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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.

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    $\begingroup$ 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? $\endgroup$ Jul 4, 2022 at 19:56
  • $\begingroup$ @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? $\endgroup$
    – user545426
    Jul 4, 2022 at 20:11

5 Answers 5

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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 $$\frac{{3\choose 2}{7\choose 4}}{{10\choose 6}}\cdot \frac14=\frac18$$

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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 $\frac{7 \choose 3}{10 \choose 3}-\frac{7 \choose 4}{10 \choose 4} = \frac18$

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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 : $\frac{1}{4}$

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

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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.

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

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    $\begingroup$ 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). $\endgroup$
    – paw88789
    Jul 4, 2022 at 20:41
  • $\begingroup$ @paw88789: I changed to a simpler formulation, thanks $\endgroup$ Jul 4, 2022 at 21:03
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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 $$\frac{3\cdot\binom7{n-3}\cdot(n-1)!\cdot(10-n)!}{10!}=\frac{(n-2)(n-1)}{240}.$$

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