Probability & Statistics Ch.2 Section 3 Question 16

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

An urn contains 4 black balls and 5 white balls. After a thorough mixing, a ball is drawn from the urn, its color is noted, and the ball is returned to the urn.

a) What is the probability that 5 black balls are observed in 15 such draws? b) What is the probability that 15 draws are required until the first black ball is observed? c) What is the probability that 15 draws are required until the fifth black ball is observed?

Solution

(a) What is the probability that 5 black balls are observed in 15 such draws?

This uses the Binomial distribution.

P(Y=y) = \binom{n}{y}\theta^y(1-\theta)^{n-y}

The probability of success is the number of black balls divided by total balls: $\theta = \frac{4}{9}$

With n = 15, y = 5:

P(Y=5) = \binom{15}{5}\left(\frac{4}{9}\right)^5\left(\frac{5}{9}\right)^{10}

= 3003 \times \frac{1024}{59049} \times \frac{9765625}{3486784401}

P(Y=5) \approx 0.1459

(b) What is the probability that 15 draws are required until the first black ball is observed?

This uses the Geometric distribution. We want 14 failures before a success:

P(X=14) = \left(\frac{5}{9}\right)^{14}\left(\frac{4}{9}\right)

P(X=14) \approx 0.0000659

(c) What is the probability that 15 draws are required until the fifth black ball is observed?

This uses the Negative-Binomial distribution.

P(K=k) = \binom{r-1+k}{k}\theta^r(1-\theta)^k

Where:

r = 5 (desired successes)
k = 15 - 5 = 10 (failures)
θ = 4/9

P(K=10) = \binom{14}{10}\left(\frac{4}{9}\right)^5\left(\frac{5}{9}\right)^{10}

= 1001 \times \frac{1024}{59049} \times \frac{9765625}{3486784401}

P(K=10) \approx 0.0486