Approximately 8 of people are lefthanded If two people are s

Approximately 8% of people are left-handed. If two people are selected at random, what is the probability of the following events?

A. P(Both are right-handed) = .8464

B. P(Both are left-handed) = .0064

C. P(One is right-handed and the other is left-handed) = ?

D. P(At least one is right-handed) = ?

This is basic probability, please show steps and clearly state the correct answer. PLEASE AND THANK YOU!!!!

C. P(One is right-handed and the other is left-handed) = ?

Given X~Binomial(n=2, p=0.08)

P(X=x)=2Cx*(0.08^x)*(0.92^(2-x))

So P(One is right-handed and the other is left-handed)

=P(Only one is left-handed)

=P(X=1)

=2C1*(0.08^1)*(0.92^(2-1))

=0.1472

---------------------------------------------------------------------------------------------------

D. P(At least one is right-handed) = ?

=1- P(no one is right-handed)

=1-P(both are left-handed)

=1-0.0064

=0.9936