There are 5 urns, the nth urn contains n − 1 red balls and 5 − n green balls (in total 4 ineach urn). We pick an urn at random and take 2 balls out of it without replacement.
(a) What is the probability the second ball is green?
(b) What is the probability the second ball is green given the first is green?
(c) What is the probability we picked either the first urn or second urn given both balls are green?
for (a), I did:
P(2nd ball is green) = 5*[{P(i th urn picked) * P(first ball drawn red/ ith urn picked) P(2nd ball drwan green/first ball red)}+ {P(i th urn picked) * P(first ball drawn green/ ith urn picked) P(2nd ball drwan green/first ball green)}]
= 5*[[{1/5 * (n-1)/4 * (5-n)/3 } + {1/5 * (5-n)/4 * (4-n)/3}] = 5* [{(n-1)(5-n)/60} + {(5-n)(4-n)/ 60}]
=[(n-1)(5-n) + (5-n)(4-n)]/12 = (5-n)(n-1+4-n)/12 = (5-n)*3/12 = (5-n)/4
And for b): 5*{P(i th urn picked) * P(first ball drawn green/ ith urn picked) P(2nd ball drwan green/first ball green)} = 5*{1/5 * (5-n)/4 * (4-n)/3} =(4-n)(5-n)/12
c), P(1st urn or 2nd urn/ both are green) = [ [P(1st or 2nd]* (4-n)(5-n)/6 ] / [{P(1st or 2nd]* (4-n)(5-n)/6} + {{P(3rd or 4th]* (4-n)(5-n)/6}] = {(4-n)(5-n)/15} / 2*{(4-n)(5-n)/15} = 1/2
But I am really uncertain about my solutions. If anyone has different solution or easier way to get the answer please let me know! Thanks a lot!
Best Answer
There are $5\cdot4\cdot3=60$ ways in which two balls can be drawn: $$\begin{array}{cr} GG&20\\ GR&10\\ RG&10\\ RR&20\\ \hline \text{Total}&60 \end{array}$$ These are the $20$ ways in which $GG$ can be drawn: $$\begin{array}{cr} GG_1&12\\ GG_2&6\\ GG_3&2\\ \hline \text{Total}&20 \end{array}$$ $$\bbox[border:2px solid green]{ \begin{array}{cc|cc|cc} \text{(a)}&P=\frac{30}{60}=\frac{1}{2}& \text{(b)}&P=\frac{20}{30}=\frac{2}{3}& \text{(c)}&P=\frac{18}{20}=\frac{9}{10} \end{array} }$$