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I came across the following probability question:

There are $50$ cards of $5$ different colors. It comprises of $10$ Red cards, $10$ blue cards, $10$ orange cards, $10$ green cards and $10$ yellow cards. Each color will have the cards numbered between $1$ to $10$. You pick $2$ cards at random. What is the probability that they are not of same color and not of same number.

So, initially, as a direct approach, I thought you would first choose the color of the first card. There are $5$ ways of doing this. Then you could choose the number of the first card. There are $10$ ways of doing this. Then you would choose the color of the second card. There are $4$ ways of doing this since it can't be the same color of the first card. Then you would choose the number of the second card. There are $9$ ways of doing this, since it cannot be the same number as the first card. Finally there are $\binom{50}{2}$ ways of choosing the $2$ cards. Thus the answer would be $$ \frac{5 \times 10 \times 4 \times 9}{\binom{50}{2}} > 1 $$ which is wrong since it is greater than $1$. The answer given is $$ \frac{\binom{5}{2} \times 10 \times 9}{\binom{50}{2}} = \frac{36}{49}$$

My question with this answer is that that it seems that you choose the two colors at the same time and then you choose the numbers. What happens if you choose them one after the other? What is wrong with my answer?

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  • $\begingroup$ Your initial reasoning is good, but you have undercounted the denominator: there are $50\times 49$ ways of choosing the first card followed by the second, so your answer ought to be $$\frac{(5\times 10)(4\times 9)}{50\times 49}=\frac{36}{49}.$$ $\endgroup$ – whuber Aug 16 '17 at 15:34
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What?

There are 49 cards left so each has the probability 1/49 that it gets drawn next.

There are 9 cards of the first card's color and 4 cards of the first card's number, so the probability that you draw a card that has neither the first card's color nor its number is:

$$(49 - (9 + 4)) \times \frac{1}{49} = \frac{36}{49}$$

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