What is the probability of having 10 or more crashes in the first 1000 rides?
$begingroup$
You have recently invested in an electric skateboard. You always wear a helmet and padding to protect yourself (of course), but are worried that at some point you'll take a tumble and get scratched up. After assessing your skating ability, you estimate your probability of a crash on any single ride to be 0.005.
I got this question while studying geometric and binomial distribution.
My current idea is
1 - (P(x=0) + P(x=1) + P(x=2) + .... + P(x=9))
where x is the number of crashes, and get each probability by binomial distribution's formula.
I came up with this equation as I thought having 10 or more crashes means same as not having 9 or less crashes.
Is my approach correct? Also, if it's correct, is there any way to simplify the calculation?
probability statistics binomial-distribution
asked Jan 3 at 1:16
Daniel KimDaniel Kim
111
$endgroup$
add a comment |
$begingroup$
You have recently invested in an electric skateboard. You always wear a helmet and padding to protect yourself (of course), but are worried that at some point you'll take a tumble and get scratched up. After assessing your skating ability, you estimate your probability of a crash on any single ride to be 0.005.
I got this question while studying geometric and binomial distribution.
My current idea is
1 - (P(x=0) + P(x=1) + P(x=2) + .... + P(x=9))
where x is the number of crashes, and get each probability by binomial distribution's formula.
I came up with this equation as I thought having 10 or more crashes means same as not having 9 or less crashes.
Is my approach correct? Also, if it's correct, is there any way to simplify the calculation?
probability statistics binomial-distribution
asked Jan 3 at 1:16
Daniel KimDaniel Kim
111
$endgroup$
1
$begingroup$
I think your approach is correct. I think one way to "simplify" the calculation is by using the Poisson approximation. If $X =$ number of crashes then $X$ is approximately Poisson with $lambda = 1000(.005)$ and then you can do $P(X ge 10) = 1-P(X=0) -P(X=1) cdots$
$endgroup$
– HJ_beginner
Jan 3 at 1:35
add a comment |
$begingroup$
You have recently invested in an electric skateboard. You always wear a helmet and padding to protect yourself (of course), but are worried that at some point you'll take a tumble and get scratched up. After assessing your skating ability, you estimate your probability of a crash on any single ride to be 0.005.
I got this question while studying geometric and binomial distribution.
My current idea is
1 - (P(x=0) + P(x=1) + P(x=2) + .... + P(x=9))
where x is the number of crashes, and get each probability by binomial distribution's formula.
I came up with this equation as I thought having 10 or more crashes means same as not having 9 or less crashes.
Is my approach correct? Also, if it's correct, is there any way to simplify the calculation?
probability statistics binomial-distribution
asked Jan 3 at 1:16
Daniel KimDaniel Kim
111
$endgroup$
You have recently invested in an electric skateboard. You always wear a helmet and padding to protect yourself (of course), but are worried that at some point you'll take a tumble and get scratched up. After assessing your skating ability, you estimate your probability of a crash on any single ride to be 0.005.
I got this question while studying geometric and binomial distribution.
My current idea is
1 - (P(x=0) + P(x=1) + P(x=2) + .... + P(x=9))
where x is the number of crashes, and get each probability by binomial distribution's formula.
I came up with this equation as I thought having 10 or more crashes means same as not having 9 or less crashes.
Is my approach correct? Also, if it's correct, is there any way to simplify the calculation?
probability statistics binomial-distribution
probability statistics binomial-distribution
asked Jan 3 at 1:16
Daniel KimDaniel Kim
111
asked Jan 3 at 1:16
Daniel KimDaniel Kim
111
asked Jan 3 at 1:16
Daniel KimDaniel Kim
111
asked Jan 3 at 1:16
Daniel KimDaniel Kim
111
asked Jan 3 at 1:16
Daniel KimDaniel Kim
111
111
1
$begingroup$
I think your approach is correct. I think one way to "simplify" the calculation is by using the Poisson approximation. If $X =$ number of crashes then $X$ is approximately Poisson with $lambda = 1000(.005)$ and then you can do $P(X ge 10) = 1-P(X=0) -P(X=1) cdots$
$endgroup$
– HJ_beginner
Jan 3 at 1:35
add a comment |
1
$begingroup$
I think your approach is correct. I think one way to "simplify" the calculation is by using the Poisson approximation. If $X =$ number of crashes then $X$ is approximately Poisson with $lambda = 1000(.005)$ and then you can do $P(X ge 10) = 1-P(X=0) -P(X=1) cdots$
$endgroup$
– HJ_beginner
Jan 3 at 1:35
1
1
$begingroup$
I think your approach is correct. I think one way to "simplify" the calculation is by using the Poisson approximation. If $X =$ number of crashes then $X$ is approximately Poisson with $lambda = 1000(.005)$ and then you can do $P(X ge 10) = 1-P(X=0) -P(X=1) cdots$
$endgroup$
– HJ_beginner
Jan 3 at 1:35
$begingroup$
I think your approach is correct. I think one way to "simplify" the calculation is by using the Poisson approximation. If $X =$ number of crashes then $X$ is approximately Poisson with $lambda = 1000(.005)$ and then you can do $P(X ge 10) = 1-P(X=0) -P(X=1) cdots$
$endgroup$
– HJ_beginner
Jan 3 at 1:35
add a comment |
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1
$begingroup$
I think your approach is correct. I think one way to "simplify" the calculation is by using the Poisson approximation. If $X =$ number of crashes then $X$ is approximately Poisson with $lambda = 1000(.005)$ and then you can do $P(X ge 10) = 1-P(X=0) -P(X=1) cdots$
$endgroup$
– HJ_beginner
Jan 3 at 1:35