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probability of x losers in n trials
The Next CEO of Stack OverflowProbability I lost when my friend told me I lost$1 bet, 1000 events, 45% chance of winning each event. What are the odds of winning?Roulette paradox - martingale probability after leaving tableRoulette: a Naive, but Fundamental Question about Probabilitycan this cashback casino promotion be exploited?Roulette probability - something doesn't add upCalculating Risk of Ruin for known profit/loss and probabilitiesWhy is the casino losing money?When gambling, do I get my money's worth? (Or: Does the amount I lose per bet determine the number of bets until I lose all my money?)Kelly Criterion for a finite number of bets
$begingroup$
I am new here and I am not sure if this is the correct place to ask this question.
It seems like a simple probability question but I am stuck on it. It is a problem I am trying to solve.
Lets say I have 43 bets.
The probability of a bet winning is 0.64.
The probability of a bet losing is 0.36.
What is the probability that 29 of those 43 bets will be losing/losers?
I worked it out like this:
29 / 43 = 67%
0.36 * 67 = 24.12% chance that 29 of those 43 bets will be losing/losers.
But this seems very simplistic and incorrect to me. Because for example - when the number of bets goes up e.g. 43 goes up to 100 bets. The probability that 29 of those 100 bets losing goes down to say 10.44%. Which seems wrong to me. Because I think if the number of bets goes up, then the probability of losing a fixed number of 29 should also go up, as there are higher "chances" of losing.
Could you list the correct answer, working out and possibly a link where I can read to learn more.
I did read up on this, but there were so many ways of achieving this. I read about Binomial distributions etc, but the answers I kept calculating just seemed very wrong to me.
Many Thanks!
probability statistics expected-value gambling
asked Mar 27 at 18:58
NathanNathan
1
New contributor
Nathan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am new here and I am not sure if this is the correct place to ask this question.
It seems like a simple probability question but I am stuck on it. It is a problem I am trying to solve.
Lets say I have 43 bets.
The probability of a bet winning is 0.64.
The probability of a bet losing is 0.36.
What is the probability that 29 of those 43 bets will be losing/losers?
I worked it out like this:
29 / 43 = 67%
0.36 * 67 = 24.12% chance that 29 of those 43 bets will be losing/losers.
But this seems very simplistic and incorrect to me. Because for example - when the number of bets goes up e.g. 43 goes up to 100 bets. The probability that 29 of those 100 bets losing goes down to say 10.44%. Which seems wrong to me. Because I think if the number of bets goes up, then the probability of losing a fixed number of 29 should also go up, as there are higher "chances" of losing.
Could you list the correct answer, working out and possibly a link where I can read to learn more.
I did read up on this, but there were so many ways of achieving this. I read about Binomial distributions etc, but the answers I kept calculating just seemed very wrong to me.
Many Thanks!
probability statistics expected-value gambling
asked Mar 27 at 18:58
NathanNathan
1
New contributor
Nathan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am new here and I am not sure if this is the correct place to ask this question.
It seems like a simple probability question but I am stuck on it. It is a problem I am trying to solve.
Lets say I have 43 bets.
The probability of a bet winning is 0.64.
The probability of a bet losing is 0.36.
What is the probability that 29 of those 43 bets will be losing/losers?
I worked it out like this:
29 / 43 = 67%
0.36 * 67 = 24.12% chance that 29 of those 43 bets will be losing/losers.
But this seems very simplistic and incorrect to me. Because for example - when the number of bets goes up e.g. 43 goes up to 100 bets. The probability that 29 of those 100 bets losing goes down to say 10.44%. Which seems wrong to me. Because I think if the number of bets goes up, then the probability of losing a fixed number of 29 should also go up, as there are higher "chances" of losing.
Could you list the correct answer, working out and possibly a link where I can read to learn more.
I did read up on this, but there were so many ways of achieving this. I read about Binomial distributions etc, but the answers I kept calculating just seemed very wrong to me.
