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.

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

Answer

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: $\theta=\frac{4}{9}$

$$P(K=k)={{r-1+k}\choose k}\theta^r(1-\theta )^k$$ $$P(K=10)={{5-1+10}\choose 10}(\frac{4}{9})^5(1-\frac{4}{9} )^10$$ $$P(K=10)={14 \choose 10}(\frac{4}{9})^5(\frac{5}{9} )^10$$ $$P(K=10)=1001 * (\frac{1024}{59049})(\frac{9765625}{3486784401} )$$ $$P(k=10)\approx 0.0486179$$