What is the % chance I will pick the correct random number in multiple drawings
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Let’s say I pick a number between 1-10. I then draw a random number between that range.. If my number doesn’t come up, I draw again. One caveat is that if a previously drawn number is drawn again, I draw again. What is the % chance my number will come up after some amount of drawings?
Let’s see an example...
I have a range of 1-10 and 5 chances to draw my number
I pick the number 6
I randomly draw a 3 (drawing 1)
I randomly draw a 4 (drawing 2)
I randomly draw a 4 (doesn’t count as a drawing because 4 was already drawn)
I randomly draw a 2 (drawing 3)
I randomly draw a 5 (drawing 4)
I randomly draw a 9 (drawing 5)
In this scenario, my number (6) was never drawn. I want to know what are the overall chances of drawing my number given the rules above.
I would assume it is something like...
First draw is 1/10
Second draw is 1/9
Third draw is 1/8
Fourth draw is 1/7
Fifth draw is 1/6
If this is true, how do I come up with a total % probability that my number will be drawn in N number of drawings?
probability
asked Jan 11 at 15:17
Matt MessingerMatt Messinger
1
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add a comment |
$begingroup$
Let’s say I pick a number between 1-10. I then draw a random number between that range.. If my number doesn’t come up, I draw again. One caveat is that if a previously drawn number is drawn again, I draw again. What is the % chance my number will come up after some amount of drawings?
Let’s see an example...
I have a range of 1-10 and 5 chances to draw my number
I pick the number 6
I randomly draw a 3 (drawing 1)
I randomly draw a 4 (drawing 2)
I randomly draw a 4 (doesn’t count as a drawing because 4 was already drawn)
I randomly draw a 2 (drawing 3)
I randomly draw a 5 (drawing 4)
I randomly draw a 9 (drawing 5)
In this scenario, my number (6) was never drawn. I want to know what are the overall chances of drawing my number given the rules above.
I would assume it is something like...
First draw is 1/10
Second draw is 1/9
Third draw is 1/8
Fourth draw is 1/7
Fifth draw is 1/6
If this is true, how do I come up with a total % probability that my number will be drawn in N number of drawings?
probability
asked Jan 11 at 15:17
Matt MessingerMatt Messinger
1
$endgroup$
add a comment |
$begingroup$
Let’s say I pick a number between 1-10. I then draw a random number between that range.. If my number doesn’t come up, I draw again. One caveat is that if a previously drawn number is drawn again, I draw again. What is the % chance my number will come up after some amount of drawings?
Let’s see an example...
I have a range of 1-10 and 5 chances to draw my number
I pick the number 6
I randomly draw a 3 (drawing 1)
I randomly draw a 4 (drawing 2)
I randomly draw a 4 (doesn’t count as a drawing because 4 was already drawn)
I randomly draw a 2 (drawing 3)
I randomly draw a 5 (drawing 4)
I randomly draw a 9 (drawing 5)
In this scenario, my number (6) was never drawn. I want to know what are the overall chances of drawing my number given the rules above.
I would assume it is something like...
First draw is 1/10
Second draw is 1/9
Third draw is 1/8
Fourth draw is 1/7
Fifth draw is 1/6
If this is true, how do I come up with a total % probability that my number will be drawn in N number of drawings?
probability
asked Jan 11 at 15:17
Matt MessingerMatt Messinger
1
$endgroup$
Let’s say I pick a number between 1-10. I then draw a random number between that range.. If my number doesn’t come up, I draw again. One caveat is that if a previously drawn number is drawn again, I draw again. What is the % chance my number will come up after some amount of drawings?
Let’s see an example...
I have a range of 1-10 and 5 chances to draw my number
I pick the number 6
I randomly draw a 3 (drawing 1)
I randomly draw a 4 (drawing 2)
I randomly draw a 4 (doesn’t count as a drawing because 4 was already drawn)
I randomly draw a 2 (drawing 3)
I randomly draw a 5 (drawing 4)
I randomly draw a 9 (drawing 5)
In this scenario, my number (6) was never drawn. I want to know what are the overall chances of drawing my number given the rules above.
I would assume it is something like...
