Question

A baseball player hit 53 home runs in a season. Of the 53 home​ runs, 17 went to right​ field, 15 went to right center​ field, 10 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 nothing. ​(Round to three decimal places as​ needed.)

Answer 1

Answer:

a) 0.321 = 32.1% probability that a randomly selected home run was hit to right​ field

b) 0.038 = 3.8% probability that a randomly selected home run was hit to left​ field

c) 3.8% probability of hitting a home run to left field, which is lower than 5%, so yes, it was unusual for this player to hit a home run to left field.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

(a) What is the probability that a randomly selected home run was hit to right​ field?

Desired outcomes:

17 home runs to right field, so $D = 17$

Total outcomes:

53 total home runs, so $T = 53$

$p = \frac{17}{53} = 0.321$

0.321 = 32.1% 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?

Desired outcomes:

2 home runs to left field, so $D = 2$

Total outcomes:

53 total home runs, so $T = 53$

$p = \frac{2}{53} = 0.038$

0.038 = 3.8% 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?

Probabilities lower than 5% means that the event is unusual.

3.8% probability of hitting a home run to left field, which is lower than 5%, so yes, it was unusual for this player to hit a home run to left field.