemarketer makes six phone calls per hour and is able to make a sale on 30 percent of these contacts. During the next two hours, find: a. The probability of making exactly four sales. b. The probability of making no sales. c. The probability of making exactly two sales. d. The mean number of sales in the two-hour period.
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