Sue owns 11 pairs of shoes: six identical black pairs, three identical brown pairs and two identical gray pairs. If she picks two shoes at random, what is the

22 shoes

12 black,  6 brown,  4 grey   (Aussie spelling)

P(Bl,Bl)   =     $$\left({\frac{{\mathtt{12}}}{{\mathtt{22}}}}{\mathtt{\,\times\,}}{\frac{{\mathtt{11}}}{{\mathtt{21}}}}\right)$$   

P(Br,Br) =     $$\left({\frac{{\mathtt{6}}}{{\mathtt{22}}}}{\mathtt{\,\times\,}}{\frac{{\mathtt{5}}}{{\mathtt{21}}}}\right)$$

P(G,G) =   $$\left({\frac{{\mathtt{4}}}{{\mathtt{22}}}}{\mathtt{\,\times\,}}{\frac{{\mathtt{3}}}{{\mathtt{21}}}}\right)$$

P(same colour)  =    $$\left({\frac{{\mathtt{12}}}{{\mathtt{22}}}}{\mathtt{\,\times\,}}{\frac{{\mathtt{11}}}{{\mathtt{21}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{6}}}{{\mathtt{22}}}}{\mathtt{\,\times\,}}{\frac{{\mathtt{5}}}{{\mathtt{21}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{22}}}}{\mathtt{\,\times\,}}{\frac{{\mathtt{3}}}{{\mathtt{21}}}}\right) = {\frac{{\mathtt{29}}}{{\mathtt{77}}}} = {\mathtt{0.376\: \!623\: \!376\: \!623\: \!376\: \!6}}$$

P(same colour)  =      $${\frac{{\mathtt{29}}}{{\mathtt{77}}}}$$

That is incorrect(I think)

The correct answer is 7/33

p(black pair) = (12/22)*(6/21) → 72/462

p(brown pair) = (6/22)*(3/21) → 18/462

p(grey pair) = (4/22)*(2/21) → 8/462

p(matching pair) = (72 + 18 + 8)/462 → 98/462 → 7/33

.

Sorry Alan

Here is another approach:

We will consider these by the cases of what our first shoe selection is 12/22 . If our first shoe is black, which happens with probability 6/21 , then our second shoe will be black and for the opposite foot with probability (6/22)(3/21) . Likewise, for brown shoes, our probability is the product (4/22)(2/21) . And for gray, . So the sum is equal to:

(12 * 6 + 6 * 3 + 4 * 2)/(22 * 21)

98 / (33 * 14)

7/33