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A Cautionary Suggestion
How to calculate the probability of this?
How to solve this probability exercise?How to calculate the probability that the average of a multinomial process exceeds some valueProbability of an event if the sample space has identical elementsProbability of multiple independent events occurring after repeated attemptsProbability Of 4 Decks of cardsProbability: problem with 300 ballsProbability: Drawing numbered lottery ticketsProbability with DistributionsProbability of drawing n elements from a set in x attempts where an element can be drawn more than onceWhat is the % chance I will pick the correct random number in multiple drawings
$begingroup$
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.
probability
$endgroup$
add a comment |
$begingroup$
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.
probability
$endgroup$
2
$begingroup$
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
$endgroup$
– Daniel Freedman
Jan 12 '12 at 13:47
1
$begingroup$
You haven't told us what the 49 numbers you have are.
$endgroup$
– Rahul
Jan 12 '12 at 13:54
$begingroup$
o...sorry, the 49 numbers are simply 1-49.
$endgroup$
– Ted Wong
Jan 12 '12 at 13:57
$begingroup$
@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 $1/7!$.
$endgroup$
– André Nicolas
Jan 12 '12 at 16:23
add a comment |
$begingroup$
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.
probability
$endgroup$
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.
probability
probability
edited Jan 12 '12 at 13:58
Ted Wong
asked Jan 12 '12 at 13:41
Ted WongTed Wong
10637
10637
2
$begingroup$
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
$endgroup$
– Daniel Freedman
Jan 12 '12 at 13:47
1
$begingroup$
You haven't told us what the 49 numbers you have are.
$endgroup$
– Rahul
Jan 12 '12 at 13:54
$begingroup$
o...sorry, the 49 numbers are simply 1-49.
$endgroup$
– Ted Wong
Jan 12 '12 at 13:57
$begingroup$
@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 $1/7!$.
$endgroup$
– André Nicolas
Jan 12 '12 at 16:23
add a comment |
2
$begingroup$
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
$endgroup$
– Daniel Freedman
Jan 12 '12 at 13:47
1
$begingroup$
You haven't told us what the 49 numbers you have are.
$endgroup$
– Rahul
Jan 12 '12 at 13:54
$begingroup$
o...sorry, the 49 numbers are simply 1-49.
$endgroup$
– Ted Wong
Jan 12 '12 at 13:57
$begingroup$
@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 $1/7!$.
$endgroup$
– André Nicolas
Jan 12 '12 at 16:23
2
2
$begingroup$
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
$endgroup$
– Daniel Freedman
Jan 12 '12 at 13:47
$begingroup$
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
$endgroup$
– Daniel Freedman
Jan 12 '12 at 13:47
1
1
$begingroup$
You haven't told us what the 49 numbers you have are.
$endgroup$
– Rahul
Jan 12 '12 at 13:54
$begingroup$
You haven't told us what the 49 numbers you have are.
$endgroup$
– Rahul
Jan 12 '12 at 13:54
$begingroup$
o...sorry, the 49 numbers are simply 1-49.
$endgroup$
– Ted Wong
Jan 12 '12 at 13:57
$begingroup$
o...sorry, the 49 numbers are simply 1-49.
$endgroup$
– Ted Wong
Jan 12 '12 at 13:57
$begingroup$
@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 $1/7!$.
$endgroup$
– André Nicolas
Jan 12 '12 at 16:23
$begingroup$
@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 $1/7!$.
$endgroup$
– André Nicolas
Jan 12 '12 at 16:23
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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!
$endgroup$
$begingroup$
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?
$endgroup$
– Ted Wong
Jan 12 '12 at 13:52
$begingroup$
@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...
$endgroup$
– user21436
Jan 12 '12 at 13:59
add a comment |
$begingroup$
The number of ways of drawing 7 distinct numbers in a given order out of 49 numbers is $^49P_7$. So, exactly one of them is a favourable even t to you. This means, The answer is $dfrac1^49P_7$ which is $dfrac42!49!$.
