Identify the parameter p in the following binomial distribution scenario. The probability of buying a movie ticket with a popcorn coupon is 0.774, and the probability of buying a movie ticket without a popcorn coupon is 0.226. If you buy 16 movie tickets, we want to know the probability that more than 10 of the tickets have popcorn coupons. (Consider tickets with popcorn coupons as successes in the binomial distribution.)

[tex]P(X=x)=^nC_xp^xq^{n-x}[/tex] , where n is the total number of trials , p is the probability of getting success in each trial and q is the probability of not getting success in each trial .

For the given scenario , Let x denotes the number of tickets have popcorn coupons as success.

As per given , the probability of buying a movie ticket with a popcorn coupon : p= 0.774

the probability of buying a movie ticket without a popcorn coupon : q= 0.226

if n= 16

Then, the probability that more than 10 of the tickets have popcorn coupons will be :-

[tex]P(x>10)=P(x=11)+P(x=12)+P(x=13)+P(x=14)+P(x=15)+P(x=16)\\\\=^{16}C_{11}(0.774)^{11}(0.226)^5+^{16}C_{12}(0.774)^{12}(0.226)^4+^{16}C_{13}(0.774)^{13}(0.226)^3+^{16}C_{14}(0.774)^{14}(0.226)^2+^{16}C_{15}(0.774)^{15}(0.226)^1+^{16}C_{16}(0.774)^{16}(0.226)^0\\\\=\dfrac{16!}{11!5!}(0.774)^{11}(0.226)^5+\dfrac{16!}{12!4!}(0.774)^{12}(0.226)^4+\dfrac{16!}{13!3!}(0.774)^{13}(0.226)^3+\dfrac{16!}{14!2!}(0.774)^{14}(0.226)^2+(16)(0.774)^{15}(0.226)^1+(1)(0.774)^{16}(0.226)^0\\\\=0.868404497984\approx0.8684[/tex]

Hence, the required probability = 0.8684