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1

$begingroup$

If I have a known probability of an event occurring, 1% chance, and I need the event to occur a number of times, 120 times, about how many times would I have to repeat the event before I can expect it to happen that number of times?

probability

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DylanDylan

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$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  How sure do you want to be that the event will have happened that number of times?
  $endgroup$
  – Jake Westfall
  7 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

If I have a known probability of an event occurring, 1% chance, and I need the event to occur a number of times, 120 times, about how many times would I have to repeat the event before I can expect it to happen that number of times?

probability

share|cite|improve this question

asked 9 hours ago

DylanDylan

61

New contributor

Dylan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  How sure do you want to be that the event will have happened that number of times?
  $endgroup$
  – Jake Westfall
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

If I have a known probability of an event occurring, 1% chance, and I need the event to occur a number of times, 120 times, about how many times would I have to repeat the event before I can expect it to happen that number of times?

probability

share|cite|improve this question

asked 9 hours ago

DylanDylan

61

New contributor

Dylan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked 9 hours ago

DylanDylan

61

New contributor

Dylan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 9 hours ago

DylanDylan

61

New contributor

Dylan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 9 hours ago

DylanDylan

61

New contributor

Dylan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 9 hours ago

DylanDylan

61

2 Answers
2

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3

$begingroup$

Consider a sequence of $n$ independent trials with success probability $p$. Let $X$ be the number of successes out of the $n$ trials. Then $X$ has a Binomial distribution with parameters $n$ and $p$. The expected value of a Binomial rv is $E(X) = np$. A simple approach is to set this equal to $120$ and solve for $n$. Since $p=0.01$, we have $n(0.01) = 120$ which means that $n=12,000$ trials are expected to obtain $120$ successes.

---

Alternatively, here is a related approach that gives the number of trials needed to observe $r=120$ successes with some probability $gamma$ (i.e. $gamma=0.95$).

Consider a sequence of independent trials with success probability $p$. Let $X$ be the number of trials required to observe $r$ successes. Then $X$ has a Negative Binomial distribution with parameters $r$ and $p$. In your case, $X sim textNegative-Binomial(120, 0.01)$, and you want to find $x$ such that $$P(X leq x) = gamma.$$

Although the Negative Binomial distribution has no closed form quantile function, this $x$ can be solved for easily. For instance, the answer can be obtained in R by typing: qnbinom(.95, 120, .01). The answer $x=13728$ indicates that $13,728$ trials are required to have a 95% chance of observing $120$ (or more) successes.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

knrumseyknrumsey

1,760416

$endgroup$

add a comment | 

1

$begingroup$

First, I'm going to assume that the experiments are independent since you said the probably of a success is always 1%. The keyword in your question is "expected," which means we'll be looking for a mean or expected value.

If you are interested in the number of trials $X$ (with common probability of success $p$), needed to obtain $r$ successes, then you can model this as a negative binomial random variable with probability mass function:

begineqnarray*
f_X(x|r,p) & = & x-1 choose r-1p^r(1-p)^x-r
endeqnarray*

for $x=r,r+1,...,$

The expected value of the negative binomial is well known as:

begineqnarray*
E(X) & = & fracrp
endeqnarray*

In your case, $p=0.01$ and $r=120$. So the expected number of independent trials of your experiment (times) needed to obtain 120 successes is $120/0.01=12,000$

share|cite|improve this answer

edited 5 hours ago

answered 8 hours ago

StatsStudentStatsStudent

6,51432245

$endgroup$

add a comment | 

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3

$begingroup$

Consider a sequence of $n$ independent trials with success probability $p$. Let $X$ be the number of successes out of the $n$ trials. Then $X$ has a Binomial distribution with parameters $n$ and $p$. The expected value of a Binomial rv is $E(X) = np$. A simple approach is to set this equal to $120$ and solve for $n$. Since $p=0.01$, we have $n(0.01) = 120$ which means that $n=12,000$ trials are expected to obtain $120$ successes.

---

Alternatively, here is a related approach that gives the number of trials needed to observe $r=120$ successes with some probability $gamma$ (i.e. $gamma=0.95$).

Consider a sequence of independent trials with success probability $p$. Let $X$ be the number of trials required to observe $r$ successes. Then $X$ has a Negative Binomial distribution with parameters $r$ and $p$. In your case, $X sim textNegative-Binomial(120, 0.01)$, and you want to find $x$ such that $$P(X leq x) = gamma.$$

Although the Negative Binomial distribution has no closed form quantile function, this $x$ can be solved for easily. For instance, the answer can be obtained in R by typing: qnbinom(.95, 120, .01). The answer $x=13728$ indicates that $13,728$ trials are required to have a 95% chance of observing $120$ (or more) successes.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

knrumseyknrumsey

1,760416

$endgroup$

add a comment | 

3

$begingroup$

Consider a sequence of $n$ independent trials with success probability $p$. Let $X$ be the number of successes out of the $n$ trials. Then $X$ has a Binomial distribution with parameters $n$ and $p$. The expected value of a Binomial rv is $E(X) = np$. A simple approach is to set this equal to $120$ and solve for $n$. Since $p=0.01$, we have $n(0.01) = 120$ which means that $n=12,000$ trials are expected to obtain $120$ successes.

