Eight percent of all college graduates hired by companies stay with the same company for more than five years. The probability that in a random sample of 12 such college graduates hired recently by companies, exactly two will stay with the same company for more than five years is

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Adetunmbiadekunle Adetunmbiadekunle · Answer 1 · 2020-04-25T03:03:47+02:00

Answer:

0.1835

Step-by-step explanation:

This question can be solved by way of binomial distribution.

Let’s have the probability of success as p which is the probability that they stay with the company for more than five years;

p = 8/100 = 0.08

probability of failure, meaning not staying is q = 1-p = 1-0.08 = 0.92

Now let’s write the expression for the binomial distribution. That would be;

probability = 12C2 * 0.08^2 * 0.92^10

= 0.1835

MrRoyal MrRoyal · Answer 2 · 2021-10-29T12:07:39+02:00

Probabilities are used to determine the chances of an event.

The probability that exactly two stays more than 5 years is 0.1835

The given parameters are:

[tex]\mathbf{n = 12}[/tex] ---- sample

[tex]\mathbf{x = 2}[/tex] --- selected

[tex]\mathbf{p = 80\%}[/tex] --- sample proportion

So, the probability that exactly two stays more than 5 years is calculated using the following binomial probability:

[tex]\mathbf{P(x) = ^nC_x p^x (1 - p)^{n -x}}[/tex]

So, we have:

[tex]\mathbf{P(x) = ^{12}C_2 (8\%)^2 (1 - 8\%)^{12 -2}}[/tex]

[tex]\mathbf{P(x) = ^{12}C_2 (8\%)^2 (92\%)^{10}}[/tex]

Solve 12C2 using a calculator

[tex]\mathbf{P(x = 2) = 66 \times (8\%)^2 \times (92\%)^{10}}[/tex]

[tex]\mathbf{P(x = 2) = 0.1835}[/tex]

Hence, the probability that exactly two stays more than 5 years is 0.1835