a. 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: two red balls
b. 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: two blue balls or two black balls
c. 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: one black ball and one red ball in any order
Verified Correct Answer: .
Step-by-Step Mathematical Solution
a. Total ball = 5 blues + 3 blacks + 2 red balls = 10 balls
Selection with replacement
Pr(1st red ball) = \(\frac{2}{10}\)
Pr(2nd red ball) = \(\frac{2}{10}\)
Therefore, Pr(two red balls) = \(\frac{2}{10}\) x \(\frac{2}{10}\) = \(\frac{4}{100}\)
= \(\frac{1}{25}\) = 0.04
b. Pr( of selecting 2 blue) = \(\frac{5}{10}\) x \(\frac{5}{10}\) = \(\frac{25}{100}\)
= \(\frac{1}{4}\)
Pr(of selecting 2 black balls) = \(\frac{3}{10}\) x \(\frac{3}{10}\) = \(\frac{9}{100}\)
Pr( 2 blue or 2 black balls) = \(\frac{1}{4}\) + \(\frac{9}{100}\) = \(\frac{25}{100}\) + \(\frac{9}{100}\) = \(\frac{34}{100}\) = 0.34.
c) 1 black, and 1 red in any other means it can be 1 black, and 1 red, and 1 black
= \(\frac{3}{10}\) x \(\frac{2}{10}\) + \(\frac{2}{10}\) + \(\frac{3}{10}\)
= \(\frac{6}{100}\) + \(\frac{6}{100}\) = \(\frac{12}{100}\) = 0.12.
Source: Myschool Past Questions Archive (Record #76113)