Mixing
Probability of multiple trials?
$begingroup$
give an event $E$ and probability that event occurs as P. If it was attempted $1000$ times. What would be the answers to the following question.
- probability of at least the event occurring one time ? is it $1-(1-P)^{1000}$
- probability of at most the event occurring one time? is it $P*(1-P)^{999}$
- probability of at least the event occurring three times? is it $P^3*(1-(1-P)^{997})$
- probability of at most the event occurring three times ? is it $P^3*(1-P)^{997}$
- probability of event occurring more than one time ? is it $P^2*(1-(1-P)^{998})$
- probability of event not occurring at all ? is it $(1-p)^{1000}$
probability probability-theory
edited Oct 4 '17 at 11:53
kimi Tanaka
414411
asked Oct 3 '17 at 11:48
user119020user119020
544
$endgroup$
|
show 7 more comments
$begingroup$
give an event $E$ and probability that event occurs as P. If it was attempted $1000$ times. What would be the answers to the following question.
- probability of at least the event occurring one time ? is it $1-(1-P)^{1000}$
- probability of at most the event occurring one time? is it $P*(1-P)^{999}$
- probability of at least the event occurring three times? is it $P^3*(1-(1-P)^{997})$
- probability of at most the event occurring three times ? is it $P^3*(1-P)^{997}$
- probability of event occurring more than one time ? is it $P^2*(1-(1-P)^{998})$
- probability of event not occurring at all ? is it $(1-p)^{1000}$
probability probability-theory
edited Oct 4 '17 at 11:53
kimi Tanaka
414411
asked Oct 3 '17 at 11:48
user119020user119020
544
$endgroup$
1
$begingroup$
"$-3$ down vote favorite"....did you cut and paste this from another question on this site?
$endgroup$
– lulu
Oct 3 '17 at 11:51
$begingroup$
What have you tried? These are all basic questions in the use of the binomial distribution
$endgroup$
– lulu
Oct 3 '17 at 11:53
$begingroup$
@lulu. Yea it was 1st posted at mathoverflow. And as for your question i have just started to learn Probability.So these are some of the questions that has popped in to my mind
$endgroup$
– user119020
Oct 3 '17 at 12:08
1
$begingroup$
Well, the second one is wrong. Let $P_i$ be the probability that it occurs exactly $i$ times, so you want $P_0+P_1$. As you point out $P_0=(1-p)^{1000}$. From the binomial distribution we get $P_1=1000times ptimes (1-p)^{999}$.
$endgroup$
– lulu
Oct 3 '17 at 23:15
1
$begingroup$
No, I wasn't saying that. Others are wrong along the same lines. For the third, say, we have $P_3=binom {1000}3times p^3times (1-p)^{997}$ and so on.
$endgroup$
– lulu
Oct 4 '17 at 10:59
|
show 7 more comments
$begingroup$
give an event $E$ and probability that event occurs as P. If it was attempted $1000$ times. What would be the answers to the following question.
- probability of at least the event occurring one time ? is it $1-(1-P)^{1000}$
- probability of at most the event occurring one time? is it $P*(1-P)^{999}$
- probability of at least the event occurring three times? is it $P^3*(1-(1-P)^{997})$
- probability of at most the event occurring three times ? is it $P^3*(1-P)^{997}$
- probability of event occurring more than one time ? is it $P^2*(1-(1-P)^{998})$
- probability of event not occurring at all ? is it $(1-p)^{1000}$
probability probability-theory
edited Oct 4 '17 at 11:53
kimi Tanaka
414411
asked Oct 3 '17 at 11:48
user119020user119020
544
$endgroup$
give an event $E$ and probability that event occurs as P. If it was attempted $1000$ times. What would be the answers to the following question.
