A box contains 5 blue balls, 3 black balls and 2 red balls of the same size. A ball is selected at random from the box and then replaced. A second ball is then selected. Find the probability of obtaining
(a) two red balls ;
(b) two blue balls or two black balls ;
(c) one black and one red ball in any order.
Explanation
Number of blue balls = 5
Number of black balls = 3
Number of red balls = 2
= 10 balls.
(a) P(two red balls) = \(\frac{2}{10} \times \frac{2}{10} = \frac{1}{25}\)
(b) P(2 blue or 2 black balls) = P(2 blue) or P(2 black)
= \((\frac{5}{10} \times \frac{5}{10}) + (\frac{3}{10} \times \frac{3}{10})\)
= \(\frac{25}{100} + \frac{9}{100}\)
= \(\frac{34}{100} = \frac{17}{50}\)
(c) P( one red and one black in any order) = P(one red, one black) or P(one black, one red)
= \((\frac{2}{10} \times \frac{3}{10}) + (\frac{3}{10} \times \frac{2}{10})\)
= \(\frac{12}{100} = \frac{3}{25}\)