Question

a large fast - food restaurant is having a promotional game where game pieces can be found on various products. customers can win food or cash prizes. according to the company, the probability of winning a prize (large or small) with any eligible purchase is 0.155. consider your next 33 purchases that produce a game piece. calculate the following: this is a binomial distribution. round your answers to at least 4 decimal places. a) what is the probability that you win 6 prizes? b) what is the probability that you win more than 8 prizes? c) what is the probability that you win between 3 and 7 (inclusive) prizes? d) what is the probability that you win 3 prizes or fewer? question help: video read message instructor

Explanation:

Step1: Recall binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successes, \(p\) is the probability of success in a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Given \(n = 33\), \(p=0.155\), \(1-p = 0.845\)

Part a)

Step1: Calculate \(C(33,6)\)

\(C(33,6)=\frac{33!}{6!(33 - 6)!}=\frac{33!}{6!×27!}=\frac{33\times32\times31\times30\times29\times28}{6\times5\times4\times3\times2\times1}=1107568\)

Step2: Calculate \(P(X = 6)\)

\(P(X = 6)=C(33,6)\times(0.155)^{6}\times(0.845)^{27}\)
\(P(X = 6)=1107568\times(0.155)^{6}\times(0.845)^{27}\approx0.1473\)

Part b)

Step1: Use complement rule

\(P(X>8)=1 - P(X\leq8)\)
\(P(X\leq8)=\sum_{k = 0}^{8}C(33,k)\times(0.155)^{k}\times(0.845)^{33 - k}\)
Using a binomial probability calculator or software:
\(C(33,0)\times(0.155)^{0}\times(0.845)^{33}+C(33,1)\times(0.155)^{1}\times(0.845)^{32}+\cdots+C(33,8)\times(0.155)^{8}\times(0.845)^{25}\)
\(P(X\leq8)\approx0.9394\)
\(P(X>8)=1 - 0.9394 = 0.0606\)

Part c)

Step1: Calculate \(P(3\leq X\leq7)\)

\(P(3\leq X\leq7)=\sum_{k = 3}^{7}C(33,k)\times(0.155)^{k}\times(0.845)^{33 - k}\)
\(P(3\leq X\leq7)=P(X = 3)+P(X = 4)+P(X = 5)+P(X = 6)+P(X = 7)\)
\(C(33,3)\times(0.155)^{3}\times(0.845)^{30}+C(33,4)\times(0.155)^{4}\times(0.845)^{29}+C(33,5)\times(0.155)^{5}\times(0.845)^{28}+C(33,6)\times(0.155)^{6}\times(0.845)^{27}+C(33,7)\times(0.155)^{7}\times(0.845)^{26}\)
\(P(3\leq X\leq7)\approx0.6570\)

Part d)

Step1: Calculate \(P(X\leq3)\)

\(P(X\leq3)=\sum_{k = 0}^{3}C(33,k)\times(0.155)^{k}\times(0.845)^{33 - k}\)
\(P(X\leq3)=C(33,0)\times(0.155)^{0}\times(0.845)^{33}+C(33,1)\times(0.155)^{1}\times(0.845)^{32}+C(33,2)\times(0.155)^{2}\times(0.845)^{31}+C(33,3)\times(0.155)^{3}\times(0.845)^{30}\)
\(P(X\leq3)\approx0.1603\)

Answer:

a) \(0.1473\)
b) \(0.0606\)
c) \(0.6570\)
d) \(0.1603\)