Question

The probability that an employee will be late to work at a large corporation is 0.21. What is the probability on a given day that in a department of 5 employees at most 2 are late ?

Answer 1

In this problem, we have a probability binomial distribution

we have that

in a department of 5 employees at most 2 are late

means

0 employees late and 5 employees not late

1 employee late and 4 employees not late

2 employees late and 3 employees not late

the formula to calculate the probability is equal to

$P(X)=\frac{n!}{x!(n-x)!}p^x\cdot q^{(n-x)}$

where

p=0.21

q=1-p=1-0.21=0.79

n=5

so

Step 1

Find out the probability when x=0 (0 employees late and 5 employees not late)

substitute given values

$P(X=0)=\frac{5!}{0!(5-0)!}0.21^0\cdot0.79^{(5-0)}$

P(x=0)=0.3077

step 2

Find out the probability when x=1 (1 employee late and 4 employees not late)

substitute given values

$P(X=1)=\frac{5!}{1!(5-1)!}0.21^1\cdot0.79^{(5-1)}$

P(x=1)=0.4090

step 3

Find out the probability when x=2 (2 employees late and 3 employees not late)

substitute

$P(X=2)=\frac{5!}{2!(5-2)!}0.21^2\cdot0.79^{(5-2)}$

P(x=2)=0.2174

therefore

the probability on a given day that in a department of 5 employees at most 2 are

late is equal to

P(x≤2)=P(x=0)+P(x=1)+P(x=2)

P(x≤2)=0.3077+0.4090+0.2174

P(x≤2)=0.9341

the answer is

P=0.9341