If there are $5,000,000$ couples in a city, and the probability that a couple matches a specific description is $1\over12,000,000$...? If there are $5,000,000$ couples in a city, and the probability that a couple matches a specific description is $1\over 12,000,000$, what are the chances that there are two couples that match the specific description given that there is at least one couple that matches the description?
I guess I'm supposed to use conditional probability, but I'm not sure how.
 A: Let the random variable $X$ be the number of matching couples. You're then looking for (I think, it is not quite clear from the problem statement)
$$ P(X \ge 2 \mid P\ge 1) $$
which clearly equals
$$ 1 - P(X=1 \mid X\ge 1) = 1 - \frac{P(X=1)}{P(X\ge 1)} $$
Using the binomial distribution, we get
$$ P(X=1) = \binom{500000}{1}\frac{1}{12000000} \bigl(\tfrac{11999999}{12000000}\bigr)^{4999999} $$
To a very good approximation this equals
$$ \tfrac{5}{12} e^{-4999999/12000000} \approx \tfrac5{12} e^{-5/12} $$
Similarly,
$$ P(X\ge 1) = 1 - P(X=0) = 1 - \bigl(\tfrac{11999999}{12000000}\bigr)^{5000000} \approx 1 - e^{-5/12} $$
Putting these two together,
$$ P(X\ge 2\mid X\ge 1) =
1 - \frac{\frac5{12} e^{-5/12}}{1-e^{-5/12}} =
1 - \frac{5/12}{e^{5/12}-1} \approx 0.1939 $$
A: $$\newcommand{\C}[2]{^{#1}C_{#2}}$$
Use 

Let p be the probability an event will happen,
  $$$$q be it not happening,
  $$$$Given n trials have taken place
  $$$$Probability that event occurred r times is
  $$P(n,r)=\C{n}{r} (p)^r (q)^{n-r}$$

So 

probability of atleast 1 couple= 1- probability of no couple$$$$

$$1-\C{5,000,000}{0}(\frac{1}{12,000,000})^0(1-\frac{1}{12,000,000})^{5,000,000}$$$$$$
Using binomial approximation,
$$1-(1-\frac{5,000,000}{12,000,000})$$$$$$
$$\frac{5}{12}$$$$$$

Probability of two couples,

$$\C{5,000,000}{2}(\frac{1}{12,000,000})^2(1-\frac{1}{12,000,000})^{4,999,998}$$$$$$
$$\frac{5,000,000×4,999,999}{2}(\frac{1}{12×10^6})^2(1-\frac{4,999,998}{12,000,000})$$$$$$
Approximately
$$\frac{25×10^{12}}{2}(\frac{1}{144×10^{12}})(1-\frac{5}{12})$$$$$$
$$\frac{25×7}{2×144×12}$$

Now using conditional probability,

$$P=\frac{\frac{25×7}{2×144×12}}{\frac{5}{12}}$$
$$P=\frac{5×7}{288}$$
