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dariober
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It seems to me that the binomial distribution is more appropriate. This is how I see it:

A given section has a $p = 1/300$ chance of being hit by a strike. There have been 213 strikes so $N = 213$. The probability of being hit by exactly 1 strike is then (in R code):

dbinom(1, size= 213, prob= 1/300)
0.3498

by exactly two strikes:

dbinom(2, size= 213, prob= 1/300)
0.124

and so on for 3, 4, ..., 213. So the probability of being hit by 1 or more strikes is the sum of the individual probabilities, about 51%:

sum(dbinom(1:213, size= 213, prob= 1/300))
0.5089

As noted by @Wolfmercury in comment, one can simply use $1 - binom(k= 0, N, p)$:

1 - dbinom(0, size= 213, prob= 1/300)
0.5089385

However, I thought the long-winded approach above makes the reasoning more explicit.

It seems to me that the binomial distribution is more appropriate. This is how I see it:

A given section has a $p = 1/300$ chance of being hit by a strike. There have been 213 strikes so $N = 213$. The probability of being hit by exactly 1 strike is then (in R code):

dbinom(1, size= 213, prob= 1/300)
0.3498

by exactly two strikes:

dbinom(2, size= 213, prob= 1/300)
0.124

and so on for 3, 4, ..., 213. So the probability of being hit by 1 or more strikes is the sum of the individual probabilities, about 51%:

sum(dbinom(1:213, size= 213, prob= 1/300))
0.5089

It seems to me that the binomial distribution is more appropriate. This is how I see it:

A given section has a $p = 1/300$ chance of being hit by a strike. There have been 213 strikes so $N = 213$. The probability of being hit by exactly 1 strike is then (in R code):

dbinom(1, size= 213, prob= 1/300)
0.3498

by exactly two strikes:

dbinom(2, size= 213, prob= 1/300)
0.124

and so on for 3, 4, ..., 213. So the probability of being hit by 1 or more strikes is the sum of the individual probabilities, about 51%:

sum(dbinom(1:213, size= 213, prob= 1/300))
0.5089

As noted by @Wolfmercury in comment, one can simply use $1 - binom(k= 0, N, p)$:

1 - dbinom(0, size= 213, prob= 1/300)
0.5089385

However, I thought the long-winded approach above makes the reasoning more explicit.

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dariober
  • 5.3k
  • 19
  • 23

It seems to me that the binomial distribution is more appropriate. This is how I see it:

A given section has a $p = 1/300$ chance of being hit by a strike. There have been 213 strikes so $N = 213$. The probability of being hit by exactly 1 strike is then (in R code):

dbinom(1, size= 213, prob= 1/300)
0.3498

by exactly two strikes:

dbinom(2, size= 213, prob= 1/300)
0.124

and so on for 3, 4, ..., 213. So the probability of being hit by 1 or more strikes is the sum of the individual probabilities, about 51%:

sum(dbinom(1:213, size= 213, prob= 1/300))
0.5089