What is the cumulative binomial distribution, on the probability of “at least one”

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I know that the probability mass function of the binomial distribution is

$f(k,n,p) = Pr(k;n,p) = Pr(X = k) = nchoose kp^k(1-p)^n-k$

This functions returns the probability of getting exactly k successes within n trials, with p being the probability of success for each trial.

The cumulative function is

$F(k;n,p) = Pr(X le k) = sum_i=0^lfloor k rfloor nchoose ip^i(1-p)^n-i$

where $displaystyle scriptstyle lfloor krfloor ,$ is the "floor" under k, i.e. the greatest integer less than or equal to k. But this cumulative function gives, as I understood, the probability of having the number of successes between 0 and k.

How to get a cumulative function which gives me, at least one success, that is, the probability of having the number of successes between 1 and n.

For example, someone throws an even dice. After throwing 1000 times, what's the probability of getting at least once, the number 6?

probability-distributions binomial-distribution

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asked Jul 18 at 19:57

João Pimentel Ferreira

1012

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why do you use the floor function? In my opinion it is not necessary, since k is an integer
  – callculus
  Jul 18 at 20:10
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

I know that the probability mass function of the binomial distribution is

$f(k,n,p) = Pr(k;n,p) = Pr(X = k) = nchoose kp^k(1-p)^n-k$

This functions returns the probability of getting exactly k successes within n trials, with p being the probability of success for each trial.

The cumulative function is

$F(k;n,p) = Pr(X le k) = sum_i=0^lfloor k rfloor nchoose ip^i(1-p)^n-i$

where $displaystyle scriptstyle lfloor krfloor ,$ is the "floor" under k, i.e. the greatest integer less than or equal to k. But this cumulative function gives, as I understood, the probability of having the number of successes between 0 and k.

How to get a cumulative function which gives me, at least one success, that is, the probability of having the number of successes between 1 and n.

For example, someone throws an even dice. After throwing 1000 times, what's the probability of getting at least once, the number 6?

probability-distributions binomial-distribution

share|cite|improve this question

asked Jul 18 at 19:57

João Pimentel Ferreira

1012

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why do you use the floor function? In my opinion it is not necessary, since k is an integer
  – callculus
  Jul 18 at 20:10
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

I know that the probability mass function of the binomial distribution is

$f(k,n,p) = Pr(k;n,p) = Pr(X = k) = nchoose kp^k(1-p)^n-k$

This functions returns the probability of getting exactly k successes within n trials, with p being the probability of success for each trial.

The cumulative function is

$F(k;n,p) = Pr(X le k) = sum_i=0^lfloor k rfloor nchoose ip^i(1-p)^n-i$

where $displaystyle scriptstyle lfloor krfloor ,$ is the "floor" under k, i.e. the greatest integer less than or equal to k. But this cumulative function gives, as I understood, the probability of having the number of successes between 0 and k.

How to get a cumulative function which gives me, at least one success, that is, the probability of having the number of successes between 1 and n.

For example, someone throws an even dice. After throwing 1000 times, what's the probability of getting at least once, the number 6?

probability-distributions binomial-distribution

share|cite|improve this question

asked Jul 18 at 19:57

João Pimentel Ferreira

1012

I know that the probability mass function of the binomial distribution is

$f(k,n,p) = Pr(k;n,p) = Pr(X = k) = nchoose kp^k(1-p)^n-k$

This functions returns the probability of getting exactly k successes within n trials, with p being the probability of success for each trial.

The cumulative function is

$F(k;n,p) = Pr(X le k) = sum_i=0^lfloor k rfloor nchoose ip^i(1-p)^n-i$

where $displaystyle scriptstyle lfloor krfloor ,$ is the "floor" under k, i.e. the greatest integer less than or equal to k. But this cumulative function gives, as I understood, the probability of having the number of successes between 0 and k.

How to get a cumulative function which gives me, at least one success, that is, the probability of having the number of successes between 1 and n.