Many Thanks!
probability statistics expected-value gambling
asked Mar 27 at 18:58
NathanNathan
1
New contributor
Nathan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am new here and I am not sure if this is the correct place to ask this question.
It seems like a simple probability question but I am stuck on it. It is a problem I am trying to solve.
Lets say I have 43 bets.
The probability of a bet winning is 0.64.
The probability of a bet losing is 0.36.
What is the probability that 29 of those 43 bets will be losing/losers?
I worked it out like this:
29 / 43 = 67%
0.36 * 67 = 24.12% chance that 29 of those 43 bets will be losing/losers.
But this seems very simplistic and incorrect to me. Because for example - when the number of bets goes up e.g. 43 goes up to 100 bets. The probability that 29 of those 100 bets losing goes down to say 10.44%. Which seems wrong to me. Because I think if the number of bets goes up, then the probability of losing a fixed number of 29 should also go up, as there are higher "chances" of losing.
Could you list the correct answer, working out and possibly a link where I can read to learn more.
I did read up on this, but there were so many ways of achieving this. I read about Binomial distributions etc, but the answers I kept calculating just seemed very wrong to me.
Many Thanks!
probability statistics expected-value gambling
probability statistics expected-value gambling
asked Mar 27 at 18:58
NathanNathan
1
New contributor
Nathan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 27 at 18:58
NathanNathan
1
New contributor
Nathan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 27 at 18:58
NathanNathan
1
New contributor
Nathan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 27 at 18:58
NathanNathan
1
asked Mar 27 at 18:58
NathanNathan
1
1
New contributor
Nathan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nathan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's indicate with $1$ a victory and with $0$ a loss. We can create so a bijection between all the possible results of a series of $43$ bet and a binary number of $43$ ciphers. In particular, we are searching all of the binary number in this form:
$$underbrace1...1_14 text timesunderbrace0...0_29 text times$$
All of the possible permutations of this binary number are:
$$P_(43,14,29)=frac43!14!29!$$
The probability of each permutation is:
$$p(P)=(0,64)^14(0,36)^29$$
This must be multiplied for the number of permutation. So the searched probability $p(E)$ is:
$$p(E)=p(P)P_(43,14,29)=(0,64)^14(0,36)^29frac43!14!29!=0.002 %$$
:)
answered Mar 27 at 19:18
EurekaEureka
438112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Hi Eureka, Thanks for your answer. I read your comment above that the above answer switched the probabilities. The funny thing is...If I switch the above answers probabilities. 43 choose 29 .36^29 (1-.36)^43-29 = 0.002% Which is the exact same answer as yours! Is this correct then? (please excuse if my math formatting is incorrect, as I am still learning how to use formatting on the site)
$endgroup$
– Nathan
Mar 27 at 19:39
$begingroup$
Hi :) , I meant that he switched the exponents because clearly $binomnn-k=binomnk$. Indeed our results are different as you see.
$endgroup$
– Eureka
Mar 27 at 19:43
$begingroup$
So are both the answers correct? or just yours above? thanks :-)
$endgroup$
– Nathan
Mar 27 at 19:47
$begingroup$
For sure they aren't both correct: math is just one ahaha. I think that he's wrong because he wrote $(0.64)^29$ , but you requested 29 losses not 29 winnings.
$endgroup$
– Eureka
Mar 27 at 19:49
$begingroup$
Yes true! haha, I will try to build your answer in Excel over next few days, thanks!
$endgroup$
– Nathan
Mar 27 at 20:49
add a comment |
$begingroup$
$$43 choose 29 (1-.64)^29 .64^43-29 = 0.0000205822$$
answered Mar 27 at 19:08
David G. StorkDavid G. Stork
11.5k41533
$endgroup$
$begingroup$
You switched the two probabilities.
$endgroup$
– Eureka
Mar 27 at 19:26
$begingroup$
Hi David, thanks for your answer. I guess it is just binomial coefficient * pk(1-p)(n-k) Which is similiar to what is explained here mathsisfun.com/data/binomial-distribution.html
$endgroup$
– Nathan
Mar 27 at 19:28
$begingroup$
I managed to replicate your answer in Excel using the COMBIN for the binomial coefficient calculation, thanks!
