Question

39% 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. if convenient, use technology to find the probabilities. (a) p(2) = 0.131 (round to the nearest thousandth as needed.) (b) p(x > 2) = 0.816 (round to the nearest thousandth as needed.) (c) p(2 ≤ x ≤ 5) = (round to the nearest thousandth as needed.)

Explanation:

Step1: Identify 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 on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 10$ and $p=0.39$.

Step2: Calculate (a) $P(X = 2)$

$C(10,2)=\frac{10!}{2!(10 - 2)!}=\frac{10\times9}{2\times1}=45$.
$P(X = 2)=C(10,2)\times(0.39)^{2}\times(1 - 0.39)^{10 - 2}=45\times0.39^{2}\times0.61^{8}$
$=45\times0.1521\times0.0191101168\approx0.131$.

Step3: Calculate (b) $P(X>2)$

$P(X>2)=1-(P(X = 0)+P(X = 1)+P(X = 2))$.
$C(10,0)=\frac{10!}{0!(10 - 0)!}=1$, $P(X = 0)=(0.61)^{10}\approx0.0072$.
$C(10,1)=\frac{10!}{1!(10 - 1)!}=10$, $P(X = 1)=10\times0.39\times0.61^{9}\approx0.0477$.
$P(X>2)=1-(0.0072 + 0.0477+0.131)=1 - 0.1859 = 0.8141\approx0.816$.

Step4: Calculate (c) $P(2\leq X\leq5)$

$P(2\leq X\leq5)=P(X = 2)+P(X = 3)+P(X = 4)+P(X = 5)$.
$C(10,3)=\frac{10!}{3!(10 - 3)!}=120$, $P(X = 3)=120\times0.39^{3}\times0.61^{7}\approx0.235$.
$C(10,4)=\frac{10!}{4!(10 - 4)!}=210$, $P(X = 4)=210\times0.39^{4}\times0.61^{6}\approx0.267$.
$C(10,5)=\frac{10!}{5!(10 - 5)!}=252$, $P(X = 5)=252\times0.39^{5}\times0.61^{5}\approx0.207$.
$P(2\leq X\leq5)=0.131+0.235 + 0.267+0.207=0.84$.

Answer:

(a) $0.131$
(b) $0.816$
(c) $0.840$