Using binomial distribution, the probability of
a. 5 will still be with the company after 1 year is 28%.
b. at most 6 will still be with the company after 1 year is 89%.
The probability of a new employee in a fast-food chain still being with the company at the end of the year is given to be 0.6, which can be taken as the success of the experiment, p.
We are finding probability for people, thus, our sample size, n = 8.
Thus, we can show the given experiment a binomial distribution, with n = 8, and p = 0.6.
(i) We are asked for the probability that 5 will still be with the company.
Thus, we take x = 5.
P(X = 5) = (8C5)(0.6⁵)((1 - 0.6)⁸⁻⁵),
or, P(X = 5) = (56)(0.07776)(0.064),
or, P(X = 5) = 0.27869184 ≈ 0.28 or 28%.
(ii) We are asked for the probability that at most 6 will still be with the company.
Thus, our x = 6, and we need to take all values below it also.
P(X ≤ 6)
= 1 - P(X > 6)
= 1 - P(X = 7) - P(X = 8)
= 1 - (8C7)(0.6)⁷((1 - 0.6)⁸⁻⁷) - (8C8)(0.6)⁸((1 - 0.6)⁸⁻⁸)
= 1 - 8*0.0279936*0.4 - 1*0.01679616*1
= 1 - 0.08957952 - 0.01679616
= 0.89362432 ≈ 0.89 or 89%.
Learn more about the binomial distribution at
https://brainly.com/question/24756209
#SPJ4
The provided question is incomplete. The complete question is:
If the probability of new employee in a fast-food chain still being with the company at the end of 1 year is 0.6, what is the probability that out of 8 newly hired people,
a. 5 will still be with the company after 1 year?
b. at most 6 will still be with the company after 1 year?
