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I got a deck with 10 red cards and 30 black cards.

Now, I draw a couple of cards 20 times. I draw the couple without replacement, but the 20 drawings of couples are with replacement (I restore the original deck at every couple drawing).

Now, what is the probability of obtaining at least one time a couple of reds?

Let me to discuss the probability of obtain a couple of reds exactly one time: It seems to me we got to suppose that this probability is hypergeometrically distributed, for a single couple is drawn without replacement. So, C(10,2)/C(40,2)=3/52. This result is confirmed by the simplier calculation (10/40)*(9/39)=3/52

Now my intuitive view of sample space is constructed associating an index to each card (reds having indexes from 1 to 10), and squaring this set of integers. Hereby, I obtain the set of all ordered couples, 1600, of which only 100 are composed by two reds. So, drawing a couple at random from this set, gives 1/16, which is conflicting with previous results. Why?

Regarding the probability of draw at least a couple of reds over 20 couple drawings, I am unsure. Any suggestion? Perhaps using limit of probabilities of unions of nested events??

Thanks.

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The probability of drawing a pair of reds is $\frac{10}{40}\cdot\frac{9}{39}=\frac{3}{52}$. Hence the probability of not drawing two reds is $1-\frac{3}{52}$. Thus, the probability of not drawing a pair of reds $20$ times is $\left(1-\frac{3}{52}\right)^{20}$ and finally, the probability of drawing a pair of reds at least once is $$1-\left(1-\frac{3}{52}\right)^{20}\approx 0.695,$$ according to Wolfram Alpha.

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  • $\begingroup$ Thanks; Further questions. There are other ways to solve this problem, without using the probability of complement set? And, what is wrong with my reasoning about taking the 100 ordered reds over the entire sample space of 1600 ordered couples? $\endgroup$ – MadHatter Oct 1 '13 at 13:44
  • $\begingroup$ You are supposed to draw couples of distinct cards, as a consequence of drawing them without replacement. On the diagonal of your square 40 by 40 the cards are not distinct. $\endgroup$ – drhab Oct 1 '13 at 17:11
  • $\begingroup$ You are right! Thanks. $\endgroup$ – MadHatter Oct 1 '13 at 22:45

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