# Probability of exactly $7$ unreplaced draws from $\{1,...,10\}$ to get $1,2,3$?

A box contains tickets numbered $$1$$ to $$10$$. Tickets are drawn at random without replacement until the numbers $$1, 2$$ and $$3$$ are drawn. Find the probability that exactly $$7$$ draws are required.

I am a little confused with this question. My first thought was to define the numbers $$1,2,$$ and $$3$$ as successes and the rest of the numbers as failures. Then,

$$P(\text{in 6 draws you get 2 successes}) = \frac{\binom{3}{2} \binom{7}{4}}{\binom{10}{6}}.$$ But I don't know where to go next or if this is even needed.

Can anyone help me? Thanks.

• Another approach: consider the $10!$ sequences of possible draws. How many of them have 1, 2, or 3 as the 7th draw and none of 1, 2, or 3 in the 8th–10th draws? Jul 4, 2022 at 19:56
• @GregMartin If I think about it in terms of sequences, the number of draws that have a 1,2, or 3 in the 7th position is the same as the number of sequences that have a 1,2, or 3 in the kth position, right?
– user545426
Jul 4, 2022 at 20:11

Actually you're off to a good start. Your calculation the probability of $$2$$ successes in the first $$6$$ draws (namely $$\frac12$$) is correct. Then for the seventh draw you have a $$\frac14$$ (conditional) probability of getting the remaining number of $$1,2$$, or $$3$$.

So your overall probability becomes $$\frac{{3\choose 2}{7\choose 4}}{{10\choose 6}}\cdot \frac14=\frac18$$

Probability $$1,2,3$$ are in the first $$7$$ draws is the probability none in last $$3$$ are $$1,2,3$$: $$\frac{7 \choose 3}{10 \choose 3}$$

Probability $$1,2,3$$ are in the first $$6$$ draws is the probability none in last $$4$$ are $$1,2,3$$: $$\frac{7 \choose 4}{10 \choose 4}$$

Probability exactly $$7$$ draws are required is the difference $$\frac{7 \choose 3}{10 \choose 3}-\frac{7 \choose 4}{10 \choose 4} = \frac18$$

In first six draws it will be two successes : $$\frac{\binom{3}{2}\times \binom{7}{4}}{\binom{10}{6}}$$

and for the seventh draw : $$\frac{1}{4}$$

$$\text{P}=\frac{\binom{3}{2}\times \binom{7}{4}}{\binom{10}{6}}\times \frac{1}{4}=\frac{1}{8}$$

Imagine $$10$$ slots. One of the three "good" numbers is to be in slot $$7$$, the other two are to be in the six preceding slots from the $$9$$ slots left.

Put them in, turn by turn.

$$Pr = \dfrac3{10}\dfrac69\dfrac58 = \dfrac18$$

• This probability seems too high...$\frac{105}{120}$ if I'm not mistaken. However the probability should be less than $\frac{3}{10}$ (the probability of getting $1, 2$, or $3$ in the seventh draw). Jul 4, 2022 at 20:41
• @paw88789: I changed to a simpler formulation, thanks Jul 4, 2022 at 21:03

Another approach: There are $$10!$$ possible sequences of draws. As @GregMartin suggests, let's count the number of good sequences.

• There are $$3$$ possibilities for what the seventh number can be (1, 2 or 3).
• The first six numbers contain the other two numbers from $$\{1,2,3\}$$ as well as four numbers from $$\{4, 5, \ldots, 10\}$$. There are $$\binom74$$ possibilites for choosing those four numbers.
• The first six numbers, now that they are fixed, can be arranged in $$6!$$ ways.
• The last three numbers are now fixed as well and can be arranged in $$3!$$ ways.

That gives us $$3\cdot\binom74\cdot6!\cdot3!$$ desirable sequences of draws, so the probability is $$\frac{3\cdot\binom74\cdot6!\cdot3!}{10!}=\frac18.$$

In general, there are $$3\cdot\binom7{n-3}\cdot(n-1)!\cdot(10-n)!$$ sequences that require exactly $$n$$ draws, so the probability for a general $$3\leq n\leq10$$ is $$\frac{3\cdot\binom7{n-3}\cdot(n-1)!\cdot(10-n)!}{10!}=\frac{(n-2)(n-1)}{240}.$$