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A box has 10 red coloured and 15 blue coloureds pens. If you were to pick out 7 pens (without replacement), what is the probability that you get exactly 2 red coloured pens? The answer is $$\frac{{10 \choose 2}{15 \choose 5}}{25 \choose 7}$$However, I am not sure the intuition behind this. I understand that you are choosing 2 red coloured pens from the 10, however, I am not sure why this is. Aren't all the pens the same colour, so why are we treating them as all different? This goes same for the other binomial coefficient chosen.

I also don't get the multiplying and dividing of the binomial coefficients. Also, is there a permutation equivalent for the answer? Thanks.

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The argument goes that there are ${25 \choose 7}$ ways of choosing $7$ pens from $25$, leading to the denominator. Of these, there are ${10 \choose 2}$ ways of choosing $2$ red pens from $10$ and ${15 \choose 5}$ ways of choosing $5$ blue pens from $15$, which you multiply together to get the number of ways of getting exactly $2$ red pens and $5$ blue pens for the numerator. That ignores order.

Alternatively you could choose, in order, the $7$ pens from $25$ in $25\cdot 24\cdot 23\cdot 22\cdot 21\cdot 20\cdot 19$ ways for the denominator. For the numerator, if you chose the $2$ red pens and then the $5$ blue pens in order there would be $10\cdot 9\cdot 15\cdot 14\cdot 13\cdot 12\cdot 11$ ways; but there are ${7 \choose 2}$ possible orders of $2$ red and $5$ blue, so you need to multiply by this. It gives the same probability, as both the numerator and the denominator are $7!$ times what they were in the earlier calculation, reflecting the number of ways of ordering $7$ pens.

Both approaches are correct and give the same probability.

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  • $\begingroup$ Okay, so for your first sentence: why do we consider each pen as different? Shouldn't we treat all the red pens the same and all the blue pens the same? Therefore, there are technically not ${25 \choose 7}$ ways for choosing 7 pens? $\endgroup$ – user12055579 Dec 14 '19 at 3:40
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    $\begingroup$ @user12055579 If you try to implement this, you have $25$ physically distinct pens even if many of them look the same. If the question was "what is the probability the first pen drawn is red?" the answer would be $\frac{10}{25} = 0.4$ not $\frac12=0.5$ $\endgroup$ – Henry Dec 14 '19 at 10:49
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    $\begingroup$ To further elaborate on your comment, it may be helpful to think of this question in another context – instead consider a deck of 25 cards were 10 of them are red and 15 of them are black. Within the colors, of course these cards are not all the same (king, jack, etc) but the probability of drawing exactly 2 red cards out of 7 is the same as the question. The numerator looks at each way you can possibly draw the respective different card combinations in a given color. $\endgroup$ – jpsmith Dec 14 '19 at 12:36
  • $\begingroup$ @Henry Oh, that last part in your comment made me understand it, thanks! $\endgroup$ – user12055579 Dec 14 '19 at 23:32

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