# Simple probability problems

This may be very simple, but I need to know correct answers.

Problem 1. There are $6$ balls in the box - $2$ black, and $4$ white. We take $3$ balls from the box. What is the probability that we took exactly 1 black ball.

Problem 2. What is the probability to score $7$ points when you throw two dices.

Problem 1: ${\binom{2}{1}\binom{4}{2} \over \binom{6}{3}}=0.6$

Problem 2: Denote by $(s,c)$ a pair where s - is a score, c - amount of possible outcomes. Then we have $(2,1) (3,1) (4,2) (5,2) (6,3) (7,3) (8,3) (9,2) (10,2) (11,1) (12,1)$. Amount of possible outcomes is 21. Amount of desired outcomes is 3, hence $p=\frac 3 {21}=\frac 17$

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Can you show some work for Problem 2? If it asked for the probability of 8, would the answer be $1/8$? – Dilip Sarwate Dec 7 '11 at 17:48
Now that you have shown how in Problem 2 you got $1/7$ (which by the way is the wrong answer, not the correct one that you asked for), let's turn to Problem 1 and analyze it according to your rules. Aren't there only $3$ outcomes WWW, WWB, WBB and so the probability of getting exactly one ball is $1/3$? I am not trying to be obtusely obstructive. Puzzling through these issues is an important part of learning how to think about probability, and I am very glad that you are making the effort. Your answer to Problem 1 is correct, by the way, but learn to write $\binom{n}{k}$ instead of $C_n^k$. – Dilip Sarwate Dec 7 '11 at 18:35
what exactly does $C_2^1 C_4^2 / C_6^3$ means? – Quixotic Dec 7 '11 at 18:38
Hm it seems like it's of the ambiguous notation mentioned here.I said ambiguous since apparently: $\scriptstyle C^{k}_{n}= \displaystyle C^{n}_{k}= \binom{n}{k}$ – Quixotic Dec 7 '11 at 18:42
I think the answer $1/3$ for the problem 1 is incorrect because all balls are different (i.e. distinguishable). P.S. I'll correct my designation (the first one is convenient in Russia) – Norbert Dec 7 '11 at 18:43

For problem 2):

Scoring 7 points means (presumably) the sum of the rolls is 7. Imagine you roll a die twice, in succession. You record the results as an ordered pair giving the result of the first roll, then the result of the second roll.

A score of 7 occurs in (and only in) the following ways: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1).

Out of the 36 equally likely outcomes that result from throwing a die twice in succession, exactly six will have a score of 7. So the probability of scoring 7 is ${6\over 36}={1\over6}$.

See Dilip' comment below. With unordered outcomes it's twice as likely to get "a one and a three" as "two threes".

Let's solve the problem using indistinguishable outcomes:

The outcomes from rolling two dice (dies?) simultaneously are $$\matrix{ \{1,1\} & \{1,2\}& \{1,3\}&\{1,4\}& \{1,5\} &\{1,6\} \cr & \{2,2\}&\{2,3\}& \{2,4\} &\{2,5\} &\{2,6\}\cr & &\{3,3\}& \{3,4\} &\{3,5\} &\{3,6\}\cr & & & \{4,4\} &\{4,5\} &\{4,6\}\cr & & & &\{5,5\} &\{5,6\}\cr & & & & &\{6,6\}\cr }$$

Now, imagine, please, that the two dice were of different colors, say taupe and ecrue.

In how many ways could you have obtained the outcome $\{1,2\}$? The answer is two ways: taupe was was 1 and ecrue was 2, or taupe was 2 and ecrue was 1.

In how many was could you have obtained the outcome $\{2,2\}$? Well, only one way: both of the dies showed "2".

A moment's reflection should convince you that an outcome that has different entries is twice as likely to occur as an outcome that has the same entries.

Further reflection reveals that any two outcomes of the "different entry" variety have the same probability of occurring and any two outcomes of the "same entry" variety have the same probability of occurring.

From this, we can calculate probabilities of individual outcomes: let the probability of an outcome with same entries be $a$. Then the probability of an outcome with different entries is $2a$.

