Probability distribution of k consecutive successes with n maximum trials
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Let $X$ be a random variable that represents the number of trials of a given experiment. The outcome of a single trial is a Bernoulli random variable, with probability of success $p$, and trials are independent.
The maximum number of trials is $n$, but if there are $k<n$ consecutive successes the experiment ends.
What is the probability distribution?
probability probability-distributions random-variables
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up vote
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Let $X$ be a random variable that represents the number of trials of a given experiment. The outcome of a single trial is a Bernoulli random variable, with probability of success $p$, and trials are independent.
The maximum number of trials is $n$, but if there are $k<n$ consecutive successes the experiment ends.
What is the probability distribution?
probability probability-distributions random-variables
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X$ be a random variable that represents the number of trials of a given experiment. The outcome of a single trial is a Bernoulli random variable, with probability of success $p$, and trials are independent.
The maximum number of trials is $n$, but if there are $k<n$ consecutive successes the experiment ends.
What is the probability distribution?
probability probability-distributions random-variables
Let $X$ be a random variable that represents the number of trials of a given experiment. The outcome of a single trial is a Bernoulli random variable, with probability of success $p$, and trials are independent.
The maximum number of trials is $n$, but if there are $k<n$ consecutive successes the experiment ends.
What is the probability distribution?
probability probability-distributions random-variables
asked Aug 17 '14 at 16:54
vladimirm
605514
605514
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Let $a_n$ be the required probability.
Then $a_n+1 = a_n+(1-a_n-k)p^k(1-p)$
Where $a_0=a_1=a_2=...=a_k-1=0,a_k=p^k$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $a_n$ be the required probability.
Then $a_n+1 = a_n+(1-a_n-k)p^k(1-p)$
Where $a_0=a_1=a_2=...=a_k-1=0,a_k=p^k$
add a comment |Â
up vote
0
down vote
Let $a_n$ be the required probability.
Then $a_n+1 = a_n+(1-a_n-k)p^k(1-p)$
Where $a_0=a_1=a_2=...=a_k-1=0,a_k=p^k$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $a_n$ be the required probability.
Then $a_n+1 = a_n+(1-a_n-k)p^k(1-p)$
Where $a_0=a_1=a_2=...=a_k-1=0,a_k=p^k$
Let $a_n$ be the required probability.
Then $a_n+1 = a_n+(1-a_n-k)p^k(1-p)$
Where $a_0=a_1=a_2=...=a_k-1=0,a_k=p^k$
edited Aug 17 at 8:25
answered Jul 17 at 4:46
Makar
566116
566116
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