Today, we will continue our discussion on why it is important to understand the workings behind seemingly miraculous shortcuts. We will use another example from probability.

**Question 1:** A bag contains 4 white balls, 2 black balls & 3 red balls. One by one three balls are drawn out with replacement (i.e. a ball is drawn and then put back. Thereafter, another ball is drawn). What is the probability that the third ball is red?

**Solution:**

The question is simple, isn’t it? When you draw balls with replacement, the probability stays the same at the beginning of every cycle. On first draw, the probability of drawing a red ball is 3/9 (since there are 3 red balls and 9 balls in all). When you draw the ball and put in back, the probability of drawing a red ball again stays the same i.e. 3/9 (since again there are 3 red balls and 9 balls in all). The situation at the beginning of every draw is the same.

Right, let’s move on to the question we actually wanted to discuss!

**Question 2:** A bag contains 4 white balls, 2 black balls & 3 red balls. One by one three balls are drawn out without replacement (the balls are not put back). What is the probability that the third ball is red?

**Solution:** This question differs from the previous one. The balls are not replaced here so every pick is not the same.

I guess, intuitively, you will say that the probability of getting a red ball in the third draw will now change. Now here is a fun fact: the probability that the third ball is red is still 3/9. Without knowing the other results, the probability of drawing a red ball will not change for the successive drawings. The probability will remain the same as it is on the first draw. It’s good to remember this i.e. it’s a shortcut. Rather than working on all the cases e.g. you pick a red first, then a red and a red again or you pick a red first, then a non red and then a red again etc, you know that the probability will not change in any successive draw.

Let’s try to understand why this is so. We will take the example of the second draw and see why the probability of picking a red stays the same as the probability of picking a red in the first draw.

Probability of picking a red in the second draw:

Let’s discuss the two cases: ‘First draw is red’ and ‘first draw is not red’

Case 1: First draw is red and the second draw is red.

Probability of first draw being red = (3/9)

Probability of second draw being red = (2/8)

The probability of picking a red first and then a red again = (3/9)*(2/8)

Case 2: First draw is non red and the second draw is red.

Probability of first draw being non red = (6/9)

Probability of second draw being red = (3/8)

The probability of picking non red first and then red = (6/9)*(3/8)

(6/9)*(3/8) = (3/9)*(6/8).

(3/9)*(6/8) is the probability of picking a red ball first and then a non red ball.

Think about it: Probability of picking a non red ball first and then a red ball will be the same as the probability of picking a red ball first and then a non red ball. Note here that we do not need to deal with white and black balls separately. They are just non-red since we don’t care about the balls of other colors.

Total probability of second draw being red = Probability in case 1 + Probability in case 2

= (3/9) * (2/8) + (3/9) * (6/8) = 3/9 * (2/8 + 6/8) = 3/9 * 1

This is just the probability of picking a red ball first and then any ball (non red or red). Probability of picking ANY ball will be 1. Hence, the probability of picking a red in the second draw will be the same as the probability of picking a red in the first draw.

Using the same concept, you can see that the probability of picking a red in any draw will be the same if the balls obtained in the previous draws are not known.

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

This is great! Thanks!

Karishma,

When will the order matter? I see that you have explained “how” the probabilities, for cases with and without replacement, are the same. “Why” would it be the same? I did the long method, and got the same result.

Do you think that probabilities will be different for : If the third ball is RED and the fourth Ball is White? Thoughts?

Thanks

Hello Karishma, Please clarify for the case where we draw three balls from the bag without replacement. Thank you.

nr – non red ball, r – red ball, – does not equal

Drawing two balls from the bag:

The second draw has to be red:(r, r) (nr, r) = 3/9 x 2/8 + 6/9 x 3/8 = 3/9 This is good.

Drawing three balls from the bag:

The third draw has to be red: (r, r, r), (nr, nr, r) & (nr, r, r) =

3/9 x 2/8 x 1/7 + 6/9 x 5/8 x 3/7 + 6/9 x 3/8 x 2/7 3/9 Answer varies here.

The ‘does not equal’ sign did not get published above in the last line. I retype the same here –

3/9 x 2/8 x 1/7 + 6/9 x 5/8 x 3/7 + 6/9 x 3/8 x 2/7

This above expression does not equal 3/9. Answer varies here.

Offhand, I see that you have missed one case (r, nr, r). Does it add up now?