These are all questions about the Binomial Distribution, where there are exactly two possible outcomes for the events under consideration.

We need to use the probability fact that P(A) = 1 - P(A-complement)

and P(X = k) where X is the number of "successes" in n trials where probability of success is p

is given by nCk p^k ((1-p)^(n-k) Here nCk is number of combinations of n objects taken k at a time.

5.2/6 p = .20, n = 6

a) k = 0,

We want P(X = 0)

which is 6C0 .20^0 .80^6 = 1 * .80^6 = .2621 (rounded to 3 decimal places is .262)

The remaining problems 5.6/3 and are

b) P(At least 1 person has done a one-time fling)

= 1 - P(no people have done a one-time fling) = 1 - P(X = 0)

= 1 - .2621 = .7379 (rounded to 3 decimal places is .738)

c) P(No more than 2 people have done a one-fling) =

1 - P(0 people have done a one-time fling OR 1 person has done a one-time fling OR 2 people have done a one-time fling)

Since these events are disjoint, this is the same as

1 - [ P(0 people have done a one-time fling) + P( 1 person has done a one-time fling) + P( 2 people have done a one-time fling)]

We have calculated P(0 people have done a one-time fling) = .2621 in part a)

Now, P( 1 person has done a one-time fling) = 6C1 .2^1 .8^5 = 6 * .2^1 * .8^5 = .3932

and P( 2 people have done a one-time fling) = 6C2 .2^2 .8^4 = 15 * .2^2 * .8^4 = .2452

So, the desired probability is 1 - [.2621 + .3932 + .2452] = 1 - .9005 = .0995 (rounded to 3 decimal places is .100)

The remaining problems 5.2/7 and 9 are done similarly.