# Simple probability, drawing of lots

Assume there is a drawer with $4$ balls of different colors. You draw lots $3$ times and put the ball back after each draw.

What is the probability that the same ball will be drawn exactly $1$ time?

My 'guess': I would use the binomial formula:

$$P_{1} = \binom{3}{1} \cdot \left(\frac{1}{4}\right)^{\mkern -4mu 1} \cdot \left(1 - \frac{1}{4}\right)^{\mkern -4mu 3 - 1} = 3 \cdot \frac{1}{4} \cdot \frac{9}{16} = \frac{27}{64}.$$

Is this correct?

Follow-up

It seems that I got the correct answer. Now to a related question:

Why is the probability of drawing the same ball exactly $0$ times the same the probability of drawing it exactly $1$ time? This is now intuitively clear to me.

• You are absolutely right. – Satish Ramanathan Jun 6 '14 at 13:43
• This is the probability that a specific ball (the blue one) is drawn exactly once. If that is what you intended to compute, the answer is correct. – André Nicolas Jun 6 '14 at 13:43
• $\frac{27}{64} \ne 0.14$ – Alex Jun 6 '14 at 14:05

From your follow up question, I consider that you mean a specific ball, that you want to draw exactly once (original question) or not of all (follow up.)

Let us first generalize the problem: We have a urn with $n$ balls. One of the ball if marked. We draw $n-1$ at put the balls we draw back into the urn.

Questions:

1. What is the probability to draw the marked ball exactly once?
2. What is the probability to not draw the marked ball at all?

Solutions:

1. $p_1 = {n-1 \choose 1} \cdot \frac{1}{n} \cdot (\frac{n-1}{n})^{n-2}$
2. $p_2 = (\frac{n-1}{n})^{n-1}$

As you noticed the two solutions coincide. This is due to following equality: $${n-1 \choose 1} \cdot \frac{1}{n} = \frac{(n-1)!}{(n-2)! 1!} \cdot \frac{1}{n} = (n-1) \cdot \frac{1}{n} = \frac{n-1}{n}$$

• Great. Thank you very much! – Svend Tveskæg Jun 6 '14 at 14:56
• where's the inequality? – Alex Jun 6 '14 at 15:02
• @Alex : Your right! My fault, it should be equality instead of inequality. I changed my answer accordingly. Thanks. – crixstox Jun 16 '14 at 9:37

This is called sampling with replacement. Assuming I correctly understand your question ('same ball exactly 1 time') what you need is the probability of draws like $ABC, ABD, ACD, BCD$ - there are totally 4 unique combinations and each combination has $3!$ representations (i.e. ABC, BCA...CBA), hence there are $3!4 = 4!$ 'positive outcomes' out of $4^3$ total outcomes. Hence your probability is $\frac{4!}{4^3}$.

• Just another reasoning with the same result: For the first draw there are no restriction, for the second one you can should any off the three remaining balls (sucess probability $\frac{3}{4}$) and in the last drawing round any off the two remaining balls. Overall you have the probability: $p = \frac{3}{4} \cdot \frac{2}{4}$, which coincides with the answer from @Alex. – crixstox Jun 6 '14 at 14:03
• no, the result is different. $\frac{3}{8} \ne 27/64 \ne 0.14$. The question is quite vague though. – Alex Jun 6 '14 at 14:05