Question

a.    What is the probability of the first selected person is a left-handed men and the second selected person is a right-handed men?

 

Answer 1

suppose there are 10 persons and 4 are left handed and 6 right handed persons.

Case 1: without replacement

P(first is left handed and second right handed)= $\frac{4}{10}*\frac{6}{9}$ = $\frac{24}{90}$ = 0.27

Case 2: With replacement

P(first is left handed and second right handed)= $\frac{4}{10}*\frac{6}{10}$ =$\frac{24}{100}$ =0.24