A box contains 2 red balls and 3 green ones. Three draws are made at random without replacement from the box. Find the chance that a red ball is never drawn.

Answer 1

There are two ways to answer this question:

(1) Using combinatorial probability.

Think about it as, the probability of the event taking place equals the number of ways to get 3 green balls in a row, divided by the number of ways to get 3 balls out of a total of 5 balls. So it's:

P(E) = |E| / |Ω|, where Ω indicates the entire sample space, in this case, all the ways to get 3 balls out of 5.

Since you're dealing with sampling without replacement here, you can use combinations and you will have:

P(E) = nCr(3, 3) / nCr(5, 3), where nCr indicates the binomial coefficient with n=total sample and r= what is being sampled without replacement. This ultimately equals 0.1, so there is a 10% probability of drawing 3 green balls and thereby never drawing a single red ball

(2) Computing the probability of each draw independently and multiplying by each other

In other words:

P(Draw green on first draw) * P(Draw green on second draw) * P(Draw green on third draw)

By accounting for the reduction in number of balls in the denominator of the individual probabilities, you thereby ensure independence which enables you to multiply the probabilities by each other. For each probability, for example, for the first draw, you know you have 3 green balls and 5 total balls. So the probability of getting green on the first draw is simply 3/5. On the second draw, you now only have 4 total balls and you're assuming you've already drawn a green one, so now you only have 2 green left out of 4, and so on, and so forth.

(3/5) * (2/4) * (1/3) = 0.1

As you can see, either method gets you the same answer!