Question

According to a study done in an office, 6 out of 10 employees were interested in doing work from home,if three employees were selected at random from the above 10 employees,what is the probability that at most two of the selected were interested in doing work from home?

Answer 1

Answer:

The probability that at most two of the selected were interested in doing work from home is 5/6 or 0.833

Step-by-step explanation:

Consider the provided information.

It is given that 6 out of 10 employees were interested in doing work from home.

The probability of person interested in doing work from home is:

P(I) = 6/10

The probability of not interested person is:

P(N.I) = 4/10

We want out of 3 at most 2 of the selected were interested in doing work from home.

That means P(x = 0) + P(x = 1) + P(x = 2)

But the above method is the longer, instead of this we can find the probability as:

1 - P(x = 3)

P(at most 2) = 1-P(3) = $1-\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}$

P(at most 2) = 1-P(3) = $1-\frac{1}{6}$

P(at most 2) = 1-P(3) = $\frac{5}{6}\approx0.833$

Hence, the probability that at most two of the selected were interested in doing work from home is 5/6 or 0.833