Answer:
The probability[tex]0.75095[/tex] and the parameter [tex]p=0.629[/tex]
Step-by-step explanation:
The formula for probability in a binomial distribution is:
[tex]b(x;n,p)=\frac{n!}{x!(n-x)!}\ast p^{x}\ast(1-p)^{n-x}[/tex]
where p is the probability of success (ticket with popcorn coupon), n is the number of trials (tickets bought) and x the number of successes desired. In this case p=0.629 (probability of buying a movie ticket with coupon), n=29, and x=17,18,19, ...29.
[tex]b(17;29,0.629)=\frac{29!}{17!(29-17)!}\ast0.629^{17}\ast(1-0.629)^{29-17}=0.133\,25\\ b(18;29,0.629)=\frac{29!}{18!(29-18)!}\ast0.629^{18}\ast(1-0.629)^{29-18}=0.150\,61[/tex]
[tex]b(19;29,0.629)=\frac{29!}{19!(29-19)!}\ast0.629^{19}\ast(1-0.629)^{29-19}=0.147\,84\\ b(20;29,0.629)=\frac{29!}{20!(29-20)!}\ast0.629^{20}\ast(1-0.629)^{29-20}=0.125\,32 \\ b(21;29,0.629)=\frac{29!}{21!(29-21)!}\ast0.629^{21}\ast(1-0.629)^{29-21}=0.091\,06 \\ b(22;29,0.629)=\frac{29!}{22!(29-22)!}\ast0.629^{22}\ast(1-0.629)^{29-22}=0.056\,14[/tex]
[tex]b(23;29,0.629)=\frac{29!}{23!(29-23)!}\ast0.629^{23}\ast(1-0.629)^{29-23}=2.896\,8\times10^{-2} \\ b(24;29,0.629)=\frac{29!}{24!(29-24)!}\ast0.629^{24}\ast(1-0.629)^{29-24}=1.227\,8\times10^{-2}\\ b(25;29,0.629)=\frac{29!}{25!(29-25)!}\ast0.629^{25}\ast(1-0.629)^{29-25}=4.163\,4\times10^{-3} \\ b(26;29,0.629)=\frac{29!}{26!(29-26)!}\ast0.629^{26}\ast(1-0.629)^{29-26}=1.085\,9\times10^{-3} \\ b(27;29,0.629)=\frac{29!}{27!(29-27)!}\ast0.629^{27}\ast(1-0.629)^{29-27}=2.045\,7\times10^{-4}[/tex]
[tex]b(28;29,0.629)=\frac{29!}{28!(29-28)!}\ast0.629^{28}\ast(1-0.629)^{29-28}=2.477\,4\times10^{-5} \\ b(29;29,0.629)=\frac{29!}{29!(29-29)!}\ast0.629^{29}\ast(1-0.629)^{29-29}=1.448\,3\times10^{-6}[/tex]
The probability of more than 16 is equal to the sum of the probability of x=17, 17,18,19, ...29.
[tex]b(x>16;29,0.629)=0.13325+0.15061+0.14784+0.12532+0.09106+0.05614+2.8968\times10^{-2}+1.2278\times10^{-2}+4.1634\times10^{-3}+1.0859\times10^{-3}+2.0457\times10^{-4}+2.4774\times10^{-5}+1.4483\times10^{-6}=0.75095[/tex]
