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five cards are drawn from a pack of 52 well-shuffled cards. What is the probability that at least two jacks are obtained?

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  • $\begingroup$ Hint: It is the same probability as 1-P("no jack or one jack are obtained")=1-P(X=0)-P(X=1). One minus converse probability. $\endgroup$ May 5, 2015 at 17:50

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Let's first calculate the probability of obtaining no and only 1 Jack.

Taking into account that there are 4 Jacks, it'll be easy to find favorable and possible cases. Order doesn't matter, so we'll take n over r, or nCr.

P(No Jacks) = 48C5 / 52C5 = 0.658842

Let's find now the probability of drafting only one Jack. For that, i'll take first 4 cards with no jack, and multiply it for the number of possible Jacks to be taken.

P(One Jack) = ( 48C4 / 52C5 ) * 4 = 0.2994736

Now you simply have to take that two probabilities from 1.

P(at least 2 Jacks) = 1 - P(No Jack) - P(One Jack) = 0.04168437

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  • $\begingroup$ Why do you have 8 Jacks? $\endgroup$
    – wythagoras
    May 5, 2015 at 18:03
  • $\begingroup$ You don't, my mistake. Let me edit. $\endgroup$
    – Masclins
    May 5, 2015 at 18:06

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