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)}](https://tex.z-dn.net/?f=%5Cfrac%7BmC2%20%2BnC2%7D%7B%28m%2Bn%29C2%7D%20%5C%5C%3D%5Cfrac%7Bm%28m-1%29%2Bn%28n-1%29%7D%7B%28m%2Bn%29%28m%2Bn-1%29%7D)
=![\frac{m^2+n^2-m-n}{(m+n)(m+n-1)}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%5E2%2Bn%5E2-m-n%7D%7B%28m%2Bn%29%28m%2Bn-1%29%7D)
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](https://tex.z-dn.net/?f=%28%5Cfrac%7Bm%7D%7Bm%2Bn%7D%20%29%5E2%20%2B%28%5Cfrac%7Bn%7D%7Bm%2Bn%7D%20%29%5E2)
=![\frac{m^2+n^2}{(m+n)^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%5E2%2Bn%5E2%7D%7B%28m%2Bn%29%5E2%7D)
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](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bm%2Bn%7D%20%3E%5Cfrac%7Bm-1%7D%7Bm%2Bn-1%7D%20%5C%5Cm%5E2%2Bmn-m%3Em%5E2%2Bmn-m-n%5C%5Cn%3E0)
Since n>0 is true always, b is greater than a.