3 distinct numbers are selected from $\{1,\ldots,9\}$. What is the probability that 9 is selected?

3 distinct digits are randomly selected from the set of nine digits [1-9]. What is the probability that 9 is selected.

I thought that the probability should be (1/9) + (1/8) + (1/7) since you have to chose 3 DISTINCT numbers from the set and therefore the first time the probability of selecting 9 is (1/9), the second time the probability of selecting 9 is 1/8 since this time you only have to chose a set of 8 digits and similarly the third time it should be (1/7).

But according to my teacher the probability should be (1/9)*3 = 1/3 OR (1/9) + (1/9) + (1/9) = (1/3).

I cannot understand why it should be (1/3) Could someone please tell me the right way to do this.

Thanks

• You need to count how many ways there are of choosing 3 distinct numbers from the set, and how many ways among them allow you to get the number 9. Then make the division – Petite Etincelle Jul 31 '14 at 13:32
• The title does not fit the text of the question. – Did Jul 31 '14 at 13:34

Think of it this way: randomly order the numbers from 1 to 9, e.g.

$$3, 2, 9, 8, 1, 4, 7, 6, 5$$

then partition the first three numbers from the remaining 6:

$$3, 2, 9 \,|\, 8, 1, 4, 7, 6, 5$$

Since the 9 has an equal probability of being in any of those 9 positions, what is the probability that it will be within those first three? Clearly $3/9=1/3$.

The $\frac{1}{9}$ counts the probability that the first one chosen is $9$.

But $\frac 18$ is the probability that the second one is $9$ if the first one is not $9$. The odds of the second being $9$ is therefore $\frac{8}{9}\frac{1}{8}=\frac{1}{9}$, where $\frac{8}9$ is the probability that the first is not a $9$.

Finally, the odds $\frac{1}{7}$ is the odds of the third being $9$ if the first two not being $9$. That is $\frac{1}{7}\frac{7}{9}=\frac{1}{9}$ is the odds of the third being $9$.

So the probability of getting a $9$ is $\frac19+ \frac19 +\frac19=\frac 13$.

Let's do a smaller example. Suppose you choose 3 distinct numbers from $\{1,2,3\}$. What is the chance that $3$ is selected?

According to your reasoning, the probability is $\frac13+\frac12+\frac11 \approx 1.83$, which is nonsense, since probabilities can't be bigger than 1.

Now what went wrong? You must have counted something twice. Can you figure out what?

(The analog of your teacher's answer for this case is $\frac13 + \frac13 + \frac13 = 1$, which is correct.)

While Ozo's answer explains very clearly why it is 1/3, I would like to explain the faulty in your way.

The probability of choosing $9$ in the first choice is indeed $1/9$, however the probability of choosing $9$ in the second choice isn't $1/8$. First you have to choose the first number, and it must be different from 9, hence $8/9$. Then you must choose 9 on the second time, and hence $8/9*1/8$. Following the same logic, the probability of choosing it the third time would be $8/9*7/8*1/7$ and not $1/7$.

You can see that $1/9+8/9*1/8+8/9*7/8*1/7=1/3$. What you did wrong was to think only about the specific choice of $9$, and the not the choices before it.

Choosing 3 elements from 1-9 in $\binom{9}{3}$ ways. Choosing 3 elements from 1-8 in $\binom{8}{3}$ ways. So the probability of getting a 9 is the complementary $1-\frac{\binom{8}{3}}{\binom{9}{3}}=1-\frac{6}{9}=\frac{1}{3}$

This is a question of basic probability. Required probability = Favourable cases/Total cases

Firstly, total cases = 9 C 3, as we know that we have to select 3 distinct digits out of the given nine digits. Secondly, favourable cases = 1 x 8 C 2, as we know that out of the three digits 9 is compulsory and we have to choose two other numbers from the set.

Therefore, P = 8 C 2/9 C 3 = 1/3