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 =
= [tex]\frac{mC2 +nC2}{(m+n)C2} \\=\frac{m(m-1)+n(n-1)}{(m+n)(m+n-1)}[/tex]
=[tex]\frac{m^2+n^2-m-n}{(m+n)(m+n-1)}[/tex]
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 = [tex](\frac{m}{m+n} )^2 +(\frac{n}{m+n} )^2[/tex]
=[tex]\frac{m^2+n^2}{(m+n)^2}[/tex]
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
[tex]\frac{m}{m+n} >\frac{m-1}{m+n-1} \\m^2+mn-m>m^2+mn-m-n\\n>0[/tex]
Since n>0 is true always, b is greater than a.
