Question

A bag contains three black balls, four white balls and five red balls. Three balls are removed without relacement, what is the probability of obtaining at most two red balls?

Answer 1

Answer:

your answer mate

Step-by-step explanation:

First we calculate the number of ways of removing any 3 balls from the 3+4+5=12

balls. This is 12C3 = 220 ways to remove any 3 balls.

We first find the number of ways we don't want, which is the number of obtaining

either no reds or just 1 red.

The number of ways of getting no reds is the number of ways of choosing 3 from

the 3+4=7 non-red balls. This is 7C3 = 35 ways.

The number of ways of getting exactly 1 red is the number of ways of:

1. choosing 2 from the 3+4=7 non-red balls. This is 7C2 = 21 ways.

and for each of these 21 ways,

2. choosing the 1 red ball. This is 5C1 = 5 ways

That's 21∙5 = 105 ways to get exactly 1 red ball.

So the total number of ways to get what we don't want is 35+105 = 140 ways.

That leaves 220-140=80 ways we want.

So the desired probability is 80 ways out of 220 or 80/220 which reduces to

4/11.