$endgroup$
– Nathan
Mar 27 at 19:29
$begingroup$
Thanks for the correction David! thanks for the answer
$endgroup$
– Nathan
Mar 27 at 20:50
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's indicate with $1$ a victory and with $0$ a loss. We can create so a bijection between all the possible results of a series of $43$ bet and a binary number of $43$ ciphers. In particular, we are searching all of the binary number in this form:
$$underbrace1...1_14 text timesunderbrace0...0_29 text times$$
All of the possible permutations of this binary number are:
$$P_(43,14,29)=frac43!14!29!$$
The probability of each permutation is:
$$p(P)=(0,64)^14(0,36)^29$$
This must be multiplied for the number of permutation. So the searched probability $p(E)$ is:
$$p(E)=p(P)P_(43,14,29)=(0,64)^14(0,36)^29frac43!14!29!=0.002 %$$
:)
answered Mar 27 at 19:18
EurekaEureka
438112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Hi Eureka, Thanks for your answer. I read your comment above that the above answer switched the probabilities. The funny thing is...If I switch the above answers probabilities. 43 choose 29 .36^29 (1-.36)^43-29 = 0.002% Which is the exact same answer as yours! Is this correct then? (please excuse if my math formatting is incorrect, as I am still learning how to use formatting on the site)
$endgroup$
– Nathan
Mar 27 at 19:39
$begingroup$
Hi :) , I meant that he switched the exponents because clearly $binomnn-k=binomnk$. Indeed our results are different as you see.
$endgroup$
– Eureka
Mar 27 at 19:43
$begingroup$
So are both the answers correct? or just yours above? thanks :-)
$endgroup$
– Nathan
Mar 27 at 19:47
$begingroup$
For sure they aren't both correct: math is just one ahaha. I think that he's wrong because he wrote $(0.64)^29$ , but you requested 29 losses not 29 winnings.
$endgroup$
– Eureka
Mar 27 at 19:49
$begingroup$
Yes true! haha, I will try to build your answer in Excel over next few days, thanks!
$endgroup$
– Nathan
Mar 27 at 20:49
add a comment |
$begingroup$
Let's indicate with $1$ a victory and with $0$ a loss. We can create so a bijection between all the possible results of a series of $43$ bet and a binary number of $43$ ciphers. In particular, we are searching all of the binary number in this form:
$$underbrace1...1_14 text timesunderbrace0...0_29 text times$$
All of the possible permutations of this binary number are:
$$P_(43,14,29)=frac43!14!29!$$
The probability of each permutation is:
$$p(P)=(0,64)^14(0,36)^29$$
This must be multiplied for the number of permutation. So the searched probability $p(E)$ is:
$$p(E)=p(P)P_(43,14,29)=(0,64)^14(0,36)^29frac43!14!29!=0.002 %$$
:)
answered Mar 27 at 19:18
EurekaEureka
438112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Hi Eureka, Thanks for your answer. I read your comment above that the above answer switched the probabilities. The funny thing is...If I switch the above answers probabilities. 43 choose 29 .36^29 (1-.36)^43-29 = 0.002% Which is the exact same answer as yours! Is this correct then? (please excuse if my math formatting is incorrect, as I am still learning how to use formatting on the site)
$endgroup$
– Nathan
Mar 27 at 19:39
$begingroup$
Hi :) , I meant that he switched the exponents because clearly $binomnn-k=binomnk$. Indeed our results are different as you see.
$endgroup$
– Eureka
Mar 27 at 19:43
$begingroup$
So are both the answers correct? or just yours above? thanks :-)
$endgroup$
– Nathan
Mar 27 at 19:47
$begingroup$
For sure they aren't both correct: math is just one ahaha. I think that he's wrong because he wrote $(0.64)^29$ , but you requested 29 losses not 29 winnings.
$endgroup$
– Eureka
Mar 27 at 19:49
$begingroup$
Yes true! haha, I will try to build your answer in Excel over next few days, thanks!
