Suppose that in a large metropolitan area, 84% of all households have cable tv. Suppose you are interested in selecting a group of six households from this area. Let X be the number of households in a group of six households from this area that have cable tv. For what proportion of groups will at most three of the households have cable tv?

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

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