First draw is 1/10
Second draw is 1/9
Third draw is 1/8
Fourth draw is 1/7
Fifth draw is 1/6
If this is true, how do I come up with a total % probability that my number will be drawn in N number of drawings?
probability
probability
asked Jan 11 at 15:17
Matt MessingerMatt Messinger
1
asked Jan 11 at 15:17
Matt MessingerMatt Messinger
1
asked Jan 11 at 15:17
Matt MessingerMatt Messinger
1
asked Jan 11 at 15:17
Matt MessingerMatt Messinger
1
asked Jan 11 at 15:17
Matt MessingerMatt Messinger
1
1
add a comment |
add a comment |
1 Answer
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$begingroup$
Let us say you have drawn $n$ cards, and one of these $n$ cards is $5$ .
The number of ways $5$ can be one of $n$ cards = ${}^{9}P_{n-1}*n$ $...bf{(1)}$
The total number of ways one can select $n$ cards $= {}^{10}P_n$ $...bf{(2)}$
Therefore , $P(E) = frac{bf{(1)}}{bf{(2)}}= frac{9!*n}{(10-n)!}*frac{(10-n)!}{10!}=boxed{frac{n}{10}}$
Note:- $n$ equals the number of drawings. % of card coming up = $frac{n}{10}*100 = 10n$
answered Jan 12 at 5:37
RahuboyRahuboy
43310
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Let us say you have drawn $n$ cards, and one of these $n$ cards is $5$ .
The number of ways $5$ can be one of $n$ cards = ${}^{9}P_{n-1}*n$ $...bf{(1)}$
The total number of ways one can select $n$ cards $= {}^{10}P_n$ $...bf{(2)}$
Therefore , $P(E) = frac{bf{(1)}}{bf{(2)}}= frac{9!*n}{(10-n)!}*frac{(10-n)!}{10!}=boxed{frac{n}{10}}$
Note:- $n$ equals the number of drawings. % of card coming up = $frac{n}{10}*100 = 10n$
answered Jan 12 at 5:37
RahuboyRahuboy
43310
$endgroup$
add a comment |
$begingroup$
Let us say you have drawn $n$ cards, and one of these $n$ cards is $5$ .
The number of ways $5$ can be one of $n$ cards = ${}^{9}P_{n-1}*n$ $...bf{(1)}$
The total number of ways one can select $n$ cards $= {}^{10}P_n$ $...bf{(2)}$
Therefore , $P(E) = frac{bf{(1)}}{bf{(2)}}= frac{9!*n}{(10-n)!}*frac{(10-n)!}{10!}=boxed{frac{n}{10}}$
Note:- $n$ equals the number of drawings. % of card coming up = $frac{n}{10}*100 = 10n$
answered Jan 12 at 5:37
RahuboyRahuboy
43310
$endgroup$
add a comment |
$begingroup$
Let us say you have drawn $n$ cards, and one of these $n$ cards is $5$ .
The number of ways $5$ can be one of $n$ cards = ${}^{9}P_{n-1}*n$ $...bf{(1)}$
The total number of ways one can select $n$ cards $= {}^{10}P_n$ $...bf{(2)}$
Therefore , $P(E) = frac{bf{(1)}}{bf{(2)}}= frac{9!*n}{(10-n)!}*frac{(10-n)!}{10!}=boxed{frac{n}{10}}$
Note:- $n$ equals the number of drawings. % of card coming up = $frac{n}{10}*100 = 10n$
answered Jan 12 at 5:37
RahuboyRahuboy
43310
$endgroup$
Let us say you have drawn $n$ cards, and one of these $n$ cards is $5$ .
The number of ways $5$ can be one of $n$ cards = ${}^{9}P_{n-1}*n$ $...bf{(1)}$
The total number of ways one can select $n$ cards $= {}^{10}P_n$ $...bf{(2)}$
Therefore , $P(E) = frac{bf{(1)}}{bf{(2)}}= frac{9!*n}{(10-n)!}*frac{(10-n)!}{10!}=boxed{frac{n}{10}}$
Note:- $n$ equals the number of drawings. % of card coming up = $frac{n}{10}*100 = 10n$
answered Jan 12 at 5:37
RahuboyRahuboy
43310
answered Jan 12 at 5:37
RahuboyRahuboy
43310
answered Jan 12 at 5:37
RahuboyRahuboy
43310
answered Jan 12 at 5:37
RahuboyRahuboy
43310
43310
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