The probability that Miguel's score on the Water Hole is at most 5 is 0.80. The expected value of his score using the short hit is 4.45. The short hit has a better expected value than the long hit at a 0.4 probability of success, but if the probability of a successful long hit exceeds 0.55, the long hit becomes the better option.
To solve these problems related to Miguel's golf scores, you need to understand probability distributions and expected value calculations.
To find the probability that Miguel's score on the Water Hole is at most 5, we add the probabilities of scoring 3, 4, or 5: P(X ≤ 5) = 0.15 + 0.40 + 0.25 = 0.80.
The expected value of X is calculated by multiplying each score by its probability and adding the results:
E(X) = 3(0.15) + 4(0.40) + 5(0.25) + 6(0.15) + 7(0.05) = 4.45.
This is the average score Miguel is expected to make on the Water Hole when taking short hits.
For the long hit, the expected value when successful is 4.2 compared to 5.4 if it fails.
Let's calculate the overall expected value given a 0.4 probability of success:
E(Long Hit) = 0.4(4.2) + 0.6(5.4) = 4.92.
Since this expected value is higher than the expected value of the short hit (4.45), the short hit is better in terms of improving the expected score.
To determine which values of p make the long hit better, solve for p where E(Long Hit) with variable p is less than the expected value of the short hit: p(4.2) + (1 - p)(5.4) < 4.45.
Solving for p, we find that p > 0.55.
So, if the probability of a successful long hit is greater than 0.55, then the long hit approach would offer a better expected value.