Answer-
The probability of meeting at least one person with the flu in 12 is 0.3062 or 30.62%
Solution-
The probability of k successes in n trials where p is the probability of success on any given trial is given by,
[tex]P(n;k,p)=\binom{n}{k}(p)^k(1-p)^{n-k}[/tex]
Here, 3 people out of 100 have the flu, so
[tex]p=0.03[/tex]
As there are total 12 number of people, so
[tex]n=12[/tex]
And
[tex]P(\text{At least one flu})=1-P(\text{0 flu})[/tex]
The probability selecting 0 number of person with the flu out of 12 is,
[tex]=\binom{12}{0}(0.03)^0(1-0.03)^{12-0}[/tex]
[tex]=1\times 1\times(0.97)^{12}[/tex]
[tex]=0.6938[/tex]
So, the probability of meeting at least one person with the flu in 12 is,
[tex]=1-0.6938[/tex]
[tex]=0.3062[/tex]
[tex]=30.62\%[/tex]
