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.)

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.