Then, since the sum of the probabilities of all outcomes is 1 $$1= 6\cdot a+15\cdot (2a).$$

So $a=1/36$, and we have

$P(\text{outcome with same entries})={1\over36}$ and $P(\text{outcome with different entries})={2\over36}$

Now we can find

$$P( \text{score of 7} ) =P(1,6)+P(2,5)+P(3,4) =3\cdot {2\over36}={1\over6}.$$

That was quite a bit of reflecting to arrive at the answer. As far as "real life" is concerned, you could imagine the two dice (dies?) were distinguishable; and in that case, it is much easier to convince yourself that there are 36 equally likely outcomes.

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I think results of rolling the dices should be treated as unordered pairs. – Norbert Dec 7 '11 at 17:56
@Norbert You could, but it's easier to think about if you impose an order. – David Mitra Dec 7 '11 at 17:58
Oh, shoot! David. I was hoping the OP would have a chance to work it out for himself.... Actually, rolling two dice is a good problem for making students think a bit. Lots of beginners would argue that there are only $21$ equally likely outcomes ($15$ outcomes when the numbers are different plus $6$ outcomes when the numbers are the same), not $36$, when two dice are rolled. – Dilip Sarwate Dec 7 '11 at 17:59
@Norbert If you have one die that is rolled twice, we have $36$ outcomes. Suppose that you have two dice that you roll simultaneously. One is colored red and the other green and so you clearly have $36$ different outcomes. But I am color-blind to red and green and cannot tell the dice apart. Should I use $21$ outcomes and you use $36$? As David Mitra says, is easier to use $36$ outcomes because they are equally likely. With $21$ outcomes as you want, they are not equally likely. $15$ have probability $\frac{2}{36}$ and $6$ have probability $\frac{1}{36}$ which makes life harder. – Dilip Sarwate Dec 7 '11 at 18:07
@norbert if you really want to stick with the unordered view, you have to use the probabilities at the end of Dilip's comment. There are 3 equally likely unordered pairs that sum to 7: (1,6),(2,7),(3,4) each with a probability of $\frac{2}{36}$. So, we sum the probabilities of those pairs and get a total probability of $\frac{6}{36} = \frac{1}{6}$ – Drew Christianson Dec 7 '11 at 18:40

The critical distinction to make in problem 1 is whether you're sampling with or without replacement. This is, when you pick a ball out of the jar, do you note its color and put in back in, or keep it it. If it's the former case, you're sampling with replacement. In that case, the number of black balls in a sample of three is binomially distributed variable (thanks to Dilip for pointing out this & that my original analysis was incorrect) with $n=3$ and $p = \frac{1}{3}$. Thus, the probability of drawing one black ball is given by: $$\binom{3}{1}\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^2 = \frac{4}{9}$$

If you're sampling with out replacement, the solution is a little different. Consider, you could pick a black ball, then two whites, a white, then a black, then a white, or a two whites and then a black. The probabilities of those situations will just be the products of the probabilities of each draw. Thus, they will be: $$\frac{2}{6}\frac{4}{5}\frac{3}{4} , \frac{4}{6}\frac{2}{5}\frac{3}{4} , \frac{4}{6}\frac{3}{5}\frac{2}{4}$$ respectively. The sum, then, will be the total probability of drawing two white balls and one black: $$\frac{2}{6}\frac{4}{5}\frac{3}{4} + \frac{4}{6}\frac{2}{5}\frac{3}{4} + \frac{4}{6}\frac{3}{5}\frac{2}{4} = \frac{3(4*3*2)}{6*5*4} = \frac{3}{5}$$ This is equivalent to the solution you posted for this problem.

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Please correct tex errors in this post. I have no such privileges – Norbert Dec 7 '11 at 18:26
Sorry about that, it checked out in the preview but apparently the full MathJax doesn't like mismatched  and \$ – Drew Christianson Dec 7 '11 at 18:28
This is so complicated to me, don't trust nobody. – Norbert Dec 7 '11 at 21:01
+1 for fixing your answer. – Dilip Sarwate Dec 13 '11 at 0:07