Question

The 1989 U.S. Open golf tournament was played on the East Course of the Oak Hills Country Club in Rochester, New York. During the second round, four golfers scored a hole in one on the par 3 sixth hole. The odds of a professional golfer making a hole in one are estimated to be 3,708 to 1, so the probability is 1/3,709. There were 156 golfers participating in the second round that day.

a. What is the probability that no one gets a hole in one on the sixth hole? (Round your answer to 5 decimal places.)
Probability
b. What is the probability that exactly one golfer gets a hole in one on the sixth hole? (Round your answer to 5 decimal places.)
Probability
c. What is the probability that four golfers score a hole in one on the sixth hole? (Round your answer to 3 decimal places.)
Probability

Answer 1

To calculate the probabilities of golf scenarios, complement and binomial probability formulas are used. The probability that no golfer scores a hole in one is found using the complement rule raised to the number of golfers, while the probabilities of one and four holes in one are determined using the binomial probability formula.

Step-by-step explanation:

To determine the probability that no one gets a hole in one on the sixth hole, we can use the complement rule. The probability of a single golfer not getting a hole in one is 1 - (1/3,709). For all 156 golfers to not get a hole in one, raise this probability to the power of 156.

Probability no one gets a hole in one = (1 - 1/3,709)^{156} ≈ (3708/3709)^{156}.

Continuing, the probability that exactly one golfer gets a hole in one is calculated using the binomial probability formula with one success and 155 failures out of 156 trials:

Probability exactly one hole in one = (156 choose 1) * (1/3,709)^1 * (3708/3709)^{155}.

The probability that four golfers score a hole in one on the sixth hole can also be found using the binomial formula, but with four successes and 152 failures:

Probability four holes in one = (156 choose 4) * (1/3,709)^4 * (3708/3709)^{152}.

Answer 2

Using the binomial probability formula, the probabilities of no hole in one, exactly one, and exactly four holes in one among 156 golfers are calculated and rounded as requested. The events are incredibly rare with probabilities that decrease significantly as the number of holes in one increases.

Step-by-step explanation:

To find the probability of various outcomes when each of 156 golfers takes a shot at a par 3 hole, we use the binomial probability formula P(X=k) = (n C k) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successful outcomes, p is the probability of success on a single trial, and n C k is the binomial coefficient.

a. The probability that no one gets a hole in one on the sixth hole is calculated using the formula with k=0: P(0) = (156 C 0) * (1/3709)^0 * (3708/3709)^156 which equals approximately 0.95915 after rounding to five decimal places.

b. The probability that exactly one golfer gets a hole in one is found with k=1: P(1) = (156 C 1) * (1/3709)^1 * (3708/3709)^155 which equals approximately 0.04069 after rounding.

c. The probability that four golfers score a hole in one is calculated using k=4: P(4) = (156 C 4) * (1/3709)^4 * (3708/3709)^152. This is a significantly smaller probability, around 0.000 when rounded to three decimal places, indicating that such an event is incredibly rare.