Question

A baseball player hit 60 home runs in a season. Of the 60 home runs, 19 went to right field, 18 went to right center field, 12 went to center field, 10 went to left center field, and 1 went left field.
1) What is the probability that a randomly selected home run was hit to the right field?
2) What is the probability that a randomly selected home run was hit to left field?
3) Was it unusual for this player to hit a home run to left field? Explain.
A. No, because Upper P left parenthesis right-center field right parenthesis greater than 0.05.No because P(right center field)>0.05.
B.Yes, because of P(right center-right center field)less than<0.5.
C. No, because the probability of an unusual event is 0.
D. No, because this player hit 19 home runs to right center field.

Answer 1

Answer:

1. 19/60 = 0.32

2. 1/60 =0.12

Step-by-step explanation:

Right field = 19

Right center field =18

Center field =12

Left center field = 10

Left field = 1

Total home runs = 60

Let P stand for Probability

Then

1. P(right field) = 19/60 =0.32

2. P(left field) = 1/60 =0.12

3. Option C is the answer because the probability of an unusual events is less than 0.05. so 0 is also less than 0.05