Probability of at least two being grey Say we have $11$ grey and $15$ white mice, so $26$ in total in a container we can't see. We want to take $5$ of them home. What is the probability of at least two of them being grey?


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*2 grey: The probability of the first being grey is $\frac{11}{26}$ the second grey is $\frac{10}{25}$, the last three white $\frac{15\cdot 14 \cdot 13}{24\cdot 23\cdot 22}$.

*3 grey: $\frac{11\cdot 10 \cdot 9 \cdot 15\cdot 14\cdot 13}{26\cdot 25\cdot 24\cdot 23\cdot 22}$

*4 grey: $\frac{11\cdot 10 \cdot 9 \cdot 8 \cdot 15\cdot 14}{26\cdot 25\cdot 24\cdot 23\cdot 22}$

*5 grey:  $\frac{11\cdot 10 \cdot 9 \cdot 8 \cdot 7\cdot 15}{26\cdot 25\cdot 24\cdot 23\cdot 22}$ 


And the probability of at least $2$ grey is the sum of these?

The probability that at least two are grey is one minus the probability that at most one is grey.

The probability that none are grey is : $\dfrac{15}{26}\dfrac{14}{25}\dfrac{13}{24}\dfrac{12}{23}\dfrac{11}{22}$ or in short $\left.\dbinom {15}5\middle/\dbinom{26}5\right.$
The probability that one is grey is : $\dfrac{11}{26}\dfrac{15}{25}\dfrac{14}{24}\dfrac{13}{23}\dfrac{12}{22}\times 5$  Since the order does not matter we count all the orders.  Or in short $\left.\dbinom 51\dbinom{11}1\dbinom {15}4\middle/\dbinom{26}5\right.$
Put it together
