Question

Miguel is a golfer, and he plays on the same course each week. The following table shows the probability distribution for his score on one particular hole, known as the Water Hole.
Score 3 4 5 6 7
Probability 0.15 0.40 0.25 0.15 0.05
Let the random variable X represent Miguel’s score on the Water Hole. In golf, lower scores are better.

(a) Suppose one of Miguel’s scores from the Water Hole is selected at random. What is the probability that Miguel’s score on the Water Hole is at most 5? Show your work.





The name of the Water Hole comes from the small lake that lies between the tee, where the ball is first hit, and the hole. Miguel has two approaches to hitting the ball from the tee, the short hit and the long hit. The short hit results in the ball landing before the lake. The values of X in the table are based on the short hit. The long hit, if successful, results in the ball traveling over the lake and landing on the other side.

A potential issue with the long hit is that the ball might land in the water, which is not a good outcome. Miguel thinks that if the long hit is successful, his expected value improves to 4.2. However, if the long hit fails and the ball lands in the water, his expected value would be worse and increases to 5.4.

(c) Suppose the probability of a successful long hit is 0.4. Which approach, the short hit or the long hit, is better in terms of improving the expected value of the score? Justify your answer.



(d) Let p represent the probability of a successful long hit. What values of p will make the long hit better than the short hit in terms of improving the expected value of the score? Explain your reasoning.

Answer 1

Answer:

  (a) 0.80

  (c) short hit is better

  (d) p > 0.7083

Step-by-step explanation:

(a) The probabilities for scores of 6 or higher are 0.15 and 0.05. Their total is 0.20, so the probability of a 5 or less is 1 -0.20 = 0.80.

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(c) The expected score using the long hit approach is ...

  p(success)·(success score) +p(failure)·(failure score) = 0.4(4.2) +0.6(5.4) = 4.92

The expected score using the short hit approach is ...

  3·0.15 +4·0.40 +5·0.25 +6·0.15 +7·0.05 = 4.55

Miguel's expected score is lower using the short hit approach (4.55 vs 4.92).

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(d) As we saw above, for the probability of a successful long hit of p, the expected value of the score is ...

  p(4.2) +(1 -p)(5.4) = 5.4 -1.2p . . . . . . expected long hit score

We want to find p such that this value is less than the short hit expected value:

  5.4 -1.2p < 4.55

  0.85 -1.2p < 0 . . . . . subtract 4.55

  0.7083 - p < 0 . . . . . divide by 1.2

  p > 0.7083 . . . . . . . . add p

If the probability of a successful long hit is p > 0.7083, then the expected long hit score is less than the expected short hit score.