Finite Math: Probability

Question

Assume that 37% of the voting population in the US considers themselves as Democrats. Ten of these voters are selected at random

(a) What is the probability that exactly 3 of the 10 selected consider themselves as Democrats?

(b) What is the probability that at most 3 of the 10 selected consider themselves as Democrats?

Answer 1

Let n be the number of voters. There are 10 voters in the sample. Let p be the sample proportion for voters who are Democrats. p = 0.37

P(X = 3) = C(10,3) = [10!/(3!*7!)]*(0.37)³(0.63)⁷ = 0.2394

P(X ≤ 3) = C(10,0) + C(10,1) + C(10,2) + C(10,3)

= [10!/(0!*10!)]*(0.37)⁰(0.63)¹⁰ + [10!/(1!*9!)]*(0.37)¹(0.63)⁹ + [10!/(2!*8!)]*(0.37)²(0.63)⁸ + [10!/(3!*7!)]*(0.37)³(0.63)⁷

= 0.0098 + 0.0578 + 0.1529 + 0.2394 = 0.45999 ≈ 0.46

The probability that exactly 3 out of 10 voters who are Democrats is 0.2394.

The probability that at most 3 out of 10 voters who are Democrats is about 0.46.

Answer 2

a) C(10,3)(0.37)^3(0.63)^7

b) C(10,3)(0.37)^3(0.63)^7 +

C(10,2)(0.37)^2(0.63)^8

C(10,1)(0.37)^1(0.63)^9

C(10,0)(0.37)^0(0.63)^10