Probability function of a discrete random variable problem

[stevecallaway]

Homework Statement

Ten cards are face down in a row on a table. Exactly one of them is an ace. You turn the cards over oen 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?

Homework Equations

P(a|b)=P(a&b) / P(B)

The Attempt at a Solution

P(X=0) = P(1st card is the ace)=1/10
P(X=1)=P(2nd card is the ace|1st card is not ace) * P(1st card is not ace)=9/10 * 1/9
P(X=2)=9/10 * 8/9 * 1/8 = 1/10
so p(x)=1/10 for x=0,1,...,9
I am having trouble understanding how the book arrived at the solution. For P(X=1), it appears to me that they manipulated the equation P(a|b)=P(a&b) / P(B) to be P(a|b) * P(b) = P(a&b). So they are solving for P(a&b). But isn't the key to solve for P(a|b)?

 

[jav]

In this case, you are trying to solve for the probability that card X is an ace (without knowing any of the other cards). If you solve for the P(a|b) you are solving for the probability of event a occurring given that you know b has occurred.

ie you want to solve for P(X is an ace and 1,2,...,X-1 are not an aces). That is what they mean when they say solve for P(a n b).