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Three cards are dealt at random from a standard deck of 52 cards. What is the probability that the first card is a 4, the second card is a \(\clubsuit\), and the third card is a 2?

TheMathCoder  Apr 21, 2018
 #1
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HEEEEEEEEEEEEEELP!

TheMathCoder  Apr 25, 2018
 #2
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I don't know much about "playing cards", but I will give it a go and if you know the answer, please let me know if I screwed up or not!!.

 

There are four 4s in a deck of cards, so the probability of drawing a 4 is:

4/52.

The probability of the 2nd card being a "club?" is ordinarily 13/52. right? But we have already drawn 1 card, which was a 4. But this 4 could very well have been a 4 of "club", which means there would be only 12 cards of club left out of 51. So, the probability of drawing a card of club would be:

12/51.

But this 2nd card could very well have been a "2 of club". Therefore, the probability of drawing a 2 of any other suite would be:

3/50.

So, the overall probability would be:4/52 x 12/51 x 3/50 =144 / 132,600 =6 / 5,525.

And that is the best I could do!!.

Guest Apr 25, 2018
edited by Guest  Apr 25, 2018
 #3
avatar+290 
+2

The answer was really \(\frac{1}{663}\) 

IDK why really...

TheMathCoder  Apr 26, 2018

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