Question 1206340
p = the probability that they use the credit cards because of the rewards program.
q = 1 - p = the probability that they don't use the credit cards because of the rewards program.


binomial probability distribution is used for this.


n = 10 = total number of possible choices.
x = 0 to 10 = the number of students who are using the credit card because of the rewards program.
p = .21
q = 1 - p = .79


find the probability that (a) exactly​ two, (b) more than​ two, (c) between two and five inclusive, use the credit cards for the rewards program.


for (a), you want p(x) for x = 2.
for (b), you want p(x) for x > 2.
for (c), you want p(x) for x = 2 to 5 inclusive.


i used excel to do the calculations.
they are shown below:


<img src = "http://theo.x10hosting.com/2024/030103.jpg">


problems are shown below.


find the probability that (a) exactly​ two, (b) more than​ two, (c) between two and five inclusive, use the credit cards for the rewards program.


answers to the problems are shown below.


p(x) for x = 2 is equal to 0.301070243.
p(x) for x > 2 is equal to 0.352558644.
p(x) for x = 2 to 5 inclusive is equal to 0.64543536.


the binomial probability formula is p(x) = p^x * q^(n-x) * c(n,x).


for example:


p(x = 2) is equal to .21^2 * .79^(10-2) * c(10,2) = .21^2 * .79^8 * 45 = .3010702433.
this agrees with the excel calculation when you round to the same number of decimal digits.


c(10,2) = 10! / (2! * 8!) = (10 * 9 * 8!) / (2! * 8!) = (10 * 9) / 2 = 45.


the general formula for c(n,x) is n! / (x! * (n-x)!)