$endgroup$
– Nathan
Mar 27 at 20:49
add a comment |
$begingroup$
Let's indicate with $1$ a victory and with $0$ a loss. We can create so a bijection between all the possible results of a series of $43$ bet and a binary number of $43$ ciphers. In particular, we are searching all of the binary number in this form:
$$underbrace1...1_14 text timesunderbrace0...0_29 text times$$
All of the possible permutations of this binary number are:
$$P_(43,14,29)=frac43!14!29!$$
The probability of each permutation is:
$$p(P)=(0,64)^14(0,36)^29$$
This must be multiplied for the number of permutation. So the searched probability $p(E)$ is:
$$p(E)=p(P)P_(43,14,29)=(0,64)^14(0,36)^29frac43!14!29!=0.002 %$$
:)
answered Mar 27 at 19:18
EurekaEureka
438112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let's indicate with $1$ a victory and with $0$ a loss. We can create so a bijection between all the possible results of a series of $43$ bet and a binary number of $43$ ciphers. In particular, we are searching all of the binary number in this form:
$$underbrace1...1_14 text timesunderbrace0...0_29 text times$$
All of the possible permutations of this binary number are:
$$P_(43,14,29)=frac43!14!29!$$
The probability of each permutation is:
$$p(P)=(0,64)^14(0,36)^29$$
This must be multiplied for the number of permutation. So the searched probability $p(E)$ is:
$$p(E)=p(P)P_(43,14,29)=(0,64)^14(0,36)^29frac43!14!29!=0.002 %$$
:)
answered Mar 27 at 19:18
EurekaEureka
438112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 27 at 19:25
answered Mar 27 at 19:18
EurekaEureka
438112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 27 at 19:18
EurekaEureka
438112
answered Mar 27 at 19:18
EurekaEureka
438112
438112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Hi Eureka, Thanks for your answer. I read your comment above that the above answer switched the probabilities. The funny thing is...If I switch the above answers probabilities. 43 choose 29 .36^29 (1-.36)^43-29 = 0.002% Which is the exact same answer as yours! Is this correct then? (please excuse if my math formatting is incorrect, as I am still learning how to use formatting on the site)
$endgroup$
– Nathan
Mar 27 at 19:39
$begingroup$
Hi :) , I meant that he switched the exponents because clearly $binomnn-k=binomnk$. Indeed our results are different as you see.
$endgroup$
– Eureka
Mar 27 at 19:43
$begingroup$
So are both the answers correct? or just yours above? thanks :-)
$endgroup$
– Nathan
Mar 27 at 19:47
$begingroup$
For sure they aren't both correct: math is just one ahaha. I think that he's wrong because he wrote $(0.64)^29$ , but you requested 29 losses not 29 winnings.
$endgroup$
– Eureka
Mar 27 at 19:49
$begingroup$
Yes true! haha, I will try to build your answer in Excel over next few days, thanks!
$endgroup$
– Nathan
Mar 27 at 20:49
add a comment |
$begingroup$
Hi Eureka, Thanks for your answer. I read your comment above that the above answer switched the probabilities. The funny thing is...If I switch the above answers probabilities. 43 choose 29 .36^29 (1-.36)^43-29 = 0.002% Which is the exact same answer as yours! Is this correct then? (please excuse if my math formatting is incorrect, as I am still learning how to use formatting on the site)
$endgroup$
– Nathan
Mar 27 at 19:39
$begingroup$
Hi :) , I meant that he switched the exponents because clearly $binomnn-k=binomnk$. Indeed our results are different as you see.
$endgroup$
– Eureka
Mar 27 at 19:43
$begingroup$
So are both the answers correct? or just yours above? thanks :-)
$endgroup$
– Nathan
Mar 27 at 19:47
$begingroup$
For sure they aren't both correct: math is just one ahaha. I think that he's wrong because he wrote $(0.64)^29$ , but you requested 29 losses not 29 winnings.