- probability of at least the event occurring one time ? is it $1-(1-P)^{1000}$
- probability of at most the event occurring one time? is it $P*(1-P)^{999}$
- probability of at least the event occurring three times? is it $P^3*(1-(1-P)^{997})$
- probability of at most the event occurring three times ? is it $P^3*(1-P)^{997}$
- probability of event occurring more than one time ? is it $P^2*(1-(1-P)^{998})$
- probability of event not occurring at all ? is it $(1-p)^{1000}$
probability probability-theory
probability probability-theory
edited Oct 4 '17 at 11:53
kimi Tanaka
414411
asked Oct 3 '17 at 11:48
user119020user119020
544
edited Oct 4 '17 at 11:53
kimi Tanaka
414411
asked Oct 3 '17 at 11:48
user119020user119020
544
edited Oct 4 '17 at 11:53
kimi Tanaka
414411
edited Oct 4 '17 at 11:53
kimi Tanaka
414411
edited Oct 4 '17 at 11:53
kimi Tanaka
414411
414411
asked Oct 3 '17 at 11:48
user119020user119020
544
asked Oct 3 '17 at 11:48
user119020user119020
544
asked Oct 3 '17 at 11:48
user119020user119020
544
544
1
$begingroup$
"$-3$ down vote favorite"....did you cut and paste this from another question on this site?
$endgroup$
– lulu
Oct 3 '17 at 11:51
$begingroup$
What have you tried? These are all basic questions in the use of the binomial distribution
$endgroup$
– lulu
Oct 3 '17 at 11:53
$begingroup$
@lulu. Yea it was 1st posted at mathoverflow. And as for your question i have just started to learn Probability.So these are some of the questions that has popped in to my mind
$endgroup$
– user119020
Oct 3 '17 at 12:08
1
$begingroup$
Well, the second one is wrong. Let $P_i$ be the probability that it occurs exactly $i$ times, so you want $P_0+P_1$. As you point out $P_0=(1-p)^{1000}$. From the binomial distribution we get $P_1=1000times ptimes (1-p)^{999}$.
$endgroup$
– lulu
Oct 3 '17 at 23:15
1
$begingroup$
No, I wasn't saying that. Others are wrong along the same lines. For the third, say, we have $P_3=binom {1000}3times p^3times (1-p)^{997}$ and so on.
$endgroup$
– lulu
Oct 4 '17 at 10:59
|
show 7 more comments
1
$begingroup$
"$-3$ down vote favorite"....did you cut and paste this from another question on this site?
$endgroup$
– lulu
Oct 3 '17 at 11:51
$begingroup$
What have you tried? These are all basic questions in the use of the binomial distribution
$endgroup$
– lulu
Oct 3 '17 at 11:53
$begingroup$
@lulu. Yea it was 1st posted at mathoverflow. And as for your question i have just started to learn Probability.So these are some of the questions that has popped in to my mind
$endgroup$
– user119020
Oct 3 '17 at 12:08
1
$begingroup$
Well, the second one is wrong. Let $P_i$ be the probability that it occurs exactly $i$ times, so you want $P_0+P_1$. As you point out $P_0=(1-p)^{1000}$. From the binomial distribution we get $P_1=1000times ptimes (1-p)^{999}$.
$endgroup$
– lulu
Oct 3 '17 at 23:15
1
$begingroup$
No, I wasn't saying that. Others are wrong along the same lines. For the third, say, we have $P_3=binom {1000}3times p^3times (1-p)^{997}$ and so on.
$endgroup$
– lulu
Oct 4 '17 at 10:59
1
1
$begingroup$
"$-3$ down vote favorite"....did you cut and paste this from another question on this site?
$endgroup$
– lulu
Oct 3 '17 at 11:51
$begingroup$
"$-3$ down vote favorite"....did you cut and paste this from another question on this site?
$endgroup$
– lulu
Oct 3 '17 at 11:51
$begingroup$
What have you tried? These are all basic questions in the use of the binomial distribution
$endgroup$
– lulu
Oct 3 '17 at 11:53
$begingroup$
What have you tried? These are all basic questions in the use of the binomial distribution
$endgroup$
– lulu
Oct 3 '17 at 11:53
$begingroup$
@lulu. Yea it was 1st posted at mathoverflow. And as for your question i have just started to learn Probability.So these are some of the questions that has popped in to my mind
$endgroup$
– user119020
Oct 3 '17 at 12:08
$begingroup$
@lulu. Yea it was 1st posted at mathoverflow. And as for your question i have just started to learn Probability.So these are some of the questions that has popped in to my mind
$endgroup$
– user119020
Oct 3 '17 at 12:08
1
1
$begingroup$
Well, the second one is wrong. Let $P_i$ be the probability that it occurs exactly $i$ times, so you want $P_0+P_1$. As you point out $P_0=(1-p)^{1000}$. From the binomial distribution we get $P_1=1000times ptimes (1-p)^{999}$.
