Chapter 2 Section 3 Question 16
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.
1. What is the probability that 5 black balls are observed in 15 such draws?
2. What is the probability that 15 draws are required until the first black ball is observed?
3. What is the probability that 15 draws are required until the fifth black ball is observed?
Author: Mohammad-Ali Bandzar| Date:Oct 19 2020
(a) What is the probability that 5 black balls are observed in 15 such draws?
This question is asking for the binomial probability as all our trials are independent.
From Example 2.3.3 in the book we know: $$p\scriptsize{Y}\normalsize{(y)=P(Y=y)={n\choose y}\theta ^y(1-\theta)^{n-y}}$$ n represents the number of trials
$\theta$ represents the probability of a success
y represents the desired number of successful outcomes

In this question the probability of success is the number of black balls divided by the total number of balls: $\theta=\frac{4}{9}$
We have 15 independent trials. Therefore $n=15$
$$Y ~ Binomial(15, 4/9)$$ $$P(Y=y)={n\choose y}\theta ^y(1-\theta)^{n-y}$$ $$P(Y=5)={15\choose 5}(\frac{4}{9}) ^5(1-\frac{4}{9})^{15-5}$$ $$P(Y=5)=3003*(\frac{4}{9}) ^5(\frac{5}{9})^{10}$$ $$P(Y=5)\approx .077*0.0028$$ $$P(Y=5)\approx 0.1459$$
(b) What is the probability that 15 draws are required until the first black ball is observed?
Since we are trying to calculate number of failures before the first success we have a geometric distribution
From Example 2.3.4 in the book we know: $$p\scriptsize{X}\normalsize{(x)=P(X=x)=(1-\theta )^x\theta}$$ $\theta$ is the probability of success per event
x represents the number of failed events before a success.

in this case we want 14 failures before a success so $x=14$
In this question the probability of success is the number of black balls divided by the total number of balls: $\theta=\frac{4}{9}$
$$P(X=x)=(1-\theta )^x\theta$$ $$P(X=14)=(1-\frac{4}{9} )^15(\frac{4}{9})$$ $$P(X=14)=(\frac{5}{9} )^15(\frac{4}{9}$$ $$P(X=14)\approx 0.000065876$$
(c) What is the probability that 15 draws are required until the fifth black ball is observed?
This is negative binomial distribution as our trials stop after a certain number of successful events. Since the desired number of successful events is greater than one this is negative-Binomial distribution instead of geometric.
From Example 2.3.5 in the book we know: $$p\scriptsize{K}\normalsize{(k)=P(K=k)={{r-1+k}\choose k}\theta^r(1-\theta )^k}$$ r represents the desired number of successes
k represents the number of failures
$\theta$ represents the probability of a success.

since we want five successes $r=5$
since we want 15 total trials, trials-successes=failures: $k=15-5=10$ In this question the probability of success is the number of black balls divided by the total number of balls: $\theta=\frac{4}{9}$
$$P(K=k)={{r-1+k}\choose k}\theta^r(1-\theta )^k$$ $$P(K=10)={{5-1+10}\choose 10}(\frac{4}{9})^5(1-\frac{4}{9} )^10$$ $$P(K=10)={14 \choose 10}(\frac{4}{9})^5(\frac{5}{9} )^10$$ $$P(K=10)=1001 * (\frac{1024}{59049})(\frac{9765625}{3486784401} )$$ $$P(k=10)\approx 0.0486179$$