Question

An urn contains colored balls;5 red balls, 8 green balls, and 10 blue balls. Suppose  If the 3 balls are drawn one after another without replacement, what is the probability that the colors observed will be Red, Green, Blue in this order?  If the three balls are drawn simultaneously from the urn (without replacement), what is the probability that the selected balls will be all different?

Answer 1

An urn contains 5 red balls, 8 green balls, and 10 blue balls.

So, in total, there are 5+8+10, ie. 23 colored balls.

Question (a)

3 balls are drawn one after another without replacement. we have to find the probability that the colors of the observed ball will be red, green, blue, in this order.

At first there are 5 red balls; and a total of 23 colored balls.

So, the first ball is red with probability 5/23.

Now, there are 8 green balls; but there are 22 colored balls, as the drawing is without replacement.

So, the probability of drawing a green ball next is 8/22.

Now, there are 10 blue balls, but a total of 21 colored balls left, because the draw is without replacement.

So, the probability of drawing a blue ball next is 10/21.

So, the required probability is

The probability of drawing a red, then a green and then a blue ball is 0.0376.

Question (b)

Three balls are drawn simultaneously from the urn, without replacement.

We have to find the probability that the colors of the selected balls will be different.

Now, from 23 total balls, 3 can be drawn without replacement, in (23 C 3) ways.

Now, if the selected balls are of different colors, then one ball has to be red, one green and one blue.

So, the red ball can come up in 5 ways, the green ball can come up in 8 ways and the blue ball in 10 ways.

So, the required probability is

So, the probability of getting balls of different colors is 0.1739.