$endgroup$
– lulu
Oct 3 '17 at 23:15
$begingroup$
Well, the second one is wrong. Let $P_i$ be the probability that it occurs exactly $i$ times, so you want $P_0+P_1$. As you point out $P_0=(1-p)^{1000}$. From the binomial distribution we get $P_1=1000times ptimes (1-p)^{999}$.
$endgroup$
– lulu
Oct 3 '17 at 23:15
1
1
$begingroup$
No, I wasn't saying that. Others are wrong along the same lines. For the third, say, we have $P_3=binom {1000}3times p^3times (1-p)^{997}$ and so on.
$endgroup$
– lulu
Oct 4 '17 at 10:59
$begingroup$
No, I wasn't saying that. Others are wrong along the same lines. For the third, say, we have $P_3=binom {1000}3times p^3times (1-p)^{997}$ and so on.
$endgroup$
– lulu
Oct 4 '17 at 10:59
|
show 7 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Hints:
The probability it happens $k$ times in $n$ independent attempts is $P(k)={n choose k}p^k (1-p)^{n-k}$. Your attempts seem to have ignored the binomial co-efficients which count the different possible orderings of successes.
To answer the questions, consider:
- $1$ minus probability $k=0$
- probability $k=0$ or $k=1$
- probability $k=0$ or $k=1$ or $k=2$ or $k=3$
- $1$ minus probability $k=0$ or $k=1$ or $k=2$
- $1$ minus probability $k=0$ or $k=1$
- probability $k=0$
answered Oct 4 '17 at 12:16
HenryHenry
101k482170
$endgroup$
add a comment |
$begingroup$
- probability of at least the event occurring one time ? is it $1-(1-P)^{1000}$
You don't have any other way to count the event not occurring in any other way.
The result is $binom{1000}{1000}$$times${$1-(1-P)^{1000}$}
- probability of at most the event occurring one time? is it $P*(1-P)^{999}$
As in Lulu's comment, it has been already explained, you need to consider not occuring and only once the event occurs. Only once the event occurs, you have many other ways to count them such as the way at 1st observaton the event happens and the rest of observations the event not happens or the way at 2nd observations the event does not happen and so on so forth.
The result is altogether not occuring and occring once.
$binom{1000}{1000}$$times${$1-(1-P)^{1000}$} + $binom{1000}{1}$$times${$ptimes(1-P)^{999}$}
- probability of at least the event occurring three times? is it $P^3*(1-(1-P)^{997})$
Now You need to add another scenario the event happens twice to the last one. And you have the probability of at most twice happens. You need to remove them from all the event 1. Now you get the probability of at least three times happens.
$1$ - {$binom{1000}{1000}$$times${$1-(1-P)^{1000}$} + $binom{1000}{1}$$times${$ptimes(1-P)^{999}$} + $binom{1000}{2}$$times$${p^{2}times(1-P)^{998}}$}
- probability of at most the event occurring three times ? is it $P^3*(1-P)^{997}$
You need to add the probaility of three times happen to the probaility of at most twice happen.
$binom{1000}{1000}$$times${$1-(1-P)^{1000}$} + $binom{1000}{1}$$times${$ptimes(1-P)^{999}$} + $binom{1000}{2}$$times$${p^{2}times(1-P)^{998}}$ + $binom{1000}{3}$$times$${p^{3}times(1-P)^{997}}$
- probability of event occurring more than one time ? is it $P^2*(1-(1-P)^{998})$
$1$ - { the probaility of at most the event occuring one time}
- the probability of event not occurring at all ? is it $(1-p)^{1000}$
Yes.
answered Oct 4 '17 at 12:59
kimi Tanakakimi Tanaka
414411
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hints:
The probability it happens $k$ times in $n$ independent attempts is $P(k)={n choose k}p^k (1-p)^{n-k}$. Your attempts seem to have ignored the binomial co-efficients which count the different possible orderings of successes.
