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 than2)equals nothing (round to the nearest thousandth as needed.) (c) p(2less than or equals less than or equals5)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.

2160ft^3 i think you could use a formala to you out with this in the future