Two urns A and B contain 100 red ball and 100 blue balls respectively. 7 balls are transferred from A to B then 7 from B to A.
if probability of drawing red ball from A is P(1) and probability of drawing red ball from B is P(2). Find P(1)+P(2).
Two urns A and B contain 100 red ball and 100 blue balls respectively. 7 balls are transferred from A to B then 7 from B to A.
if probability of drawing red ball from A is P(1) and probability of drawing red ball from B is P(2). Find P(1)+P(2).
After the swapping there are still 100 balls in each urn.
If $\displaystyle R_A$ is the number of red balls in urn A and $\displaystyle R_B$ is the number of red balls in urn B.
$\displaystyle P(1)+P(2)= \frac{R_A}{100}+ \frac{R_B}{100}$
Can you finish this off?
actually my own attempt is too weird. Let me explain what I did and correct me on it.
First we will have 93 balls in A and 107 in B (7 red +100 blue) (after first transfer of 7 balls)
then we will transfer 7 balls from B to A that can all be red or some be red.
suppose all ball are red and probability for this is 7C7/107C7 i.e choosing 7 red balls out of 107 balls in which only 7 are red
so P(a7)= (7C7/107C7)(100/100) i.e probability of drawing a red ball from A if all 7 balls from B got transferred to A.
similarly if only 6 balls are transferred
P(a6)=(7C6/107C7)(99/100)
and so on...........
so finally P(1)=P(a7)+P(a6)+.........P(a0).
similarly for B
probability of drawing a red ball when 7 red balls are transferred to A i.e P(7b)=(7C7/107C7)(0/100)
probability of drawing a red ball when 6 red balls are transferred to A i.e P(6b)=(6C7/107C7)(1/100)
so finally P(2)=P(7b)+P(6b)......P(b0).
then finally we can evaluate P(1)+P(2)