1
$\begingroup$

We have 10 uniquely numbered tickets from 0 to 9. We draw randomly 3 tickets, one at a time.

What is the probability of drawing the tickets in numerally ascending order?

By ascending order I mean any possible combination, where the numeral number of a ticket of the next draw is bigger than the last (i.e 2, 7, 9 or 3, 5, 8). Also it is to be noted that once a ticket is drawn, it will not be put back to be drawn again.

I gather that the actual number of the tickets, assuming that they all have an unique number, doesn't really matter. That said I would approach this by the permutation of the number of tickets to be drawn. In this case it would be

3! which would give us 3*2*1 = 6 = 1/6 = ~16,6 %

Bonus: If it would be mandatory to draw consecutive numbers like 3, 4, 5 or 6, 7, 8, this would then be dependent of the total number tickets available. In this bonus example we only have 8 different variations of consecutive numbers to be drawn

0, 1, 2
1, 2, 3
2, 3, 4
3, 4, 5
4, 5, 6
5, 6, 7
6, 7, 8
7, 8, 9

Which would give me

8 * 9 * 8 = 576 = 1/576

Am I right in this one?

$\endgroup$
1
  • $\begingroup$ Welcome to MSE. Please use MathJax to format math on this site. To begin with, surround all math expressions, including numbers with $ signs. $\endgroup$
    – saulspatz
    Mar 20 at 16:04
1
$\begingroup$

The first one is right, but not the second. There are $8$ admissible draws out of $10\cdot9\cdot8$ possible draws, so the probability is $\frac1{90}$.

$\endgroup$
5
  • $\begingroup$ I don't get it. First you have 8 different variations of consecutive numbers as I stated above. Let us assumet that I draw 4. Next I have only 9 tickets left and only one of those 9 tickets left will do so (that is #5) for the second ticket it is 1/9. The third ticket is drawn from the 8 there is left so it is 1/8. So my math would be 8 * 9 * 8. Or is it actually (8/10) * (1/9) * 1/8) = which would give us 1/90 $\endgroup$ Mar 20 at 16:08
  • $\begingroup$ @KasperiKoski So why does your question say that the probability is $\frac1{576}$? $\endgroup$
    – saulspatz
    Mar 20 at 16:10
  • $\begingroup$ I just multiplied all the probabilitys of each round together wich would give me 8*9*8. Like I said, I just get the first 8 from the fact that there are 8 different possibilities of consecutive numbers to be drawn. Gosh, how I am not getting this. $\endgroup$ Mar 20 at 16:18
  • $\begingroup$ @KasperiKoski I don't understand either. $8\cdot9\cdot8=576$ is not a probability. Probabilities are between $0$ and $1$. Then in a previous comment you say this is "actually" $\frac8{10}\cdot\frac19\cdot\frac18$ No, $576$ is not actually $\frac1{90}$. I'm not being picky; I can't make head of tail of this. $\endgroup$
    – saulspatz
    Mar 20 at 16:25
  • $\begingroup$ You're right. By 576 I meant 1/576 which stangely is the same that my books aswser key tells... $\endgroup$ Mar 20 at 16:26
0
$\begingroup$

The sample space in this case will be all possible draws. $\\ n(S) = ^{10}P_3$ Let, the event of getting three consecutive increasing numbers be $E$. Than, we can find $n(E) $ by selecting 3 numbers from 0-9. As there is only one way to arrange them, we don't need to permute them. Thus, $n(E) =^{10}C_3$

Now, as we have to select three cards in all the cases in the sample space, we can directly say,

$$P(E) = \frac{n(E) }{n(S) } \\ P(E) = \frac{1}{3! } \\ P(E) = \frac{1}{6}$$

A more intuitive way to think about it is that, suppose you draw any three cards, there will be 3! ways in which you could have taken them out but only one of them would have satisfied the condition provided.

Hope this is helpful!!!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.