Question 129457
Suppose that you take all of the black cards out of a standard deck of 52 cards
and thoroughly shuffle the remaining 26 red cards. From this deck of 26 red
cards you will select 3 cards, one at a time, WITHOUT replacement, and record
weather each card is picked is a face card (a jack, queen, or king), or not a
face card.
<pre><font size = 4 color = "indigo"><b>
The sample space is

AH,2H,3H,4H,5H,6H,7H,8H,9H,10H,JH,QH,KH
AD,2D,3D,4D,5D,6D,7D,8D,9D,10D,JD,QD,KD

It contains 6 face cards and 20 non-face cards.

"AND" means "multiply"
"OR" means "add".

</pre></font></b>
a.) What is the probability that none of the 3 cards picked in this way is a
face card?
<pre><font size = 4 color = "indigo"><b>
P(picking a non-face card first AND a non-face card second AND a non-face card
  third) =

P(picking a non-face card first) x P(picking a non-face card second) x
P(picking a non-face card third) =

{{{(20/26)(19/25)(18/24)}}} = {{{57/130}}}

</pre></font></b>
b.) what is the probability that exactly one of the three cards picked is a
face card? 
<pre><font size = 4 color = "indigo"><b>
P[(face card 1st AND non-face card 2nd AND non-face card 3rd) OR
  (non-face card 1st AND face card 2nd AND non-face card 3rd) OR
  (non-face card 1st AND non-face card 2nd AND face card 3rd)] =

P(face card 1st) x P(non-face card 2nd) x P(non-face card 3rd) +
P(non-face card 1st) x P(face card 2nd) x P(non-face card 3rd) +
P(non-face card 1st) x P(non-face card 2nd) x P(face card 3rd) =

{{{(6/26)(20/25)(19/24) + (20/26)(6/25)(19/24) + (20/26)(19/25)(6/24)}}} =

= {{{57/130}}}
  
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Notice that the two answers happened to come out the same.

Edwin</pre>