A probability question about choosing 6 shoes are chosen randomly from the 20 distinct pairs of shoes drawer. $\mathbf{Question:}$
A drawer has 20 distinct pairs of shoes. 6 shoes are chosen randomly from the drawer. The drawer contain only one my favorite pair.
(a) what is the probability that my favorite pair is chosen ?
(b) what is the probability that I get no complete pair ?
(c) what is the probability that I get exactly one complete pair ?
(d) what is the probability that I get at two complete pairs ?
$\mathbf{My~Attempt:}$
(a) The probability that my favorite pair is shosen $= \frac{2}{40} \cdot \frac{1}{39} \approx 0.0013$.
(b) The probability that I get no complete pair $= \frac{40 \cdot 38 \cdot 36 \cdot 34 \cdot 32 \cdot 30}{40 \cdot 39 \cdot 38 \cdot 37 \cdot 36 \cdot 35} = \frac{34 \cdot 32 \cdot 30}{39 \cdot 37 \cdot 35} \approx 0.6463$.
(c) The probability that I get exactly one complete pair $= \frac{\binom{20}{1} \cdot \binom{18}{4} \cdot 2^4}{\binom{40}{6}} \approx 0.2551$.
(d) The probability that I get at least one complete pair $= 1 - \text{The probability I get no complete pair} = 0.3537$.
$~~~~~$ So, The probability that I get at least two complete pairs
$~\hspace{11mm}$ $= \text{The probability I get at least one complete pair}$
$~\hspace{15mm}$ $- \text{The probability I get exactly one complete pair}$
$~\hspace{11mm}$ $= 0.3537 - 0.2551 = 0.0986$
$\textbf{Is that my attempt of (a), (b), (c) and (d) correct ?}$
$\textbf{If they are all correct, then are there any other ways to think about (d)}$
$\textbf{without using the answer of (b) and (c) ?}$
 A: Not all of your answers are correct.
(a) If you choose both shoes from your favorite pair, you must also choose four shoes from the remaining $40 - 2 = 38$ shoes.  Thus, the number of favorable cases is
$$\binom{2}{2}\binom{38}{4}$$
You failed to account for the fact that you are choosing six shoes, not two.  Selecting more shoes increases your chances of choosing both shoes from your favorite pair.
(b) Your answer is correct.  Another way to count the favorable cases is to observe we must choose six different pairs, and draw one of the two shoes from each pair.  Hence, the number of favorable cases is
$$\binom{20}{6}2^6$$
(c) You made a minor error.  There are $20$ ways to choose the pair from which both shoes are taken.  This leaves $20 - 1 = 19$ pairs.  To ensure that exactly one complete pair is chosen, we must select four of these $19$ pairs from which to extract one shoe each.  There are two ways to choose a shoe from each of these four pairs.  Hence, the number of favorable cases is
$$\binom{20}{1}\dbinom{19}{4}2^4$$
(d) Your method is sound.  However, with the above observations, the answer should be
$$1 - \frac{\dbinom{20}{6}2^6 + \dbinom{20}{1}\dbinom{19}{4}2^4}{\dbinom{40}{6}}$$
You could also add the probabilities of obtaining exactly two complete pairs and exactly three complete pairs.
Exactly two complete pairs:  Choose from which two of the $20$ pairs of shoes both shoes will be taken.  You must pick two additional shoes. To ensure that exactly two pairs are selected, you must select two of the remaining $20 - 2 = 18$ pairs and choose one of the two shoes from each of those pairs.

 $$\binom{20}{2}\binom{18}{2}2^2$$

Exactly three complete pairs:  Choose from which three of the $20$ pairs both shoes will be taken.

 $$\binom{20}{3}$$

Thus, the desired probability is

 $$\frac{\dbinom{20}{2}\dbinom{18}{2}2^2 + \dbinom{20}{3}}{\dbinom{40}{6}}