And the fraction is nothing but $dfrac143cdot44cdot45cdot45cdot46cdot47cdot48cdot49$.
Hope this helps.
$endgroup$
add a comment |
$begingroup$
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: $$ P(A)= textnumber of
ways $A$ can happenovertexttotal number of outcomes. $$
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 $n_1cdot n_2$.
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 $n_1cdot n_2cdot
> n_3cdot,cdots,cdot n_m$.
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
$$eqalign
T&=textstyleBigl(textnumber of outcomes atop textfor the first number Bigr)cdot
Bigl(textnumber of outcomes atop textfor the second number Bigr)cdot cdots cdot
Bigl(textnumber of outcomes atop textfor the seventh number Bigr)cr
& cr
&=49cdot48cdot47cdot46cdot45cdot44cdot43.
$$
So your probability is $1over49cdot48cdot47cdot46cdot45cdot44cdot43 $.
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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!
$endgroup$
$begingroup$
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?
$endgroup$
– Ted Wong
Jan 12 '12 at 13:52
$begingroup$
@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...
$endgroup$
– user21436
Jan 12 '12 at 13:59
add a comment |
$begingroup$
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!
$endgroup$
$begingroup$
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?
$endgroup$
– Ted Wong
Jan 12 '12 at 13:52
$begingroup$
@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...
$endgroup$
– user21436
Jan 12 '12 at 13:59
add a comment |
$begingroup$
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!
$endgroup$
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!
answered Jan 12 '12 at 13:49
SashaSasha
60.9k5110181
60.9k5110181
$begingroup$
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?
$endgroup$
– Ted Wong
Jan 12 '12 at 13:52
$begingroup$
@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...
$endgroup$
– user21436
Jan 12 '12 at 13:59
add a comment |
$begingroup$
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?
$endgroup$
– Ted Wong
Jan 12 '12 at 13:52
$begingroup$
@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...
$endgroup$
– user21436
Jan 12 '12 at 13:59
$begingroup$
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?
$endgroup$
– Ted Wong
Jan 12 '12 at 13:52
$begingroup$
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?
$endgroup$
– Ted Wong
Jan 12 '12 at 13:52
$begingroup$
@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...
$endgroup$
– user21436
Jan 12 '12 at 13:59
$begingroup$
@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...
$endgroup$
– user21436
Jan 12 '12 at 13:59
add a comment |
$begingroup$
The number of ways of drawing 7 distinct numbers in a given order out of 49 numbers is $^49P_7$. So, exactly one of them is a favourable even t to you. This means, The answer is $dfrac1^49P_7$ which is $dfrac42!49!$.
And the fraction is nothing but $dfrac143cdot44cdot45cdot45cdot46cdot47cdot48cdot49$.
Hope this helps.
$endgroup$
add a comment |
$begingroup$
The number of ways of drawing 7 distinct numbers in a given order out of 49 numbers is $^49P_7$. So, exactly one of them is a favourable even t to you. This means, The answer is $dfrac1^49P_7$ which is $dfrac42!49!$.
And the fraction is nothing but $dfrac143cdot44cdot45cdot45cdot46cdot47cdot48cdot49$.
Hope this helps.
$endgroup$
add a comment |
$begingroup$
The number of ways of drawing 7 distinct numbers in a given order out of 49 numbers is $^49P_7$. So, exactly one of them is a favourable even t to you. This means, The answer is $dfrac1^49P_7$ which is $dfrac42!49!$.
And the fraction is nothing but $dfrac143cdot44cdot45cdot45cdot46cdot47cdot48cdot49$.
Hope this helps.
$endgroup$
The number of ways of drawing 7 distinct numbers in a given order out of 49 numbers is $^49P_7$. So, exactly one of them is a favourable even t to you. This means, The answer is $dfrac1^49P_7$ which is $dfrac42!49!$.
