Prob. red 3, face card were two selected w/o replacement

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agmoore

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Mar 1, 2009
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I need help with this problem. What am I doing wrong? :?:

Select a card from the deck. Do NOT replace the card. Select another card. What is the probability that a red 3 and a face card are the 2 cards you selected?

(2/52)(12/51) = 2/221(2/1)
1/221
 
Re: Probabilyt and statistics

Your probability is simply (2/52)(12/51), which reduces to 2/221. Are you doing something wrong? You could also calculate it as (12/52)(2/51).
 
Re: Probabilty and statistics

Thanks when I get that answer my teacher told me I had the wrong numerator I was losing my mind trying to check the answer over and over.
 
Hello, agmoore!

Select a card from the deck. Do NOT replace the card. Select another card.
What is the probability that a red 3 and a face card are the 2 cards you selected?

(2/52)(12/51) = 2/221(2/1)
1/221
I don't understand your steps.
Click to expand...

We draw two cards without replacement.
We want a red 3 and a face card . . . in some order.

\(\displaystyle \text{Red 3, then a face card: }\:\frac{2}{52}\cdot\frac{12}{51} \:=\:\frac{2}{221}\)

\(\displaystyle \text{Face card, then a red 3: }\:\frac{12}{52}\cdot\frac{2}{31} \:=\:\frac{2}{221}\)


\(\displaystyle \text{Therefore: }\:p(\text{red 3, face card}) \:=\:\frac{2}{221} + \frac{2}{221} \:=\:\frac{4}{221}\)

 
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