Odds of run of successes in Bernoulli trial
0
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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
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add a comment |
0
$begingroup$
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
$endgroup$
4
$begingroup$
Have you made any attempt to compute the answer yourself?
$endgroup$
– Aditya Dua
Jan 4 at 22:45
1
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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.
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– Steve Kass
Jan 4 at 23:13
1
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This is a fairly difficult problem to solve in general - see here for solutions for given $k$ and $n$: stats.stackexchange.com/questions/21825/…
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– Math1000
Jan 4 at 23:27
add a comment |
0
0
0
$begingroup$
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
$endgroup$
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
asked Jan 4 at 22:42
Lucian cahilLucian cahil
11
edited Jan 4 at 23:03
Lucian cahil
asked Jan 4 at 22:42
Lucian cahilLucian cahil
11
asked Jan 4 at 22:42
Lucian cahilLucian cahil
11
asked Jan 4 at 22:42
Lucian cahilLucian cahil
11
11
4
$begingroup$
Have you made any attempt to compute the answer yourself?
$endgroup$
– Aditya Dua
Jan 4 at 22:45
1
$begingroup$
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.
$endgroup$
– Steve Kass
Jan 4 at 23:13
1
$begingroup$
This is a fairly difficult problem to solve in general - see here for solutions for given $k$ and $n$: stats.stackexchange.com/questions/21825/…
$endgroup$
– Math1000
Jan 4 at 23:27
add a comment |
4
$begingroup$
Have you made any attempt to compute the answer yourself?
$endgroup$
– Aditya Dua
Jan 4 at 22:45
1
$begingroup$
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.
$endgroup$
– Steve Kass
Jan 4 at 23:13
1
$begingroup$
This is a fairly difficult problem to solve in general - see here for solutions for given $k$ and $n$: stats.stackexchange.com/questions/21825/…
$endgroup$
– Math1000
Jan 4 at 23:27
4
4
$begingroup$
Have you made any attempt to compute the answer yourself?
$endgroup$
– Aditya Dua
Jan 4 at 22:45
$begingroup$
Have you made any attempt to compute the answer yourself?
$endgroup$
– Aditya Dua
Jan 4 at 22:45
1
1
$begingroup$
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.
$endgroup$
– Steve Kass
Jan 4 at 23:13
$begingroup$
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.
$endgroup$
– Steve Kass
Jan 4 at 23:13
1
1
$begingroup$
This is a fairly difficult problem to solve in general - see here for solutions for given $k$ and $n$: stats.stackexchange.com/questions/21825/…
$endgroup$
– Math1000
Jan 4 at 23:27
$begingroup$
This is a fairly difficult problem to solve in general - see here for solutions for given $k$ and $n$: stats.stackexchange.com/questions/21825/…
$endgroup$
– Math1000
Jan 4 at 23:27
add a comment |
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4
$begingroup$
Have you made any attempt to compute the answer yourself?
$endgroup$
– Aditya Dua
Jan 4 at 22:45
1
$begingroup$
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.
$endgroup$
– Steve Kass
Jan 4 at 23:13
1
$begingroup$
This is a fairly difficult problem to solve in general - see here for solutions for given $k$ and $n$: stats.stackexchange.com/questions/21825/…
$endgroup$
– Math1000
Jan 4 at 23:27