Two balls are drawn simultaneously at random; 6 red, 4 white & 5 blues. What is P(one is white and one is red)?

Total number of balls = 6 + 4 + 5 = 15

P(one is white and one is red)
= P(the first is white and the last is red) + P(the first is red and the last is white)
= (4/15) × (6/14) + (6/15) × (4/14)
= 8/35

Total possible outcomes N = 15 = n(R) + n(W) + n(B) = 6 + 4 + 5 from a single draw of two

We are looking for p(R and W) = p(W) + p(R) + p(B) - p(W and R) - p(W and B) - p(B and R) - p(W and R and B); where p(W and R and B) = 0 because we're drawing only two balls. DRAW a Venn diagram to confirm this. [See source.]

So we have p(R and W) = 1 - (4/15)*(6/15) - (4/15)*(5/15) - (5/15)*(5/15) = 0.693 ANS.

NOTE: p(W) + p(R) + p(B) = n(W)/N + n(R)/N + n(B)/N = (n(R) + n(W) + n(B))/N = N/N = 1

Here is an alternate argument:

(ways to choose 1 red ball from 6)
(ways to choose 1 white ball from 4) /
(ways to choose 2 balls from 15)

C(6,1)C(4,1)/C(15,2)
= (6*4)/(15*14/2) = 8/35

(6/15)(4/14)2! = 8/35

P(one is white and one is red) = P(RW)+P(WR) = 2(4/15)(6/14)=...

1/15 * 1/14 = 1/210