Answer:
0.05606
Step-by-step explanation:
This is a binomial probability distribution problem.
84% of all households have cable TV, thus;
p = 0.84
There are six households from this area, thus; n = 6
Formula for this binomial distribution is;
P(X) = C(n, x) × p^(x) × (1 - p)^(n - x)
We want to find the proportion of groups where at most three of the households have cable TV.
Thus, it's is;
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
P(X = 0) = C(6, 0) × 0.84^(0) × (1 - 0.84)^(6 - 0) = 0.00001678
P(X = 1) = C(6, 1) × 0.84^(1) × (1 - 0.84)^(6 - 1) = 0.000528
P(X = 2) = C(6, 2) × 0.84^(2) × (1 - 0.84)^(6 - 2) = 0.006963
P(X = 3) = C(6, 3) × 0.84^(3) × (1 - 0.84)^(6 - 3) = 0.04855
Thus;
P(X ≤ 3) = 0.00001678 + 0.000528 + 0.006963 + 0.04855 = 0.05606
