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We have a standard $52$-card deck. What is the probability of three cards drawn containing no Queens, but at least one Ace or one King?

The total no of draws: $\binom{52}{3}$

no. of draws without any Queen: $p_1= \binom{48}{3}/\binom{52}{3} = 0.78$.

no. of draws without any A or K: $p_2 = \binom{40}{3}/\binom{52}{3} = 0.45$, so $1-p_2 = 0.55$.

So, will I be correct to conclude, that:

$P(\text{no Q, but at least 1 of A,K}) = p_1(1-p_2)$

??

I have a feeling I am missing something, could someone verify, please?

Thanks

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Let:

  • $A$ denote an ace
  • $K$ denote a king
  • $Z$ denote anything but ace, king or queen

Then add up the number of ways to choose each one of the following combinations:

  • $\binom{4}{3}\cdot\binom{4}{0}\cdot\binom{40}{0}\implies{AAA}$
  • $\binom{4}{2}\cdot\binom{4}{1}\cdot\binom{40}{0}\implies{AAK}$
  • $\binom{4}{2}\cdot\binom{4}{0}\cdot\binom{40}{1}\implies{AAZ}$
  • $\binom{4}{1}\cdot\binom{4}{2}\cdot\binom{40}{0}\implies{AKK}$
  • $\binom{4}{1}\cdot\binom{4}{1}\cdot\binom{40}{1}\implies{AKZ}$
  • $\binom{4}{1}\cdot\binom{4}{0}\cdot\binom{40}{2}\implies{AZZ}$
  • $\binom{4}{0}\cdot\binom{4}{3}\cdot\binom{40}{0}\implies{KKK}$
  • $\binom{4}{0}\cdot\binom{4}{2}\cdot\binom{40}{1}\implies{KKZ}$
  • $\binom{4}{0}\cdot\binom{4}{1}\cdot\binom{40}{2}\implies{KZZ}$

Then divide the result by the total number of ways to choose $3$ out of $52$ cards, which is $\binom{52}{3}$.

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  • $\begingroup$ There is no need to distinguish aces and kings in this context. $\endgroup$ – drhab Feb 26 '15 at 9:14
  • $\begingroup$ @drhab: Do you mean that we can use the same letter in order to denote both, and then choose out of $8$ instead of out of $4$, and get the same result? Or do you mean that there is something wrong with the answer? $\endgroup$ – barak manos Feb 26 '15 at 10:04
  • $\begingroup$ Nothing wrong with the answer, but yes: choose from $8$ in stead of twice from $4$. $\endgroup$ – drhab Feb 26 '15 at 10:07
  • $\begingroup$ @drhab: Yep, good point. Not sure whether it's worth the effort changing it, as OP hasn't replied to either one of the answers given here (mine and yours). $\endgroup$ – barak manos Feb 26 '15 at 10:09
  • $\begingroup$ The decision is yours of course. If I were you I would not change it. The combination of your answer and my comment on it is also educational. I did not mean to stimulate you in changing. $\endgroup$ – drhab Feb 26 '15 at 10:12
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The probability of getting no queens is: $$\binom{4}{0}\binom{48}{3}\binom{52}{3}^{-1}=\binom{48}{3}\binom{52}{3}^{-1}$$

The probability of getting no queens, no aces and no kings is: $$\binom{12}{0}\binom{40}{3}\binom{52}{3}^{-1}=\binom{40}{3}\binom{52}{3}^{-1}$$

Consequently the probability of getting no queens but at least one ace or king is: $$\binom{48}{3}\binom{52}{3}^{-1}-\binom{40}{3}\binom{52}{3}^{-1}=\left[\binom{48}{3}-\binom{40}{3}\right]\binom{52}{3}^{-1}$$

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