Combination Probability

Question

A bag contains 21 balls, 7 red balls, 8 black balls and 6 yellow balls. How many ways can 3 balls drawn from the bag without replacement, if at least one red ball is to be included in the draw?

It will be great if steps are shown on how to answer this kind of question.

Answer 1

Hi Ahmed,

We have to use the laws of compliments for this one. The compliment of at least one red marble, would be no red marbles. So P(at least one red) = 1 - P(no reds).

How do we calculate the probability of no red marbles? Well there are 14 non-red marbles. So we have 14/21, next marble is (14-1)/(21-1) because we are taking the marble without replacement, similarly the third will be (13-1)/(20-1). To find the probability of three draws with no red marbles, multiply the three we just got. This gives us: 0.2737. It is a compliment, so we subtract this probability from 1. So, 1 - 0.2737 = 0.7263.

Hope this helps!

-Kevin