Question

a baseball player hit 64 home runs in a season. of the 64 home runs, 20 went to right field, 22 went to right center field, 11 went to center field, 9 went to left center field, and 2 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 0.313
(round to three decimal places as needed.)
(b) the probability that a randomly selected home run was hit to left field is 0.031
(round to three decimal places as needed.)
(c) was it unusual for this player to hit a home run to left field?
a. no, because the probability of an unusual event is 0
b. no, because this player hit 2 home runs to left field
c. yes, because p(left field) < 0.5
d. yes, because p(left field) < 0.05

Explanation:

Step1: Calculate probability formula

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

Step2: Solve part (a)

For a home - run hit to right field, the number of favorable outcomes \(n = 20\), and the total number of outcomes \(N=64\). Then \(P(\text{right field})=\frac{20}{64}=\frac{5}{16}=0.3125\approx0.313\)

Step3: Solve part (b)

For a home - run hit to left field, the number of favorable outcomes \(n = 2\), and the total number of outcomes \(N = 64\). Then \(P(\text{left field})=\frac{2}{64}=\frac{1}{32}=0.03125\approx0.031\)

Step4: Solve part (c)

An event is considered unusual if its probability \(P < 0.05\). Since \(P(\text{left field})\approx0.031<0.05\)

Answer:

(a) \(0.313\)
(b) \(0.031\)
(c) D. Yes because \(P(\text{left field})<0.05\)