And the fraction is nothing but $dfrac143cdot44cdot45cdot45cdot46cdot47cdot48cdot49$.
Hope this helps.
answered Jan 12 '12 at 13:55
user21436
add a comment |
add a comment |
$begingroup$
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: $$ P(A)= textnumber of
ways $A$ can happenovertexttotal number of outcomes. $$
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 $n_1cdot n_2$.
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 $n_1cdot n_2cdot
> n_3cdot,cdots,cdot n_m$.
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
$$eqalign
T&=textstyleBigl(textnumber of outcomes atop textfor the first number Bigr)cdot
Bigl(textnumber of outcomes atop textfor the second number Bigr)cdot cdots cdot
Bigl(textnumber of outcomes atop textfor the seventh number Bigr)cr
& cr
&=49cdot48cdot47cdot46cdot45cdot44cdot43.
$$
So your probability is $1over49cdot48cdot47cdot46cdot45cdot44cdot43 $.
$endgroup$
add a comment |
$begingroup$
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: $$ P(A)= textnumber of
ways $A$ can happenovertexttotal number of outcomes. $$
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 $n_1cdot n_2$.
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 $n_1cdot n_2cdot
> n_3cdot,cdots,cdot n_m$.
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
$$eqalign
T&=textstyleBigl(textnumber of outcomes atop textfor the first number Bigr)cdot
Bigl(textnumber of outcomes atop textfor the second number Bigr)cdot cdots cdot
Bigl(textnumber of outcomes atop textfor the seventh number Bigr)cr
& cr
&=49cdot48cdot47cdot46cdot45cdot44cdot43.
$$
So your probability is $1over49cdot48cdot47cdot46cdot45cdot44cdot43 $.
$endgroup$
add a comment |
$begingroup$
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: $$ P(A)= textnumber of
ways $A$ can happenovertexttotal number of outcomes. $$
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 $n_1cdot n_2$.
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 $n_1cdot n_2cdot
> n_3cdot,cdots,cdot n_m$.
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
$$eqalign
T&=textstyleBigl(textnumber of outcomes atop textfor the first number Bigr)cdot
Bigl(textnumber of outcomes atop textfor the second number Bigr)cdot cdots cdot
Bigl(textnumber of outcomes atop textfor the seventh number Bigr)cr
& cr
&=49cdot48cdot47cdot46cdot45cdot44cdot43.
$$
So your probability is $1over49cdot48cdot47cdot46cdot45cdot44cdot43 $.
$endgroup$
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: $$ P(A)= textnumber of
ways $A$ can happenovertexttotal number of outcomes. $$
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 $n_1cdot n_2$.
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 $n_1cdot n_2cdot
> n_3cdot,cdots,cdot n_m$.
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
$$eqalign
T&=textstyleBigl(textnumber of outcomes atop textfor the first number Bigr)cdot
Bigl(textnumber of outcomes atop textfor the second number Bigr)cdot cdots cdot
Bigl(textnumber of outcomes atop textfor the seventh number Bigr)cr
& cr
&=49cdot48cdot47cdot46cdot45cdot44cdot43.
$$
So your probability is $1over49cdot48cdot47cdot46cdot45cdot44cdot43 $.
edited Jan 12 '12 at 15:01
answered Jan 12 '12 at 13:53
David MitraDavid Mitra
63.5k6102164
63.5k6102164
add a comment |
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2
$begingroup$
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
$endgroup$
– Daniel Freedman
Jan 12 '12 at 13:47
1
$begingroup$
You haven't told us what the 49 numbers you have are.
$endgroup$
– Rahul
Jan 12 '12 at 13:54
$begingroup$
o...sorry, the 49 numbers are simply 1-49.
$endgroup$
– Ted Wong
Jan 12 '12 at 13:57
$begingroup$
@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 $1/7!$.
$endgroup$
– André Nicolas
Jan 12 '12 at 16:23