We have 16 balls , of which 4 are white , 3 are blue, and the rest that is 16-7 = 9 balls are yellow.
We are drawing 2 balls, and we need to find the probability of them being one white and one blue.
This can happen in two ways.
First way is drawing a white ball first, and the blue ball second.
So, the probability of drawing the white ball first is = 4/16 = 1/4
The probability of drawing the blue ball second is = 3/15 = 1/5
So, multiplying the both we get, = (1/4)x(1/5) = 1/20 (let this be way 1)
Second way is drawing the blue ball first and drawing the white ball second.
So, the probability of drawing the blue ball first is = 3/16
The probability of drawing the white ball second is = 4/15
Multiplying the above both we get , = (3/16)x(4/15) = 1/20 (let this be way 2)
Adding both ways we get, = (1/20)+(1/20) = 2/20 = 1/10.
So the correct answer is C.

(4C1 x 3C1)/(16C2)=1/10

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4/16 * 3/15 * 2 = 1/10

Since we are not replacing the first ball drawn, the probability of a white ball followed by a blue ball is:

4/16 x 3/15 = 1/4 x 1/5 = 1/20

The probability of a blue ball followed by a white ball is:

3/16 x 4/15 = 1/20

Thus the probability that one ball is white and the other is blue is the sum of the two probabilities:

1/20 + 1/20 = 2/20 = 1/10

Answer: C
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I made the mistake of just calculating the probability of the first ball being white and the second ball blue. Should both scenarios be considered unless explicitly stated that the first ball is white and the second ball is blue?