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)equals nothing ​(round to the nearest thousandth as​ needed.) ​(b) ​p(x greater than​2)equals nothing ​(round to the nearest thousandth as​ needed.) ​(c) ​p(2less than or equals less than or equals​5)equals nothing ​(round to the nearest thousandth as​ needed.)

Answer 1

We use the binomial distribution, which states that:
Probability(r out of n) = (nCr) (p)^r (q)^(n-r)
In this case, n = 10 students, p = 39% = 0.39, and q = 1 - 0.39 = 0.61
a) For r = 2: Probability(2/10) = (10C2) (0.39)^2 (0.61)^(10-2) = 45(0.39)^2 (0.61)^8 = 0.1312, which is the probability for exactly 2.
b) We can first find the probability for r = 0 and r = 1, then subtract that from 1 to determine the probability of at least 2.
For r = 0: Probability(0/10) = (10C0) (0.39)^0 (0.61)^(10-0) = (1)(1)(0.61)^10 = 0.0071
For r = 1: Probability(1/10) = (10C1) (0.39)^1 (0.61)^(10-1) = (10)(0.39)(0.61)^9 = 0.0456
Then P(0/10) + P(1/10) = 0.0527, so P(at least 2/10) = 1 - 0.0527 = 0.9473.
(c) We have P(2/10) = 0.1312, and we can calculate for the rest similarly:
For r = 3: Probability(3/10) = (10C3) (0.39)^3 (0.61)^(10-3) = 0.2237
For r = 4: Probability(4/10) = (10C4) (0.39)^4 (0.61)^(10-4) = 0.2503
For r = 5: Probability(5/10) = (10C5) (0.39)^5 (0.61)^(10-5) = 0.1920
Therefore the sum of P(2) up to P(5) is 0.7972, so this is the probability of having between 2 to 5 inclusive.