Given that about 20% of adults do one time fling.
That is probability of a person doing fling is [tex]\frac{20}{100}=0.2[/tex] . Let it be p.
Then we have to use binomial distribution formula for the given problems.
b(x;n,p)=[tex]n_{C_{x}}p^{x}(1-p)^{n-x}[/tex]
A)Probability of no one has done one time fling means x is 0 here.
Hence [tex]b(0;9,0.2)=9_{C_{0} }(0.2)^{0}(1-0.2)^{9}[/tex]
[tex]=1X1X0.8^{9}=0.1342[/tex]
b) Probability of at least one person has done fling=1-(probability of no one has done)
=1-0.1342=0.8658
c)Probability of no more than two people have done one time fling means we need to add the probabilities for x=1,x=2 along with x=0.
[tex]b(1;9,0.2)=9_{C_{1}}(0.2)^{1}(1-0.2)^{8}[/tex]
[tex]=9X0.2X0.8^{8}=0.302[/tex]
[tex]b(2;9,0.2)=9_{C_{2}}0.2^{2}(1-0.2)^{7}[/tex]
[tex]=36X0.04X0.8^{7}=0.302[/tex]
Hence probability = 0.1342+0.302+0.302 = 0.7382
