Odds of run of successes in Bernoulli trial
0
What is the formula that gives the odds I will have a run of k successes in a row, where each trial has a p chance of success, if I ran the trial n times?
probability
asked Jan 4 at 22:42
Lucian cahilLucian cahil
11
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What is the formula that gives the odds I will have a run of k successes in a row, where each trial has a p chance of success, if I ran the trial n times?
probability
asked Jan 4 at 22:42
Lucian cahilLucian cahil
11
New contributor
Lucian cahil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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4
Have you made any attempt to compute the answer yourself?
– Aditya Dua
Jan 4 at 22:45
1
I suggest you try working it out by hand for a simple case, such as $k=2$, $p=0.4$, and $n=5$. There are only $32$ possible outcomes of running the trial $5$ times. You can list each outcome with its probability of occurring, and then add up the probabilities for the outcomes that contain two consecutive successes. If you do this for a few simple cases, you might get some ideas for how to answer the general question.
– Steve Kass
Jan 4 at 23:13
1
This is a fairly difficult problem to solve in general - see here for solutions for given $k$ and $n$: stats.stackexchange.com/questions/21825/…
– Math1000
Jan 4 at 23:27
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What is the formula that gives the odds I will have a run of k successes in a row, where each trial has a p chance of success, if I ran the trial n times?
probability
asked Jan 4 at 22:42
Lucian cahilLucian cahil
11
New contributor
Lucian cahil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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What is the formula that gives the odds I will have a run of k successes in a row, where each trial has a p chance of success, if I ran the trial n times?
probability
probability
asked Jan 4 at 22:42
Lucian cahilLucian cahil
11
New contributor
Lucian cahil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 22:42
Lucian cahilLucian cahil
11
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Lucian cahil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Jan 4 at 23:03
Lucian cahil
asked Jan 4 at 22:42
Lucian cahilLucian cahil
11
New contributor
Lucian cahil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 22:42
Lucian cahilLucian cahil
11
asked Jan 4 at 22:42
Lucian cahilLucian cahil
11
11
New contributor
Lucian cahil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Lucian cahil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Lucian cahil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
4
Have you made any attempt to compute the answer yourself?
– Aditya Dua
Jan 4 at 22:45
1
I suggest you try working it out by hand for a simple case, such as $k=2$, $p=0.4$, and $n=5$. There are only $32$ possible outcomes of running the trial $5$ times. You can list each outcome with its probability of occurring, and then add up the probabilities for the outcomes that contain two consecutive successes. If you do this for a few simple cases, you might get some ideas for how to answer the general question.
– Steve Kass
Jan 4 at 23:13
1
This is a fairly difficult problem to solve in general - see here for solutions for given $k$ and $n$: stats.stackexchange.com/questions/21825/…
– Math1000
Jan 4 at 23:27
add a comment |
4
Have you made any attempt to compute the answer yourself?
– Aditya Dua
Jan 4 at 22:45
1
I suggest you try working it out by hand for a simple case, such as $k=2$, $p=0.4$, and $n=5$. There are only $32$ possible outcomes of running the trial $5$ times. You can list each outcome with its probability of occurring, and then add up the probabilities for the outcomes that contain two consecutive successes. If you do this for a few simple cases, you might get some ideas for how to answer the general question.
– Steve Kass
Jan 4 at 23:13
1
This is a fairly difficult problem to solve in general - see here for solutions for given $k$ and $n$: stats.stackexchange.com/questions/21825/…
– Math1000
Jan 4 at 23:27
4
4
Have you made any attempt to compute the answer yourself?
– Aditya Dua
Jan 4 at 22:45
Have you made any attempt to compute the answer yourself?
– Aditya Dua
Jan 4 at 22:45
1
1
I suggest you try working it out by hand for a simple case, such as $k=2$, $p=0.4$, and $n=5$. There are only $32$ possible outcomes of running the trial $5$ times. You can list each outcome with its probability of occurring, and then add up the probabilities for the outcomes that contain two consecutive successes. If you do this for a few simple cases, you might get some ideas for how to answer the general question.
– Steve Kass
Jan 4 at 23:13
I suggest you try working it out by hand for a simple case, such as $k=2$, $p=0.4$, and $n=5$. There are only $32$ possible outcomes of running the trial $5$ times. You can list each outcome with its probability of occurring, and then add up the probabilities for the outcomes that contain two consecutive successes. If you do this for a few simple cases, you might get some ideas for how to answer the general question.
– Steve Kass
Jan 4 at 23:13
1
1
This is a fairly difficult problem to solve in general - see here for solutions for given $k$ and $n$: stats.stackexchange.com/questions/21825/…
– Math1000
Jan 4 at 23:27
This is a fairly difficult problem to solve in general - see here for solutions for given $k$ and $n$: stats.stackexchange.com/questions/21825/…
– Math1000
Jan 4 at 23:27
add a comment |
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4
Have you made any attempt to compute the answer yourself?
– Aditya Dua
Jan 4 at 22:45
1
I suggest you try working it out by hand for a simple case, such as $k=2$, $p=0.4$, and $n=5$. There are only $32$ possible outcomes of running the trial $5$ times. You can list each outcome with its probability of occurring, and then add up the probabilities for the outcomes that contain two consecutive successes. If you do this for a few simple cases, you might get some ideas for how to answer the general question.
– Steve Kass
Jan 4 at 23:13
1
This is a fairly difficult problem to solve in general - see here for solutions for given $k$ and $n$: stats.stackexchange.com/questions/21825/…
– Math1000
Jan 4 at 23:27