To answer the questions, consider:
- $1$ minus probability $k=0$
- probability $k=0$ or $k=1$
- probability $k=0$ or $k=1$ or $k=2$ or $k=3$
- $1$ minus probability $k=0$ or $k=1$ or $k=2$
- $1$ minus probability $k=0$ or $k=1$
- probability $k=0$
answered Oct 4 '17 at 12:16
HenryHenry
101k482170
$endgroup$
add a comment |
$begingroup$
Hints:
The probability it happens $k$ times in $n$ independent attempts is $P(k)={n choose k}p^k (1-p)^{n-k}$. Your attempts seem to have ignored the binomial co-efficients which count the different possible orderings of successes.
To answer the questions, consider:
- $1$ minus probability $k=0$
- probability $k=0$ or $k=1$
- probability $k=0$ or $k=1$ or $k=2$ or $k=3$
- $1$ minus probability $k=0$ or $k=1$ or $k=2$
- $1$ minus probability $k=0$ or $k=1$
- probability $k=0$
answered Oct 4 '17 at 12:16
HenryHenry
101k482170
$endgroup$
add a comment |
$begingroup$
Hints:
The probability it happens $k$ times in $n$ independent attempts is $P(k)={n choose k}p^k (1-p)^{n-k}$. Your attempts seem to have ignored the binomial co-efficients which count the different possible orderings of successes.
To answer the questions, consider:
- $1$ minus probability $k=0$
- probability $k=0$ or $k=1$
- probability $k=0$ or $k=1$ or $k=2$ or $k=3$
- $1$ minus probability $k=0$ or $k=1$ or $k=2$
- $1$ minus probability $k=0$ or $k=1$
- probability $k=0$
answered Oct 4 '17 at 12:16
HenryHenry
101k482170
$endgroup$
Hints:
The probability it happens $k$ times in $n$ independent attempts is $P(k)={n choose k}p^k (1-p)^{n-k}$. Your attempts seem to have ignored the binomial co-efficients which count the different possible orderings of successes.
To answer the questions, consider:
- $1$ minus probability $k=0$
- probability $k=0$ or $k=1$
- probability $k=0$ or $k=1$ or $k=2$ or $k=3$
- $1$ minus probability $k=0$ or $k=1$ or $k=2$
- $1$ minus probability $k=0$ or $k=1$
- probability $k=0$
answered Oct 4 '17 at 12:16
HenryHenry
101k482170
answered Oct 4 '17 at 12:16
HenryHenry
101k482170
answered Oct 4 '17 at 12:16
HenryHenry
101k482170
answered Oct 4 '17 at 12:16
HenryHenry
101k482170
101k482170
add a comment |
add a comment |
$begingroup$
- probability of at least the event occurring one time ? is it $1-(1-P)^{1000}$
You don't have any other way to count the event not occurring in any other way.
The result is $binom{1000}{1000}$$times${$1-(1-P)^{1000}$}
- probability of at most the event occurring one time? is it $P*(1-P)^{999}$
As in Lulu's comment, it has been already explained, you need to consider not occuring and only once the event occurs. Only once the event occurs, you have many other ways to count them such as the way at 1st observaton the event happens and the rest of observations the event not happens or the way at 2nd observations the event does not happen and so on so forth.
The result is altogether not occuring and occring once.
$binom{1000}{1000}$$times${$1-(1-P)^{1000}$} + $binom{1000}{1}$$times${$ptimes(1-P)^{999}$}
- probability of at least the event occurring three times? is it $P^3*(1-(1-P)^{997})$
Now You need to add another scenario the event happens twice to the last one. And you have the probability of at most twice happens. You need to remove them from all the event 1. Now you get the probability of at least three times happens.
$1$ - {$binom{1000}{1000}$$times${$1-(1-P)^{1000}$} + $binom{1000}{1}$$times${$ptimes(1-P)^{999}$} + $binom{1000}{2}$$times$${p^{2}times(1-P)^{998}}$}
- probability of at most the event occurring three times ? is it $P^3*(1-P)^{997}$
You need to add the probaility of three times happen to the probaility of at most twice happen.
