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Subject: "Probability problem"     Previous Topic | Next Topic
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rabinca
Member since Jan-6-10
Jan-06-10, 02:20 PM (EST)
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"Probability problem"
 
   Greetings!

This is my first time here. I'm long ago graduated from school. My knowlwedge of probability theory is not terribly advanced and I've been wrestling with a probability problem and have no one else to turn to for help. I'm hoping someone here can help me. Here's the problem...

Imagine a game in which the object is to draw a specific card, say the Queen of Hearts, from a standard deck of playing cards. Each time a card is drawn it is removed from the deck if it is not the Queen of Hearts and set aside before the next card is drawn so the deck gets smaller with each unsuccessful draw.

My question is, "What is the probability that the Queen of Hearts will be the very last card drawn, i.e. it will not be drawn until all of the other 51 cards have been?" More generally, how would I go about computing the probability that the Queen of Hearts is not drawn until the nth trial?

I've been reading what I can find on probability theory on the internet but, given my level of math knowledge and skills, the stuff is either too basic or too advanced for me to understand well enough to extrapolate from it to solve the problem. I thank you in advance for any help you can provide.

Best regards,

Carl


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alexbadmin
Charter Member
2467 posts
Jan-06-10, 02:26 PM (EST)
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1. "RE: Probability problem"
In response to message #0
 
   The probability that your card won't be drawn on the first try is 51/52. That on the second is 50/51; on the third 49/50, etc. Your card won't be drawn on the 50th draw with probability 2/3 and on the 51st with the probability 1/2.

The product of these probabilties is the probability that the card will be left over:

51/52·50/51·49/50·...·2/3·1/2 = 1/52.

Come to think of it. The cards are left over with equal probabilities and one is always there with probabilitiy 1. For a specific card then it is 1/52.


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