<s>[INST] 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?
 [/INST] 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
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