MATH SOLVE

4 months ago

Q:
# A baseball player hit 6060 home runs in a season. Of the 6060 home runs, 1919 went to right field, 2020 went to right center field, 99 went to center field, 1010 went to left center field, and 22 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.

Accepted Solution

A:

Answer: (a) 0.317 (b) 0.167 (c) Not unusualStep-by-step explanation:Probability is defined as the chance of an event occurring.Let S be a sample space and A be an event in the sample space, the probability of event A occurring can be found using the following:P(A) = n(A) divided by n(S) where n(A) = number of elements in event A and n(S) = number of elements in the sample space.(a) P(Right field) = Number of hits to the right field divided by total number of hits = 1919/6060 = 19/60 which is approximately equal to 0.317(b) P(Left field) = 1010/6060 = 1/6 which is approximately equal to 0.167(c) The probability that the home run was hit to the left field is 0.167. The probability of an unusual event is 0, therefore, it is possibly for the player to hit a home run to the left.