Question

5. Thirty percent of the employees of a large company are minorities. A random sample of 8 employees is selected.

(Total: 7 marks; a-c: 2 marks each, d: 1 mark)

a. What is the probability that the sample contains exactly 7 minorities?

b. What is the probability that the sample contains less than 5 minorities?

c. What is the probability that the sample contains exactly 1 non-minority?

d. What is the expected number of minorities in the sample

Answer 1

In this probability problem, we will use Binomial Distribution

Using the formula nCr ar bn-r

Let, the total number of employees selected be 8

Therefore, probability of minority = 0.30

a) Exactly 7 minorities =P(X = 7) = 8C7 * (0.30)7  * (0.70)1

= 0.0012

Required Probability = 0.0012

b)Less than 5 minorities = (X<5 ) = P( X=0) + P( X=1) + P( X=2) + P( X=3) + P( X=4)

= 8C0 * (0.30)0  * (0.70)8 + 8C1 * (0.30)1  * (0.70)7 + 8C2  * (0.30)2  * (0.70)6 + 8C3  * (0.30)3  * (0.70)5 +  8C4 * (0.30)4  * (0.70)4

= 0.058 + 0.197 + 0.296 + 0.254 + 0.136

= 0.941

Required Probability = 0.941

c) 1 Non Minority (X=1) = 8C1 * (0.70)1 * (0.30)7

= 0.0012

Required Probability = 0.0012

d) Expected Value = n*p

where, n = number of samples

p = probability of minorities

Expected Value = 8* 0.30

=2.4

So, expected number of minorities = 2.4