---

Alternatively, here is a related approach that gives the number of trials needed to observe $r=120$ successes with some probability $gamma$ (i.e. $gamma=0.95$).

Consider a sequence of independent trials with success probability $p$. Let $X$ be the number of trials required to observe $r$ successes. Then $X$ has a Negative Binomial distribution with parameters $r$ and $p$. In your case, $X sim textNegative-Binomial(120, 0.01)$, and you want to find $x$ such that $$P(X leq x) = gamma.$$

Although the Negative Binomial distribution has no closed form quantile function, this $x$ can be solved for easily. For instance, the answer can be obtained in R by typing: qnbinom(.95, 120, .01). The answer $x=13728$ indicates that $13,728$ trials are required to have a 95% chance of observing $120$ (or more) successes.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

knrumseyknrumsey

1,760416

$endgroup$

add a comment | 

$begingroup$

Consider a sequence of $n$ independent trials with success probability $p$. Let $X$ be the number of successes out of the $n$ trials. Then $X$ has a Binomial distribution with parameters $n$ and $p$. The expected value of a Binomial rv is $E(X) = np$. A simple approach is to set this equal to $120$ and solve for $n$. Since $p=0.01$, we have $n(0.01) = 120$ which means that $n=12,000$ trials are expected to obtain $120$ successes.

---

Alternatively, here is a related approach that gives the number of trials needed to observe $r=120$ successes with some probability $gamma$ (i.e. $gamma=0.95$).

Consider a sequence of independent trials with success probability $p$. Let $X$ be the number of trials required to observe $r$ successes. Then $X$ has a Negative Binomial distribution with parameters $r$ and $p$. In your case, $X sim textNegative-Binomial(120, 0.01)$, and you want to find $x$ such that $$P(X leq x) = gamma.$$

Although the Negative Binomial distribution has no closed form quantile function, this $x$ can be solved for easily. For instance, the answer can be obtained in R by typing: qnbinom(.95, 120, .01). The answer $x=13728$ indicates that $13,728$ trials are required to have a 95% chance of observing $120$ (or more) successes.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

knrumseyknrumsey

1,760416

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

knrumseyknrumsey

1,760416

answered 8 hours ago

knrumseyknrumsey

1,760416

answered 8 hours ago

knrumseyknrumsey

1,760416

1

$begingroup$

First, I'm going to assume that the experiments are independent since you said the probably of a success is always 1%. The keyword in your question is "expected," which means we'll be looking for a mean or expected value.

If you are interested in the number of trials $X$ (with common probability of success $p$), needed to obtain $r$ successes, then you can model this as a negative binomial random variable with probability mass function:

begineqnarray*
f_X(x|r,p) & = & x-1 choose r-1p^r(1-p)^x-r
endeqnarray*

for $x=r,r+1,...,$

The expected value of the negative binomial is well known as:

begineqnarray*
E(X) & = & fracrp
endeqnarray*

In your case, $p=0.01$ and $r=120$. So the expected number of independent trials of your experiment (times) needed to obtain 120 successes is $120/0.01=12,000$

share|cite|improve this answer

edited 5 hours ago

answered 8 hours ago

StatsStudentStatsStudent

6,51432245

$endgroup$

add a comment | 

1

$begingroup$

First, I'm going to assume that the experiments are independent since you said the probably of a success is always 1%. The keyword in your question is "expected," which means we'll be looking for a mean or expected value.

If you are interested in the number of trials $X$ (with common probability of success $p$), needed to obtain $r$ successes, then you can model this as a negative binomial random variable with probability mass function:

begineqnarray*
f_X(x|r,p) & = & x-1 choose r-1p^r(1-p)^x-r
endeqnarray*

for $x=r,r+1,...,$

The expected value of the negative binomial is well known as:

begineqnarray*
E(X) & = & fracrp
endeqnarray*

In your case, $p=0.01$ and $r=120$. So the expected number of independent trials of your experiment (times) needed to obtain 120 successes is $120/0.01=12,000$

share|cite|improve this answer

edited 5 hours ago

answered 8 hours ago

StatsStudentStatsStudent

6,51432245

$endgroup$

add a comment | 

$begingroup$

First, I'm going to assume that the experiments are independent since you said the probably of a success is always 1%. The keyword in your question is "expected," which means we'll be looking for a mean or expected value.

If you are interested in the number of trials $X$ (with common probability of success $p$), needed to obtain $r$ successes, then you can model this as a negative binomial random variable with probability mass function:

begineqnarray*
f_X(x|r,p) & = & x-1 choose r-1p^r(1-p)^x-r
endeqnarray*

for $x=r,r+1,...,$

The expected value of the negative binomial is well known as:

begineqnarray*
E(X) & = & fracrp
endeqnarray*

In your case, $p=0.01$ and $r=120$. So the expected number of independent trials of your experiment (times) needed to obtain 120 successes is $120/0.01=12,000$

share|cite|improve this answer

edited 5 hours ago

answered 8 hours ago

StatsStudentStatsStudent

6,51432245

$endgroup$

share|cite|improve this answer

edited 5 hours ago

answered 8 hours ago

StatsStudentStatsStudent

6,51432245

answered 8 hours ago

StatsStudentStatsStudent

6,51432245

answered 8 hours ago

StatsStudentStatsStudent

6,51432245