# How to calculate the probability of this?

ok, I have 49 numbers, from 1-49. I will draw 7 numbers. Each time I draw the number out, it won't appear again. For example, the first one is "1", and I won't put the "1" back, so that it won't appear again. I would like to draw 7 numbers randomly, but I would like to draw it in this order:

1,2,3,4,5,6,7...

If the draw is like this

1,3,2,4,5,6,7

It will regards invalid. The number MUST be in this order..... How to cal the probability of drawing

1,2,3,4,5,6,7

? Thanks.

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To draw 1,2,3,4,5,6,7, you first need to draw 1. What's the probability of this? It's 1/49. Having drawn 1, you then need to draw 2. What's the probability of this? 1/48, since you say you cannot draw the same number again. Continue like this –  Daniel Freedman Jan 12 '12 at 13:47
You haven't told us what the 49 numbers you have are. –  Rahul Jan 12 '12 at 13:54
o...sorry, the 49 numbers are simply 1-49. –  Ted Wong Jan 12 '12 at 13:57
@Ted Wong: Here is an interesting related problem. Call a draw good if the numbers are in increasing order. So $(2,5,11,35,36,40,44)$ is good but $(35,44,11,2,36,40,5)$ is not. Then the probability of a good draw is $1/7!$. –  André Nicolas Jan 12 '12 at 16:23

The probability that the first picked number is one in 49. The next is one in 48, and so on. The probability to draw the sequence is the (49-7)!/49!

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so, do u means that the probability of drawing : 1,2,3,4,5,6,7 and 1,3,2,4,5,6,7 are equal? –  Ted Wong Jan 12 '12 at 13:52
@TedWong The answer to your question is YES. Note that the fact that we are drawing some specific numbers never comes into the answer. What matters is how many numbers are chosen from how many of them... –  user21436 Jan 12 '12 at 13:59

The number of ways of drawing 7 distinct numbers in a given order out of 49 numbers is $^{49}P_7$. So, exactly one of them is a favourable even t to you. This means, The answer is $\dfrac{1}{^{49}P_7}$ which is $\dfrac{42!}{49!}$.

And the fraction is nothing but $\dfrac{1}{43\cdot44\cdot45\cdot45\cdot46\cdot47\cdot48\cdot49}$.

Hope this helps.

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First, note that the outcomes from drawing seven numbers in succession and without replacement are equally likely; that is, the probability of drawing any particular sequence of seven numbers is the same as the probability of drawing any other particular sequence of seven numbers.

While this may not be easy to see, it is true. (It may be helpful to observe that if you just drew one number, it is equally likely that it is any one of the 49 numbers. From this, it is easy to see that if you drew two numbers in succession and without replacement, the various outcomes $(a,b)$ are equally likely. Think about this... One can extend to the present case with seven drawings.)

Now if one has equally likely outcomes, then to find the probability that an event $A$ happens, one computes: $$P(A)={ \text{number of ways A can happen}\over\text{total number of outcomes}}.$$

Assuming that your 49 numbers include $1$, $2$, $3$, $4$, $5$, $6$, and $7$, you need to compute the total number of different outcomes, $T$, that arise from selecting seven numbers in succession and without replacement; and then the probability of drawing seven specific numbers will be $1/T$ ($A$ in the formula above consists of one element: namely, drawing the sequence ${ 1, 2, 3, 4,5, 6, 7 }$).

To determine $T$, we use the Multiplication Principle:

Multiplication Principle:

Suppose two experiments are to be performed in succession. Suppose that the first experiment has exactly $n_1$ possible outcomes. Suppose that the second experiment always has exactly $n_2$ outcomes (regardless of what happened in experiment 1). Then the total number of outcomes from performing both experiments is $n_1\cdot n_2$.

There is an obvious generalization of the above to the case where $m$ experiments are performed in succession.

Generalized Multiplication Principle: Suppose that $m$ experiments are to be performed in succession. If, for each admissable $i$, the $i^{\rm th}$ experiment has $n_i$ outcomes regardless of what occurred in the experiments before, then the total number of outcomes from performing all the experiments is $n_1\cdot n_2\cdot > n_3\cdot\,\cdots\,\cdot n_m$.

So, for the matter at hand, we have seven experiments (drawing one number, then the second, then the third...):

There are 49 possible outcomes for drawing the first number.

Regardless of what the first number drawn was, there are 48 possible outcomes for the second number.

Regardless of what the first two numbers were, there are 47 possible outcomes for the third number.

$$\vdots$$

Regardless of what the first six numbers were, there are 43 possible outcomes for the seventh number.

The multiplication priciple states that $T$ is \eqalign{ T&=\textstyle\Bigl({\text{number of outcomes }\atop \text{for the first number }}\Bigr)\cdot \Bigl({\text{number of outcomes }\atop \text{for the second number }}\Bigr)\cdot\ \cdots\ \cdot \Bigl({\text{number of outcomes }\atop \text{for the seventh number }}\Bigr)\cr &\ \cr &=49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43. }

So your probability is $1\over49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43$.

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+1 for your effort! –  user21436 Jan 12 '12 at 15:13