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Given that i have 6 pairs of balls where each pair contains 2 balls of the same colour in the bag. So the bag would contain RR, GG, BB, PP, YY, WW (red, green, blue, pink, yellow, white).

Whats the probability of finding atleast one pair when

a) 1 Ball has been drawn out

b) 2 Balls have been drawn out

c) 3 Balls have been drawn out

d) 4 Balls have been drawn out

etc up to 12 balls in total,

I'm not necessarily looking for the answer but rather the methodology behind solving a problem like this.

Thanks!

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  • $\begingroup$ Methodology: We see the keyword "at least one" and that hints that it might make sense to do "complementary counting": that means we instead find the probability when drawing out $k$ balls of getting no pairs. To find this, we count the number of sets of $k$ balls with no pairs, divided by the total number of sets of $k$ balls ($12$ choose $k$). See Nick Peterson's good answer. $\endgroup$ – 6005 Dec 5 '16 at 6:56
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The idea: you can come up with the probability in each case by way of $$ P(\text{at least one pair in $k$ drawn})=\frac{\text{# of ways of choosing $k$ balls w/ a pair of same color}}{\text{# of ways of choosing $k$ balls}}. $$

As a hint: note that this is the same as $$ P(\text{at least one pair in $k$ drawn})=1-\frac{\text{# of ways of choosing $k$ balls of different colors}}{\text{# of ways of choosing $k$ balls}}, $$ which might be much easier to compute.

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