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From an ordinary deck of $52$ cards, five are drawn randomly. What is the probability of drawing exactly three face cards? (assume no replacement)

I wrote the probability as a fraction with denominator $\binom{52}{5}$. For the numerator I wrote $\binom{12}{3}\binom{40}{2}$. My answer was approximately $.0660$.

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    $\begingroup$ You are correct. Please read this MathJax tutorial, which explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Sep 11 '18 at 1:02
  • $\begingroup$ Thank you...I will gladly view that! $\endgroup$ – Murph Jones Sep 11 '18 at 1:03
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While pure math is great, I prefer figuring it out logically (which is exactly the same thing as mathematically, but more intuitive). You can forget an equation, but you can't forget logic.

The chance of drawing the first face card: 12/52 The chance of drawing the second face card: 11/51 The chance of drawing the third face card: 10/50 The chance of drawing the fourth card (not face): 40/49 The chance of drawing the fifth card (not face): 39/48

Total chance of one possible combination = 0.00660264105

Total combination of 3 cards out of 5 is 10 (11100,11010,11001,10110,10101,10011,01110,01101,01011,00111)

Grand total is 0.00660 * 10 = 0.066

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I wrote the probability as a fraction with denominator $\binom {52} 5$. For the numerator I wrote $\binom {12}3\binom {40}{2}$. My answer was approximately $.0660$.

Yes.   $\left.\binom {12}3\binom{40}{2}\middle/\binom{52}{5}\right.$ is the probability for selecting three from the twelve face cards and two from the forty non-face cards, when drawing any five from all fifty-two cards without replacement.   That is the probability for the event you sought, as there are indeed those counts for face and non-face cards among a standard deck.

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