A box contains two red balls and four blue balls. A ball is drawn at random from the box and then replaced before a second ball is drawn. Find the probability of drawing two red balls.
- \(\frac{2}{3}\)
- \(\frac{1}{3}\)
- \(\frac{1}{4}\)
-
\(\frac{1}{9}\)
The correct answer is: D
Explanation
Total number of balls = 2 + 4 = 6
P(of picking a red ball) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
P(of picking a blue ball) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
With replacement,
P( picking two red balls) = \(\frac{1}{3}\) × \(\frac{1}{3}\) = \(\frac{1}{9}\)
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