Question

32​% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and ask each to name the reason he or she uses credit cards. Find the probability that the number of college students who say they use credit cards because of the rewards program is​ (a) exactly​ two, (b) more than​ two, and​ (c) between two and five inclusive. Of​ convenient, use technology to find the probabilities.

Answer 1

a)
`P(X=2)=(10C2)(0.32)^2 (1-0.32)^(10-2)=0.211`

b)
`P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668`

c)
`P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816`

Explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

`X \sim Binom(n=10, p=0.32)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

Part a

`P(X=2)=(10C2)(0.32)^2 (1-0.32)^(10-2)=0.211`

Part b

`P(X> 2)=1-P(X\leq 2)=1-[P(X=0)+P(X=1)+P(X=2)]`

`P(X=0)=(10C0)(0.32)^0 (1-0.32)^(10-0)=0.0211`

`P(X=1)=(10C1)(0.32)^1 (1-0.32)^(10-1)=0.0995`

`P(X=2)=(10C2)(0.32)^2 (1-0.32)^(10-2)=0.211`

`P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668`

Part c

`P(2 \leq x \leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)`

`P(X=2)=(10C2)(0.32)^2 (1-0.32)^(10-2)=0.211`

`P(X=3)=(10C3)(0.32)^3 (1-0.32)^(10-3)=0.264`

`P(X=4)=(10C4)(0.32)^4 (1-0.32)^(10-4)=0.218`

`P(X=5)=(10C5)(0.32)^5 (1-0.32)^(10-5)=0.123`

`P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816`