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Did I solve this probability problem correctly?

The question goes like this:

In every table in Earl’s Diner, there are exactly four pairs of chopsticks, and each pair is uniquely colored. If two random customers pick four chopsticks at random, what is the probability that they pick one for each color?

The answer is in the form $\alpha/\beta$ for some coprime integers $\alpha$ and $\beta$. Determine the value of $\alpha + \beta$.

My approach went like this: there are four times that the customers will pick a chopstick, and since there are $2$ chopsticks for every color, then by FCP the number of ways that they will pick one of each color will be $$ 2 \times 2 \times 2 \times 2 = 16 $$ And then the total number of ways you can select four chopsticks would be $8C4$, which would be equivalent to $70$.

Then the probability would be $16:70$ or $8:35$ giving the answer of $43$. I am unsure if this is the correct solution and answer, can someone please verify it?

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  • $\begingroup$ What kind of diner would even have chopsticks? $\endgroup$
    – Toby Mak
    Nov 15 '19 at 10:56
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    $\begingroup$ A Chinese diner? lol. I wasn't the one to create the question. $\endgroup$ Nov 15 '19 at 11:00
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Yes this is correct.

Here is another way to do it.

The first customer picks a chopstick from the set of eight, this will always be only one color so the "distinct color" probability for this pick is $8/8=1$. Then he picks the second chopstick, this time six out of the remaiming seven obey the distinct color rule so this gives a factor of $6/7$. For the second customer his first chopstick gives a factor of $4/6$ (four "good" choices out of six total) and then $2/5$ for the last selection. So we have the overall probability

$\dfrac{8×6×4×2}{8×7×6×5}=\dfrac{8}{35}$

as claimed.

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I preassume that both customers will sit at the same table.

Alternative with same outcome.

Select one of the customers and let him/her pick one by one.

For $i=2,3,4$ let $E_i$ be the event that the $i$-th chosen stick has a color that differs from the colors of the sticks that are already chosen.

Then: $$P(E_2\cap E_3\cap E_4)=P(E_2)P(E_3\mid E_2)P(E_4\mid E_2\cap E_3)=\frac67\frac46\frac25=\frac8{35}$$

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