Question

In an urn, there are Eights balls undistinguishable by touch. Three balls are red
labeled by 1;1;0, three balls are green labeled by 2;0;1, and two balls are white labeled
by 0;1.
We draw at random three balls at the same time and consider the following events:
A: Draw three balls with different color
B: Draw three balls with same label
a. Find the probability of each event: ? , ? and ? ∩ ?
b. Determine whether the events ? and ? are mutually exclusive
c. Find the probability of drawing three balls with different color and labeled with
same number.

Answer 1

A: Draw three balls with different color
B: Draw three balls with the same label

There are 8 balls and we take out 3 balls. Total combinations we can get are 8C3 = 56.

a. P(A) = P(Draw three balls with a different color) = P(one red + one green + one white)

=> 3C1*3C1*2C1 / 8C3 = 18/56 = 9/28

P(B) = P(Draw three balls with the same label).

There are four balls with the label as 1, three balls with the label as 0, and one with the label as 2.

For three balls with different labels, we can either choose three from the four label 1 balls or all three with label 0 balls. No, of ways = 4C3 + 3C3

P(B) = ( 4C3 + 3C3 )/ 8C3 = 5/56

P(A B ) = P(three balls with different color and with the same label)

So we have only options as {0,0,0} and {1,1,1}x2 because there are two red balls with the same label. Where numbers in bracket represent label for {Red, green, white} respectively.

There are only 3 choices. So P(A B ) = 3/56

b. There is no event E mentioned. I think its a typo, it should be B. Assuming it to be B.

The question is now to Determine whether the events A and B are mutually exclusive.

If A and B are mutually exclusive then P(A B ) should be zero. But it's not the case here.

So they are not mutually exclusive.

c. Find the probability of drawing three balls with different color and labeled with same number.

On looking carefully, it is the probability of A and B or P(A B ) which is 3/56

.