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Lets Imagine that we have 8 balls of each color: black, red, yellow, white, green, orange, purple, blue, and we pick ranomply 2 balls, what are the odds that they are the same color? So there are 64 balls, I calculated this like this $\frac{8}{64} \times \frac{7}{63} \approx 1.4\%$, this is apparently wrong according to my book, what am I doing wrong?

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    $\begingroup$ I think you've just calculated the probability that both balls are black (say). The first ball can be any color. The second just has to match it. $\endgroup$
    – Brian Tung
    Commented Feb 10, 2023 at 17:33

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Well, if you want to do it the long way,

[Choose a color]$\times$[Choose two from that color]/[Choose any 2]

$= \Large\frac{\binom81\cdot\binom82}{\binom{64}2} = \frac19$

And the short way is that the first can be any, the next any $7$ from the remaining $63$, thus simply $\Large\frac 7{63}= \frac19$

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  • $\begingroup$ Ig I calculate using the formula you gave the result is $56/2016$ which is not $1/9$? $\endgroup$
    – James
    Commented Feb 10, 2023 at 17:39
  • $\begingroup$ @James: Sorry, I have corrected, i was simultaneously thinking of two approaches ! $\endgroup$ Commented Feb 10, 2023 at 18:13

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