Question

a baseball player hit 63 home runs in a season. of the 63 home runs, 22 went to right field, 17 went to right center field, 12 went to center field, 11 went to left center field, and 1 went to left field
(a) what is the probability that a randomly selected home run was hit to right field?
(b) what is the probability that a randomly selected home run was hit to left field?
(c) was it unusual for this player to hit a home run to left field? explain.
(a) the probability that a randomly selected home run was hit to right field is 0.349
(round to three decimal places as needed.)
(b) the probability that a randomly selected home run was hit to left field is 0.016
(round to three decimal places as needed.)
(c) was it unusual for this player to hit a home run to left field?
a. yes, because p(left field) < 0.5
b. no, because this player hit 1 home runs to left field
c. yes, because p(left field) < 0.05
d. no, because the probability of an unusual event is 0

Explanation:

Step1: Recall the definition of probability

The probability formula is \(P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\).

Step2: Calculate probability for part (a)

For a home - run hit to right field, the number of favorable outcomes \(n = 22\) and the total number of home - runs \(N=63\). So \(P(\text{right field})=\frac{22}{63}\approx0.349\)

Step3: Calculate probability for part (b)

For a home - run hit to left field, the number of favorable outcomes \(n = 1\) and the total number of home - runs \(N = 63\). So \(P(\text{left field})=\frac{1}{63}\approx0.016\)

Step4: Determine if the event is unusual for part (c)

An event is considered unusual if its probability \(P(A)<0.05\). Since \(P(\text{left field})\approx0.016<0.05\)

Answer:

(a) \(0.349\)
(b) \(0.016\)
(c) C. Yes, because \(P(\text{left field})<0.05\)