$binom{1000}{1000}$$times${$1-(1-P)^{1000}$} + $binom{1000}{1}$$times${$ptimes(1-P)^{999}$} + $binom{1000}{2}$$times$${p^{2}times(1-P)^{998}}$ + $binom{1000}{3}$$times$${p^{3}times(1-P)^{997}}$
- probability of event occurring more than one time ? is it $P^2*(1-(1-P)^{998})$
$1$ - { the probaility of at most the event occuring one time}
- the probability of event not occurring at all ? is it $(1-p)^{1000}$
Yes.
answered Oct 4 '17 at 12:59
kimi Tanakakimi Tanaka
414411
$endgroup$
add a comment |
$begingroup$
- probability of at least the event occurring one time ? is it $1-(1-P)^{1000}$
You don't have any other way to count the event not occurring in any other way.
The result is $binom{1000}{1000}$$times${$1-(1-P)^{1000}$}
- probability of at most the event occurring one time? is it $P*(1-P)^{999}$
As in Lulu's comment, it has been already explained, you need to consider not occuring and only once the event occurs. Only once the event occurs, you have many other ways to count them such as the way at 1st observaton the event happens and the rest of observations the event not happens or the way at 2nd observations the event does not happen and so on so forth.
The result is altogether not occuring and occring once.
$binom{1000}{1000}$$times${$1-(1-P)^{1000}$} + $binom{1000}{1}$$times${$ptimes(1-P)^{999}$}
- probability of at least the event occurring three times? is it $P^3*(1-(1-P)^{997})$
Now You need to add another scenario the event happens twice to the last one. And you have the probability of at most twice happens. You need to remove them from all the event 1. Now you get the probability of at least three times happens.
$1$ - {$binom{1000}{1000}$$times${$1-(1-P)^{1000}$} + $binom{1000}{1}$$times${$ptimes(1-P)^{999}$} + $binom{1000}{2}$$times$${p^{2}times(1-P)^{998}}$}
- probability of at most the event occurring three times ? is it $P^3*(1-P)^{997}$
You need to add the probaility of three times happen to the probaility of at most twice happen.
$binom{1000}{1000}$$times${$1-(1-P)^{1000}$} + $binom{1000}{1}$$times${$ptimes(1-P)^{999}$} + $binom{1000}{2}$$times$${p^{2}times(1-P)^{998}}$ + $binom{1000}{3}$$times$${p^{3}times(1-P)^{997}}$
- probability of event occurring more than one time ? is it $P^2*(1-(1-P)^{998})$
$1$ - { the probaility of at most the event occuring one time}
- the probability of event not occurring at all ? is it $(1-p)^{1000}$
Yes.
answered Oct 4 '17 at 12:59
kimi Tanakakimi Tanaka
414411
$endgroup$
add a comment |
$begingroup$
- probability of at least the event occurring one time ? is it $1-(1-P)^{1000}$
You don't have any other way to count the event not occurring in any other way.
The result is $binom{1000}{1000}$$times${$1-(1-P)^{1000}$}
- probability of at most the event occurring one time? is it $P*(1-P)^{999}$
As in Lulu's comment, it has been already explained, you need to consider not occuring and only once the event occurs. Only once the event occurs, you have many other ways to count them such as the way at 1st observaton the event happens and the rest of observations the event not happens or the way at 2nd observations the event does not happen and so on so forth.
The result is altogether not occuring and occring once.
$binom{1000}{1000}$$times${$1-(1-P)^{1000}$} + $binom{1000}{1}$$times${$ptimes(1-P)^{999}$}
- probability of at least the event occurring three times? is it $P^3*(1-(1-P)^{997})$
Now You need to add another scenario the event happens twice to the last one. And you have the probability of at most twice happens. You need to remove them from all the event 1. Now you get the probability of at least three times happens.
$1$ - {$binom{1000}{1000}$$times${$1-(1-P)^{1000}$} + $binom{1000}{1}$$times${$ptimes(1-P)^{999}$} + $binom{1000}{2}$$times$${p^{2}times(1-P)^{998}}$}
- probability of at most the event occurring three times ? is it $P^3*(1-P)^{997}$
You need to add the probaility of three times happen to the probaility of at most twice happen.