$endgroup$
– Eureka
Mar 27 at 19:49
$begingroup$
Yes true! haha, I will try to build your answer in Excel over next few days, thanks!
$endgroup$
– Nathan
Mar 27 at 20:49
$begingroup$
Hi Eureka, Thanks for your answer. I read your comment above that the above answer switched the probabilities. The funny thing is...If I switch the above answers probabilities. 43 choose 29 .36^29 (1-.36)^43-29 = 0.002% Which is the exact same answer as yours! Is this correct then? (please excuse if my math formatting is incorrect, as I am still learning how to use formatting on the site)
$endgroup$
– Nathan
Mar 27 at 19:39
$begingroup$
Hi Eureka, Thanks for your answer. I read your comment above that the above answer switched the probabilities. The funny thing is...If I switch the above answers probabilities. 43 choose 29 .36^29 (1-.36)^43-29 = 0.002% Which is the exact same answer as yours! Is this correct then? (please excuse if my math formatting is incorrect, as I am still learning how to use formatting on the site)
$endgroup$
– Nathan
Mar 27 at 19:39
$begingroup$
Hi :) , I meant that he switched the exponents because clearly $binomnn-k=binomnk$. Indeed our results are different as you see.
$endgroup$
– Eureka
Mar 27 at 19:43
$begingroup$
Hi :) , I meant that he switched the exponents because clearly $binomnn-k=binomnk$. Indeed our results are different as you see.
$endgroup$
– Eureka
Mar 27 at 19:43
$begingroup$
So are both the answers correct? or just yours above? thanks :-)
$endgroup$
– Nathan
Mar 27 at 19:47
$begingroup$
So are both the answers correct? or just yours above? thanks :-)
$endgroup$
– Nathan
Mar 27 at 19:47
$begingroup$
For sure they aren't both correct: math is just one ahaha. I think that he's wrong because he wrote $(0.64)^29$ , but you requested 29 losses not 29 winnings.
$endgroup$
– Eureka
Mar 27 at 19:49
$begingroup$
For sure they aren't both correct: math is just one ahaha. I think that he's wrong because he wrote $(0.64)^29$ , but you requested 29 losses not 29 winnings.
$endgroup$
– Eureka
Mar 27 at 19:49
$begingroup$
Yes true! haha, I will try to build your answer in Excel over next few days, thanks!
$endgroup$
– Nathan
Mar 27 at 20:49
$begingroup$
Yes true! haha, I will try to build your answer in Excel over next few days, thanks!
$endgroup$
– Nathan
Mar 27 at 20:49
add a comment |
$begingroup$
$$43 choose 29 (1-.64)^29 .64^43-29 = 0.0000205822$$
answered Mar 27 at 19:08
David G. StorkDavid G. Stork
11.5k41533
$endgroup$
$begingroup$
You switched the two probabilities.
$endgroup$
– Eureka
Mar 27 at 19:26
$begingroup$
Hi David, thanks for your answer. I guess it is just binomial coefficient * pk(1-p)(n-k) Which is similiar to what is explained here mathsisfun.com/data/binomial-distribution.html
$endgroup$
– Nathan
Mar 27 at 19:28
$begingroup$
I managed to replicate your answer in Excel using the COMBIN for the binomial coefficient calculation, thanks!
$endgroup$
– Nathan
Mar 27 at 19:29
$begingroup$
Thanks for the correction David! thanks for the answer
$endgroup$
– Nathan
Mar 27 at 20:50
add a comment |
$begingroup$
$$43 choose 29 (1-.64)^29 .64^43-29 = 0.0000205822$$
answered Mar 27 at 19:08
David G. StorkDavid G. Stork
11.5k41533
$endgroup$
$begingroup$
You switched the two probabilities.
$endgroup$
– Eureka
Mar 27 at 19:26
$begingroup$
Hi David, thanks for your answer. I guess it is just binomial coefficient * pk(1-p)(n-k) Which is similiar to what is explained here mathsisfun.com/data/binomial-distribution.html
$endgroup$
– Nathan
Mar 27 at 19:28
$begingroup$
I managed to replicate your answer in Excel using the COMBIN for the binomial coefficient calculation, thanks!
