Question

Exercise 1.28. We have an urn with m green balls and n yellow balls. Two balls are drawn at random. What is the probability that the two balls have the same color? (a) Assume that the balls are sampled without replacement. (b) Assume that the balls are sampled with replacement. (c) When is the answer to part (b) larger than the answer to part (a)? Justify your answer. Can you give an intuitive explanation for what the calculation tells you?

Answer 1

Answer:

Step-by-step explanation:

given that we  have an urn with m green balls and n yellow balls. Two balls are drawn at random.

a) Assume that the balls are sampled without replacement.

m green and n yellow balls

For 2 balls to be drawn at the same colour

no of ways = either 2 green or 2 blue = mC2+nC2

Total no of ways = (m+n)C2

Prob =

= $\frac{mC2 +nC2}{(m+n)C2} \\=\frac{m(m-1)+n(n-1)}{(m+n)(m+n-1)}$

=$\frac{m^2+n^2-m-n}{(m+n)(m+n-1)}$

B) Assume that the balls are sampled with replacement

In this case, probability for any draw for yellow or green will be constant as

n/M+n or m/m+n respectively

Reqd prob = $(\frac{m}{m+n} )^2 +(\frac{n}{m+n} )^2$

=$\frac{m^2+n^2}{(m+n)^2}$

c) Part B prob will be more than part a because with replacement prob is more than without replacement.

II time drawing same colour changes to m-1/.(m+n-1) if with replacement but same as m/(m+n) without replacement

$\frac{m}{m+n} >\frac{m-1}{m+n-1} \\m^2+mn-m>m^2+mn-m-n\\n>0$

Since n>0 is true always, b is greater than a.