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Got a two part question given to me (I used binomial Distribution to solve)

If the probability that an individual moves outside of his or her country of residence in a given year is $0.12$, what is the probability that less than $3$ out of a sample of $15$ move outside the country?

I approached it like this:

Less than three means $0$, $1$, or $2$. The calculation for $0$ is: $$ \frac{15!}{0!(15-0)!}0.12^0(1-.012)^15 $$

The calculation for $1$ is: $$ \frac{15!}{1!(15-1)!}0.12^1(1-.012)^14 $$

The calculation for $2$ is: $$ \frac{15!}{2!(15-2)!}0.12^2(1-.012)^13 $$

Now:

$$ P(X=0) \approx 0.147\\ P(X=1) \approx 0.300\\ P(X=2) \approx 0.287\\ P(X=0)+P(X=1)+P(X=2) \approx .3+.287+.147 \approx .734 $$ I am assuming that unless I botched the math then it should be correct.

The second question however asks

"At least two move out of the country"

I am struggling to see the difference here, I know they are different however my question is, how do I go about solving the second one? Do I just use $P(X=2)$ as my answer?

Any help would be appreciated,

Regards

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1 Answer

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At least two means $2,3,4,\ldots,13,14$, or $15$; you could work it the same way, but you’d have $14$ terms to compute and add up. It’s easier to notice that at least two covers all possibilities except $0$ and $1$. Calculate the probability that fewer than $2$ move, and subtract that from $1$.

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