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2 answers:
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Answer:
a) 0.0797 = 7.97% probability that the next home run will be on his fifth at-bat.
b) The expected number of at-bats until the next home run is 6.25.
Step-by-step explanation:
For each at bat, there are two possible outcomes. Either it is a home run, or it is not. The probability of an at bat resulting in a home run is independent of any other at-bat, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
The probability that Barry Bonds hits a home run on any given at-bat is 0.16
This means that
Part A: What is the probability that the next home run will be on his fifth at-bat?
0 on his next 4(P(X = 0) when n = 4)
Home run on his 5th at-bat, with 0.16 probability. So
0.49787136 *0.16 = 0.0797.
0.0797 = 7.97% probability that the next home run will be on his fifth at-bat.
Part B: What is the expected number of at-bats until the next home run?
The expected number of trials for n successes is given by:
In this question, . So
The expected number of at-bats until the next home run is 6.25.
Answer:
65
Step-by-step explanation:
First let's multiply both the numerator and denominator by 100 so that the decimal number becomes a whole number.
15.6 * 100 / 0.24 * 100
= 1560 / 24
= 65
