Q1
a - To maximize the chance of picking 2 white (W) balls, the best way
is to put 2 W balls in the urn U1 and 48 W balls + 50 B (black) balls in
the urn U2.
There are 4 four possibilities to choose the two urns : U1 & U1, U1 &
U2, U2 & U1, U2 & U2, each of them having the same probability 1/4.
So the probability P(1a) to pick the two W balls is equal to:
P(1a) = 1/4 * [2/2 * 1/1 + 2/2 * 48/98 + 48/98*2/2 + 48/98 * 47/97 ] =
10537 / 19012 = 0,554228...
It is easy to check that if the urn U1 contained a higher number of
W balls or a certain number of B balls, the probability P(1a) should
decrease.
b- Same reasoning: we put one W ball and one B ball in the urn U1. So
the urn U2 contains 49 W balls and 49 B balls.
So the probability to pick one W ball and one B ball (in this order) is
equal to:
1/4 * [1/2 * 1/1 + 1/2 * 49/98 + 49/98 * 1/2 + 49/98 * 49/97]
The probability to pick one B ball and one W ball (in this order) is the
same.
So the requested probability to pick one W ball and one B ball (in
any order) is equal to:
P(1b) = 1/4 * [ 1 + 2 * 49/98 + 2 * 49/98 * 49/97] = 243 /
388 = 0,626288....
Q2
We have two possible ways to choose the urns: U1 & U1 ot U2 & U2,
the corresponding probabilities being 1/2 and 1/2.
The compositions of the urns are respectively the same as in Q1.
Therefore:
a - P(2a) = 1/2*[ 2/2 * 1/1 + 48/98 * 47/97 ] = 5881 / 9506
= 0,618661...
b - P(2b) = 1/2 *[1 + 2* 49/98 * 49/97] =73 / 97 =
0,752577....
Q3
The smallest number of balls to make chance P of pulling out 2 W
balls without replacement each from a random urn exactly 1/2, is
equal to 4 balls with a distribution of 2 W balls
in the urn U1 and 1 W ball and 1 B ball in the urn U2.
Indeed P = 1/4 * [ 2/2 * 1/1 + 2/2 * 1/2 + 1/2 * 2/2 + 1/2 * 0] =
1/2
If P = 1/3, we put 2 W balls in the urn U1, 1 W ball and n B balls
in the urn U2 such that
1/4 * [2/2 * 1/1 + 2/2 * 1/(n+1) + 1/(n+1) * 2/2 + 1/(n+1) * 0 ] =
1/3
Therefore n = 5. Then there are 8 balls with 3 W
balls (2 of then in U1, the 3rd in U2) and 5 B balls all of them in
U2.