$endgroup$
– Nathan
Mar 27 at 19:29
$begingroup$
Thanks for the correction David! thanks for the answer
$endgroup$
– Nathan
Mar 27 at 20:50
add a comment |
$begingroup$
$$43 choose 29 (1-.64)^29 .64^43-29 = 0.0000205822$$
answered Mar 27 at 19:08
David G. StorkDavid G. Stork
11.5k41533
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$$43 choose 29 (1-.64)^29 .64^43-29 = 0.0000205822$$
answered Mar 27 at 19:08
David G. StorkDavid G. Stork
11.5k41533
edited Mar 27 at 19:49
answered Mar 27 at 19:08
David G. StorkDavid G. Stork
11.5k41533
answered Mar 27 at 19:08
David G. StorkDavid G. Stork
11.5k41533
answered Mar 27 at 19:08
David G. StorkDavid G. Stork
11.5k41533
11.5k41533
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You switched the two probabilities.
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– Eureka
Mar 27 at 19:26
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Hi David, thanks for your answer. I guess it is just binomial coefficient * pk(1-p)(n-k) Which is similiar to what is explained here mathsisfun.com/data/binomial-distribution.html
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– Nathan
Mar 27 at 19:28
$begingroup$
I managed to replicate your answer in Excel using the COMBIN for the binomial coefficient calculation, thanks!
$endgroup$
– Nathan
Mar 27 at 19:29
$begingroup$
Thanks for the correction David! thanks for the answer
$endgroup$
– Nathan
Mar 27 at 20:50
add a comment |
$begingroup$
You switched the two probabilities.
$endgroup$
– Eureka
Mar 27 at 19:26
$begingroup$
Hi David, thanks for your answer. I guess it is just binomial coefficient * pk(1-p)(n-k) Which is similiar to what is explained here mathsisfun.com/data/binomial-distribution.html
$endgroup$
– Nathan
Mar 27 at 19:28
$begingroup$
I managed to replicate your answer in Excel using the COMBIN for the binomial coefficient calculation, thanks!
$endgroup$
– Nathan
Mar 27 at 19:29
$begingroup$
Thanks for the correction David! thanks for the answer
$endgroup$
– Nathan
Mar 27 at 20:50
$begingroup$
You switched the two probabilities.
$endgroup$
– Eureka
Mar 27 at 19:26
$begingroup$
You switched the two probabilities.
$endgroup$
– Eureka
Mar 27 at 19:26
$begingroup$
Hi David, thanks for your answer. I guess it is just binomial coefficient * pk(1-p)(n-k) Which is similiar to what is explained here mathsisfun.com/data/binomial-distribution.html
$endgroup$
– Nathan
Mar 27 at 19:28
$begingroup$
Hi David, thanks for your answer. I guess it is just binomial coefficient * pk(1-p)(n-k) Which is similiar to what is explained here mathsisfun.com/data/binomial-distribution.html
$endgroup$
– Nathan
Mar 27 at 19:28
$begingroup$
I managed to replicate your answer in Excel using the COMBIN for the binomial coefficient calculation, thanks!
$endgroup$
– Nathan
Mar 27 at 19:29
$begingroup$
I managed to replicate your answer in Excel using the COMBIN for the binomial coefficient calculation, thanks!
$endgroup$
– Nathan
Mar 27 at 19:29
$begingroup$
Thanks for the correction David! thanks for the answer
$endgroup$
– Nathan
Mar 27 at 20:50
$begingroup$
Thanks for the correction David! thanks for the answer
$endgroup$
– Nathan
Mar 27 at 20:50
add a comment |
Nathan is a new contributor. Be nice, and check out our Code of Conduct.
Nathan is a new contributor. Be nice, and check out our Code of Conduct.
Nathan is a new contributor. Be nice, and check out our Code of Conduct.
Nathan is a new contributor. Be nice, and check out our Code of Conduct.
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