Question

Assume that you are drawing two balls without replacement from an urn that contains 13 green balls, 10 blue balls, and 1 red balls. What is the probability that you will draw a blue ball and a red ball? (Hint there is two ways this can happen).

Answer 1

Number of green balls = 13.

Number of blue balls = 10.

Number of red balls = 1.

Total number of balls  = 13 + 10 + 1 = 24.

The sequence of events are 1) drawing a blue ball and 2) drawing a red ball.

1) Sample space is a group of 24 balls, 10 of which are blue balls.

The probability of drawing a blue ball is 10/24 = 5/12.

2) Since the blue ball is removed from an urn, the sample space will now include 23 balls, 1 of which is a red ball.

Therefore the probability of drawing a red ball is 1/23.

Therefore the propability of complete sequence of events is (5/12) * (1/ 23) = 5 / (12*23) = 5/276 = 0.0181.

Probability of drawing blue and red balls without replacement is 0.0181.