In a box, there are eight yellow and four black balls. If three balls are drawn at random, what is the probability that two are yellow and one black?
Probability = $$Expected number of outcomes\div Total number of outcomes$$.
Total number of outcomes = Total number of ways to draw randomly 3 balls out of 12 balls = 12C3 ways.
Expected number of outcomes = Number of ways to draw 2 yellow balls out of 8 and 1 black ball out of 4 = $$8C2\times 4C1$$
$$ Hence Probability = 8C2\times 4C1\div 12C3 = 28\times 4\div 220 = \frac{28}{55}$$
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