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 the second round that day.
Estimate the probability that four golfers would score a hole in one on the sixth hole. Use the poisson approximation to the binomial. (Round the mean value to 3 decimal places and round your final answer to 9 decimal places.)
Well, I may be a clown bot, but I can still crunch some numbers for you!
To estimate the probability of four golfers scoring a hole in one on the sixth hole, we can use the Poisson approximation to the binomial distribution. The mean value for a Poisson distribution is equal to the product of the number of trials and the probability of success.
In this case, the number of trials is 155 (the number of golfers) and the probability of success is 1/3,709 (the probability of making a hole in one). So the mean value would be:
Mean = 155 * (1/3,709) = 0.04185
Now, let's use this mean value to calculate the probability of exactly four golfers scoring a hole in one on the sixth hole using the Poisson distribution. The formula for the probability is:
P(X = k) = (e^(-λ) * λ^k) / k!
where λ is the mean value and k is the number of successes. In this case, we want k to be 4.
P(X = 4) = (e^(-0.04185) * 0.04185^4) / 4!
Now, let's plug in the values and calculate the probability:
P(X = 4) = (e^(-0.04185) * (0.04185^4)) / (4!)
Calculating this gives us a probability of approximately 0.000018717.
So, the estimated probability that four golfers would score a hole in one on the sixth hole is approximately 0.000018717, or about 0.0018717%.
Although, it's worth mentioning that four hole-in-ones in one round is quite a remarkable feat! It's more likely to find a unicorn walking on a golf course than witnessing that.
To estimate the probability of four golfers scoring a hole in one on the sixth hole, we can use the Poisson approximation to the binomial distribution.
The mean value for a Poisson distribution is equal to the product of the number of trials (n) and the probability of success (p).
In this case, the number of trials is 155 (the number of golfers participating in the second round), and the probability of success (the probability of a hole in one) is 1/3,709.
So, the mean value (λ) for the Poisson distribution is given by:
λ = n * p
= 155 * (1/3,709)
≈ 0.0419 (rounded to 3 decimal places)
Now, we can use the Poisson probability formula to find the probability of exactly four golfers scoring a hole in one on the sixth hole:
P(X = k) = (e^(-λ) * λ^k) / k!
Where X is the random variable representing the number of golfers scoring a hole in one (which is equal to 4 in this case), e is the base of the natural logarithm (approximately 2.71828), λ is the mean value calculated earlier, and k! is the factorial of k.
Substituting the values:
P(X = 4) = (e^(-0.0419) * 0.0419^4) / 4!
Now, let's calculate the estimated probability:
P(X = 4) = (2.71828^(-0.0419) * 0.0419^4) / (4 * 3 * 2 * 1)
≈ 0.000000251 (rounded to 9 decimal places)
Therefore, the estimated probability that four golfers would score a hole in one on the sixth hole is approximately 0.000000251.
To estimate the probability that four golfers would score a hole in one on the sixth hole, we can use the Poisson approximation to the binomial distribution. The Poisson distribution is commonly used to model rare events occurring independently over a fixed interval.
First, we need to calculate the mean value (λ) for the Poisson distribution. The mean value is equal to the product of the number of trials (n) and the probability of success (p) for each trial in the binomial distribution.
In this case, the number of trials (n) is 155 (the number of golfers participating in the second round), and the probability of success (p) is 1/3,709 (the estimated probability of a hole in one for a professional golfer).
λ = n * p = 155 * 1/3,709 = 0.0417 (rounded to 3 decimal places)
Next, we can calculate the probability of exactly four hole-in-one scores using the Poisson distribution formula:
P(x = 4) = (e^(-λ) * λ^x) / x!
where e is the base of the natural logarithm (approximately 2.71828), λ is the mean value, and x is the number of hole-in-one scores (in this case, 4).
P(x = 4) = (e^(-0.0417) * 0.0417^4) / 4!
To calculate this probability, we need to use the value of e, which is approximately 2.71828. Let's go ahead and calculate it:
P(x = 4) = (2.71828^(-0.0417) * 0.0417^4) / (4 * 3 * 2 * 1)
P(x = 4) = (0.959560034 * 0.000283174321) / 24
P(x = 4) = 0.000011919338
Therefore, the estimated probability that four golfers would score a hole in one on the sixth hole, using the Poisson approximation to the binomial, is approximately 0.000011919338 (rounded to 9 decimal places).