Question

On the leeward side of the island of Oahu, in the small village of Nanakuli, about 70% of the residents are of Hawaiian ancestry. Let n = 1, 2, 3, ... represent the number of people you must meet until you encounter the first person of Hawaiian ancestry in the village of Nanakuli. (a) Write out a formula for the probability distribution of the random variable n. (Use p and n in your answer.) P(n) = Incorrect: Your answer is incorrect. (b) Compute the probabilities that n = 1, n = 2, and n = 3. (Use 3 decimal places.) P(1) P(2) P(3) (c) Compute the probability that n ≥ 4. (Use 3 decimal places.) (d) In Waikiki, it is estimated that about 7% of the residents are of Hawaiian ancestry. Repeat parts (a), (b), and (c) for Waikiki. (Use 3 decimal places.) (a) P(n) = (b) P(1) P(2) P(3) (c)

Answer 1

(a)

Probability of Hawaiian ancestry in the village of Nanakuli, p = 0.7

Probability of non-Hawaiian ancestry in the village of Nanakuli = 1- p = 1 - 0.7 = 0.3

If the nth person is the first person of Hawaiian ancestry in the village of Nanakuli, then first (n-1) persons are of non-Hawaiian ancestry.

P(n) = (1-p)^n-1p

P(n) =0.7 x 0.3^n-1

(b)

P(1) = 0.7 * 0.3^1-1 = 0.7

P(2) = 0.7 * 0.3^2-1 = 0.21

P(3) = 0.7 * 0.3^3-1 = 0.063

(c)

P(n ≥ 4) = Probability that first 3 persons are non-Hawaiian ancestry = 0.3³ = 0.027

(d)

Probability of Hawaiian ancestry in the village of Waikiki, p = 0.07

Probability of non-Hawaiian ancestry in the village of Waikiki = 1- p = 1 - 0.07 = 0.93

If the nth person is the first person of Hawaiian ancestry in the village of Waikiki, then first (n-1) persons are of non-Hawaiian ancestry.

P(n) = (1-p)^n-1p

P(n) =0.07 x 0.93^n-1

P(1) = 0.07 * 0.93^1-1 = 0.07

P(2) = 0.07 * 0.93^2-1 = 0.065

P(3) = 0.07 * 0.93^3-1 = 0.061

P(n ≥ 4) = Probability that first 3 persons are non-Hawaiian ancestry = 0.93³ = 0.804