# In a touring group of `n ` international travelers, what is the probability that at least two will stick with the tour group if the probability that a traveler independently ` `chooses to stay...

In a touring group of `n ` international travelers, what is the probability that at least two will stick with the tour group if the probability that a traveler independently ` `chooses to stay with the group is `p `?

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If the probability that a traveler stays with the tour group is defined as `p` and the eventuality of whether each traveler stays with the tour group is independent of whether the other travelers in the group stay with the tour group (not likely if there are people traveling together, but this is a simplifying assumption), then the number of travelers staying with the group is binomially distributed with mean `np` where `n` is the number of travelers in the group and `p` is the probability of a traveler *staying *with the group.

Now, note that the probability that at least two stay with the group is equal to 1 minus the probability that none stay with the group or only 1 stays with the group.

The probability that `x` stay with the group is given by the binomial formula:

`P(X=x) = ((n),(x))p^x(1-p)^(n-x)`

where

`((n),(x)) = (n!)/((n-x)!)` or "n choose x" (from Pascale's triangle)

So that the probability that none and only 1 stay with the group respectively is

`P(X=0) = (n!)/(n!)p^0(1-p)^n = (1-p)^n` ,

`P(X=1) = (n!)/((n-1)!) p^1(1-p)^(n-1) = np(1-p)^(n-1)`

Therefore the probability that at least 2 stay with the group is given by

`P(X>=2) = 1- (P(X=0) + P(X=1)) = 1 - (1-p)^n - np(1-p)^(n-1)`

If for example `p=0.2` and `n=10`, then

`P(X>=2) = 1- 0.8^10 - 10(0.2)(0.8^9) = 1 - 0.107 - 0.268 = 0.624` to 3sf

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