I have 16 cards total, 4 cards from each suit of diamonds, hearts, clubs, spades. If I draw 5 cards without replacement, what is the probability that the 5th card drawn is a club? Is there any easy short cut to do this? Or do I need to consider every scenario and sum probabilities?
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4 Answers
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The amount of calculation needed is minimal: the probability is $\dfrac{4}{16}$. All sequences of length $5$ made up of distinct cards are equally likely. Thus "fifth is club" has the same probability as "first is club."
answered Sep 29, 2013 at 20:29
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You can argue the probability is $1/4$ by symmetry: Since you start with an equal number of cards of each suit, how could the probability of getting a club on the 5th card be different from the probability of any of the other suits for the 5th card? Thus all 4 suits are equally likely for the 5th card, so they all have probability $1/4$.
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Yet another way to see that the probability that the fifth card drawn will be a club is $1/4$ is to notice that the identity of the fifth card drawn is determined in advance: it is the fifth card from the top. Then you use your hypothesis on the cards being randomly ordered (which you did not state) to see that the probability of the card in any given position being a club is $4/16$ because $4$ of the $16$ cards are clubs.
answered Sep 29, 2013 at 20:36
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I suggest the following "long way" to the beginning student who wants practice applying techniques. Divide the ways it can happen into cases that you can apply counting arguments to.
First, how many ways can you draw 5 cards?
Then, calculate the number of ways 5th card can be a club. This is the sum of
- the number of ways if no club comes before the 5th card
- the number of ways if exactly one club comes before the 5th card
and so on.
This will build character.