For example, someone throws an even dice. After throwing 1000 times, what's the probability of getting at least once, the number 6?

probability-distributions binomial-distribution

share|cite|improve this question

asked Jul 18 at 19:57

João Pimentel Ferreira

1012

share|cite|improve this question

share|cite|improve this question

asked Jul 18 at 19:57

João Pimentel Ferreira

1012

asked Jul 18 at 19:57

João Pimentel Ferreira

1012

asked Jul 18 at 19:57

João Pimentel Ferreira

1012

1012

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why do you use the floor function? In my opinion it is not necessary, since k is an integer
  – callculus
  Jul 18 at 20:10
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why do you use the floor function? In my opinion it is not necessary, since k is an integer
  – callculus
  Jul 18 at 20:10
  

  
  
  
  

Why do you use the floor function? In my opinion it is not necessary, since k is an integer
– callculus
Jul 18 at 20:10

Why do you use the floor function? In my opinion it is not necessary, since k is an integer
– callculus
Jul 18 at 20:10

add a comment | 

1 Answer
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accepted

Hint

$$P(Xgeq k) = 1-P(X lt k)$$

Solution

$$P(Xgeq 1) = 1-P(Xlt 1) = 1- 1000choose 0left(1over 6right)^0left(1-1over 6right)^999 = 1-(5over 6)^999approx 1$$ This "trick" is pretty obvious as it stands and it's pretty common to use it when dealing with calculating the cumulative probability of at least $m$ successes of big data sets when $mll n$

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edited Jul 18 at 20:34

answered Jul 18 at 20:23

Davide Morgante

1,830220

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Hint

$$P(Xgeq k) = 1-P(X lt k)$$

Solution

$$P(Xgeq 1) = 1-P(Xlt 1) = 1- 1000choose 0left(1over 6right)^0left(1-1over 6right)^999 = 1-(5over 6)^999approx 1$$ This "trick" is pretty obvious as it stands and it's pretty common to use it when dealing with calculating the cumulative probability of at least $m$ successes of big data sets when $mll n$

share|cite|improve this answer

edited Jul 18 at 20:34

answered Jul 18 at 20:23

Davide Morgante

1,830220

add a comment | 

up vote
1
down vote



accepted

Hint

$$P(Xgeq k) = 1-P(X lt k)$$

Solution

$$P(Xgeq 1) = 1-P(Xlt 1) = 1- 1000choose 0left(1over 6right)^0left(1-1over 6right)^999 = 1-(5over 6)^999approx 1$$ This "trick" is pretty obvious as it stands and it's pretty common to use it when dealing with calculating the cumulative probability of at least $m$ successes of big data sets when $mll n$

share|cite|improve this answer

edited Jul 18 at 20:34

answered Jul 18 at 20:23

Davide Morgante

1,830220

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Hint

$$P(Xgeq k) = 1-P(X lt k)$$

Solution

$$P(Xgeq 1) = 1-P(Xlt 1) = 1- 1000choose 0left(1over 6right)^0left(1-1over 6right)^999 = 1-(5over 6)^999approx 1$$ This "trick" is pretty obvious as it stands and it's pretty common to use it when dealing with calculating the cumulative probability of at least $m$ successes of big data sets when $mll n$

share|cite|improve this answer

edited Jul 18 at 20:34

answered Jul 18 at 20:23

Davide Morgante

1,830220

Hint

$$P(Xgeq k) = 1-P(X lt k)$$

Solution

$$P(Xgeq 1) = 1-P(Xlt 1) = 1- 1000choose 0left(1over 6right)^0left(1-1over 6right)^999 = 1-(5over 6)^999approx 1$$ This "trick" is pretty obvious as it stands and it's pretty common to use it when dealing with calculating the cumulative probability of at least $m$ successes of big data sets when $mll n$

share|cite|improve this answer

edited Jul 18 at 20:34

answered Jul 18 at 20:23

Davide Morgante

1,830220

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 20:34

edited Jul 18 at 20:34

edited Jul 18 at 20:34

answered Jul 18 at 20:23

Davide Morgante

1,830220

answered Jul 18 at 20:23

Davide Morgante

1,830220

answered Jul 18 at 20:23

Davide Morgante

1,830220

1,830220

add a comment | 

add a comment | 

 

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