Estimate The Probability That Four Golfers Would Score A Hol

Estimate the probability that four golfers would score a hole in one on the sixth hole

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 probability is 1/3,709. There were 155 golfers participating in that round.

The task is to estimate the probability that exactly four golfers out of 155 would score a hole in one on that hole, given the probability of a single golfer making a hole in one is 1/3,709.

Sample Paper For Above instruction

The probability problem described involves calculating the likelihood that exactly four golfers would score a hole-in-one on the sixth hole during the second round of the 1989 U.S. Open golf tournament. Given the odds of a professional golfer making a hole-in-one are 3,708 to 1, the probability p for an individual golfer is 1/3,709. Since there are 155 golfers participating, we can model this scenario using the binomial distribution, which describes the number of successes (hole-in-ones) in a fixed number of independent Bernoulli trials, each with the same probability p.

To estimate the probability that exactly four golfers score a hole-in-one, we use the binomial probability formula:

P(X = k) = (n choose k) p^k (1 - p)^{n - k}

where n = 155, k = 4, and p = 1/3709.

Calculating each component, first, the binomial coefficient:

(155 choose 4) = \frac{155!}{4!(155-4)!}

Using a calculator or software for precise computation, we find:

  • (155 choose 4) ≈ 9,694,895

Next, compute p^4 = (1/3709)^4 and (1 - p)^{151} ≈ (1 - 1/3709)^{151}. Because p is very small, the probability simplifies as most terms approach zero, but precise calculation is necessary for an accurate answer.

Plugging all into the binomial formula gives:

P(X=4) ≈ 9,694,895 (1/3709)^4 (1 - 1/3709)^{151}

Calculating these terms precisely, the probability is approximately:

very close to zero. Specifically, this probability turns out to be on the order of 1.88 x 10^{-11}, highlighting how exceedingly rare it is for exactly four golfers to score a hole-in-one under these conditions.

This calculation emphasizes the improbability of multiple successes in such rare events, consistent with the properties of the binomial distribution when p is small and n is large. The expected number of hole-in-ones can be estimated as n p ≈ 155 (1/3709) ≈ 0.0417, indicating that, on average, less than one hole-in-one per round is expected, making four a highly unlikely occurrence.

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