Probability of at least two being grey

Question

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?

1. 2 grey: The probability of the first being grey is $\frac{11}{26}$ the second grey is $\frac{10}{25}$, the last three white $\frac{15\cdot 14 \cdot 13}{24\cdot 23\cdot 22}$.
2. 3 grey: $\frac{11\cdot 10 \cdot 9 \cdot 15\cdot 14\cdot 13}{26\cdot 25\cdot 24\cdot 23\cdot 22}$
3. 4 grey: $\frac{11\cdot 10 \cdot 9 \cdot 8 \cdot 15\cdot 14}{26\cdot 25\cdot 24\cdot 23\cdot 22}$
4. 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?

Answer 1

Let us do the computations explicitly. Fix some $g,w\ge 0$, $g+w=5$, and the question is which is the exact probability $$P(g,w)$$ to choose exactly $g$ Gray and $w$ White mice. First, each such choice depends on the positions we make the choices, for instance, for $g=2$, $w=3$ we have $$\binom 52=\binom 53=\frac{4\cdot 5}{1\cdot 2}=10$$ possibilities to consider the "positions" we extract the corresponding color, the list would be

WWWGG
WWGWG
WWGGW
WGWWG
WGWGW
WGGWW
GWWWG
GWWGW
GWGWW
GGWWW

Let us fix one of the above patterns. Then we have $11\cdot 10$ possibilities to extract the two $G$ at the two places they appear, and $15\cdot 14\cdot 13$ possibilities for the $WWW$. So explicitly, the probabilities are: $$ \begin{aligned} P(5G,0W) &= \binom 50\cdot \frac {\color{red}{11\cdot10\cdot 9\cdot 8\cdot 7}} {26\cdot 25\cdot 24\cdot 23\cdot 22}\ , \\ P(4G,1W) &= \binom 51\cdot \frac {\color{red}{11\cdot10\cdot 9\cdot 8} \cdot \color{blue}{15}} {26\cdot 25\cdot 24\cdot 23\cdot 22}\ , \\ P(3G,2W) &= \binom 52\cdot \frac {\color{red}{11\cdot10\cdot 9} \cdot \color{blue}{15\cdot 14}} {26\cdot 25\cdot 24\cdot 23\cdot 22}\ , \\ P(2G,3W) &= \binom 53\cdot \frac {\color{red}{11\cdot10}\cdot \color{blue}{15\cdot 14\cdot 13}} {26\cdot 25\cdot 24\cdot 23\cdot 22}\ , \\ P(1G,4W) &= \binom 54\cdot \frac {\color{red}{11} \cdot \color{blue}{15\cdot 14\cdot 13\cdot12}} {26\cdot 25\cdot 24\cdot 23\cdot 22}\ , \\ P(0G,5W) &= \binom 55\cdot \frac {\color{blue}{15\cdot 14\cdot 13\cdot12\cdot 11}} {26\cdot 25\cdot 24\cdot 23\cdot 22}\ . \end{aligned} $$ Let us use computer aid to see these numbers, here sage...

sage: P = {}
sage: d = factorial(26)/factorial(26-5)
sage: for w in [0..5]:
....:     g = 5-w
....:     numer = binomial(5,w) \
....:           * factorial(11)/factorial(11-g) \
....:           * factorial(15)/factorial(15-w)
....:     P[w] = numer/d
....:      
sage: for w in [0..5]:
....:     print "P( %sG , %sW ) = %s ~ %f" % ( 5-w, w, P[w], P[w].n() )
....:     
P( 5G , 0W ) = 21/2990 ~ 0.007023
P( 4G , 1W ) = 45/598 ~ 0.075251
P( 3G , 2W ) = 315/1196 ~ 0.263378
P( 2G , 3W ) = 35/92 ~ 0.380435
P( 1G , 4W ) = 21/92 ~ 0.228261
P( 0G , 5W ) = 21/460 ~ 0.045652
sage: sum( P.values() )
1

The needed probability is thus

sage: P[0] + P[1] + P[2] + P[3]
167/230
sage: _.n()
0.726086956521739

Or (humanly much) simpler

sage: 1 - P[4] - P[5]
167/230
sage: _.n()
0.726086956521739

Answer 2

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

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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