Probability function of a discrete random variable


by stevecallaway
Tags: discrete, function, probability, random, variable
stevecallaway
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#1
May17-11, 01:05 PM
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1. The problem statement, all variables and given/known data
10 face cards are face down in a row on a table. Exactly one of them is an ace. You turn the cards over one at a time, moving from left to right. Let X be the random variable for the number of cards turned before the ace is turned over. What is the probability function for X?


2. Relevant equations
I know the answer is p(x)=1/10 x=0,1,...,9


3. The attempt at a solution
The first card's probability of being an ace is 1/10. The second card is 1/9. The third card is 1/8. etc... I don't understand where my thinking is flawed. Starting from left to right, you are drawing without replacement. If the first card isn't an ace, you know it's one of the last nine. So from that point on, it's 1 out of 9. etc...
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LCKurtz
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May17-11, 01:35 PM
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Quote Quote by stevecallaway View Post
1. The problem statement, all variables and given/known data
10 face cards are face down in a row on a table. Exactly one of them is an ace. You turn the cards over one at a time, moving from left to right. Let X be the random variable for the number of cards turned before the ace is turned over. What is the probability function for X?


2. Relevant equations
I know the answer is p(x)=1/10 x=0,1,...,9


3. The attempt at a solution
The first card's probability of being an ace is 1/10. The second card is 1/9. The third card is 1/8. etc... I don't understand where my thinking is flawed. Starting from left to right, you are drawing without replacement. If the first card isn't an ace, you know it's one of the last nine. So from that point on, it's 1 out of 9. etc...
The probability that the first card turned is the ace is 1/10 as you know. For the second card turned to be the ace, you need the first to fail (probability 9/10) and the second to succeed (probability 1/9) giving P(X=1) = 9/10*1/9. Continue this line of reasoning.


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