Question

15% of applicants for a job possess the right skills. A company has three positions to fill, and they interview applicants one at a time until they fill all three positions.. Use R commands to answer the following questions. (a) What is the probability that they will interview exactly fifteen applicants? (b) What is the probability that they will interview at least fifteen applicants? (c) What is the probability that they will interview at most fifteen applicants?

Answer 1

a) 0.0437

b) 0.6479

c) 0.3958

Explanation:

This is a negative binomial distribution problem. A negative binomial distribution problem describes a sequence of trials, each of which can have two outcomes (success or failure). We continue the trials indefinitely until we get r succeses.

The probability of y trials before obtaining r successes is given as

P(y) = ʸ⁻¹Cᵣ₋₁ pʳ qʸ⁻ʳ

p = 0.15

q = 1 - p = 0.85

r = 3

y = 15, ≥15 and ≤15

a) Probability of interviewing exactly 15 applicants.

p = 0.15

q = 1 - p = 0.85

r = 3

y = 15,

P(y=15) = ¹⁵⁻¹C₃₋₁ (0.15)³ (0.85)¹⁵⁻³

P(y=15) = ¹⁴C₂ (0.15)³ (0.85)¹² = 0.0436859997 = 0.0437

b) Probability of interviewing at least 15 applicants = P(y ≥ 15)

P(y ≥ 15) = 1 - P(y < 15)

P(y < 15) = P(y=3) + P(y=4) + P(y=5) + P(y=6) + P(y=7) + P(y=8) + P(y=9) + P(y=10) + P(y=11) + P(y=12) + P(y=13) + P(y=14)

= 0.003375 + 0.00860625 + 0.014630625 + 0.0207267188 + 0.0264265664 + 0.031447614 + 0.0356406292 + 0.0389501162 + 0.0413844985 + 0.0429938957 + 0.0438537736 + 0.0440531089

P(y < 15) = 0.3520887963

P(y ≥ 15) = 1 - P(y < 15) = 1 - 0.3520887963 = 0.6479112037 = 0.6479

c) Probability of interviewing at most 15 applicants

P(y ≤ 15) = P(y=3) + P(y=4) + P(y=5) + P(y=6) + P(y=7) + P(y=8) + P(y=9) + P(y=10) + P(y=11) + P(y=12) + P(y=13) + P(y=14) + P(y=15)

(y ≤ 15) = 0.003375 + 0.00860625 + 0.014630625 + 0.0207267188 + 0.0264265664 + 0.031447614 + 0.0356406292 + 0.0389501162 + 0.0413844985 + 0.0429938957 + 0.0438537736 + 0.0440531089 + 0.0436859997 = 0.395774796 = 0.3958

Hope this Helps!!!