$binom{1000}{1000}$$times${$1-(1-P)^{1000}$} + $binom{1000}{1}$$times${$ptimes(1-P)^{999}$} + $binom{1000}{2}$$times$${p^{2}times(1-P)^{998}}$ + $binom{1000}{3}$$times$${p^{3}times(1-P)^{997}}$
- probability of event occurring more than one time ? is it $P^2*(1-(1-P)^{998})$
$1$ - { the probaility of at most the event occuring one time}
- the probability of event not occurring at all ? is it $(1-p)^{1000}$
Yes.
answered Oct 4 '17 at 12:59
kimi Tanakakimi Tanaka
414411
$endgroup$
- probability of at least the event occurring one time ? is it $1-(1-P)^{1000}$
You don't have any other way to count the event not occurring in any other way.
The result is $binom{1000}{1000}$$times${$1-(1-P)^{1000}$}
- probability of at most the event occurring one time? is it $P*(1-P)^{999}$
As in Lulu's comment, it has been already explained, you need to consider not occuring and only once the event occurs. Only once the event occurs, you have many other ways to count them such as the way at 1st observaton the event happens and the rest of observations the event not happens or the way at 2nd observations the event does not happen and so on so forth.
The result is altogether not occuring and occring once.
$binom{1000}{1000}$$times${$1-(1-P)^{1000}$} + $binom{1000}{1}$$times${$ptimes(1-P)^{999}$}
- probability of at least the event occurring three times? is it $P^3*(1-(1-P)^{997})$
Now You need to add another scenario the event happens twice to the last one. And you have the probability of at most twice happens. You need to remove them from all the event 1. Now you get the probability of at least three times happens.
$1$ - {$binom{1000}{1000}$$times${$1-(1-P)^{1000}$} + $binom{1000}{1}$$times${$ptimes(1-P)^{999}$} + $binom{1000}{2}$$times$${p^{2}times(1-P)^{998}}$}
- probability of at most the event occurring three times ? is it $P^3*(1-P)^{997}$
You need to add the probaility of three times happen to the probaility of at most twice happen.
$binom{1000}{1000}$$times${$1-(1-P)^{1000}$} + $binom{1000}{1}$$times${$ptimes(1-P)^{999}$} + $binom{1000}{2}$$times$${p^{2}times(1-P)^{998}}$ + $binom{1000}{3}$$times$${p^{3}times(1-P)^{997}}$
- probability of event occurring more than one time ? is it $P^2*(1-(1-P)^{998})$
$1$ - { the probaility of at most the event occuring one time}
- the probability of event not occurring at all ? is it $(1-p)^{1000}$
Yes.
answered Oct 4 '17 at 12:59
kimi Tanakakimi Tanaka
414411
answered Oct 4 '17 at 12:59
kimi Tanakakimi Tanaka
414411
answered Oct 4 '17 at 12:59
kimi Tanakakimi Tanaka
414411
answered Oct 4 '17 at 12:59
kimi Tanakakimi Tanaka
414411
414411
add a comment |
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1
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"$-3$ down vote favorite"....did you cut and paste this from another question on this site?
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– lulu
Oct 3 '17 at 11:51
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What have you tried? These are all basic questions in the use of the binomial distribution
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– lulu
Oct 3 '17 at 11:53
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@lulu. Yea it was 1st posted at mathoverflow. And as for your question i have just started to learn Probability.So these are some of the questions that has popped in to my mind
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– user119020
Oct 3 '17 at 12:08
1
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Well, the second one is wrong. Let $P_i$ be the probability that it occurs exactly $i$ times, so you want $P_0+P_1$. As you point out $P_0=(1-p)^{1000}$. From the binomial distribution we get $P_1=1000times ptimes (1-p)^{999}$.
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– lulu
Oct 3 '17 at 23:15
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No, I wasn't saying that. Others are wrong along the same lines. For the third, say, we have $P_3=binom {1000}3times p^3times (1-p)^{997}$ and so on.
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– lulu
Oct 4 '17 at 10:59