Probability to select all 3 male mouses from 10 selected at random In a cage there are 100 mouses from which 3 are male.
Compute the probability of selecting all 3 males from a group of 10 mouses selected at random.
I have this intuition:
$$
P(male)=0.03
$$
and number of all possibilities of selecting all 3 males mouses out of 10:
$$
C(10, 3) = \frac{10!}{3! \times 7!}
$$
Resulting:
$$
P(all\ 3\ males\ from\ 10) = 0.03^3 \times 0.97^7 \times C(10,3) 
$$
I feel there is something totally wrong with it but I have no idea what.
 A: How many ways can you choose a group of 10 mice without restrictions? 
$^{100}C_{10}$
How many ways can you choose a group of 10 mice including all three male mice? 
$^{3}C_{3}\,^{97}C_{7}=\,^{97}C_{7}$ i.e. choose all three males and seven other mice.
Hence the probabolity is $\frac{^{97}C_{7}}{^{100}C_{10}}$
Your calculation is more suited to independent events, where we'd put each mouse back after choosing it and then just see the genders of 10 mice chosen in this way.
A: You're doing a binomial distribution, but this is not a binomial distribution, for two reasons: you are not doing this in multiple trials, and the random variable is not how many trials it will take, but whether or not all three male mice will be drawn in one trial.
That said, this is a simple combinatorics problem. This will be a simple fraction $\frac{n}{d}$. The denominator will be the number of $10$ mouse combinations out of $100$ there are, and the numerator will be the total number of $10$ mouse combinations out of $100$ that contain the three male mice.
A: $$P=\frac {\binom {97} {7}} {\binom {100} {10}}=\frac{\frac {97!}{90!7!}}{\frac{100!}{90!10!}}=\frac{8\cdot9\cdot10}{98\cdot99\cdot100}=0.00074212$$
