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You have an urn that contains 10 balls numbered 1-10. They are drawn one at a time without replacement.

a.) Find the probability that the second ball drawn is the ball numbered 3. Intuitively, I know that the answer is 1/10 but I am not sure why. The probability that "the second ball is drawn" implies that there are only 9 balls left, so wouldn't the answer be 1/9?

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You have to adjust for the fact that this is only possible if the first ball does not have the number 3. –  Michael Greinecker Jan 22 '12 at 15:30

3 Answers 3

In such questions, it is often very good to write out the probability tree (or possibilities, or whatnot). Or to compute the possibilities.

Suppose, for example, I wanted to know how many ways the 3 is drawn first. There is 1 way to draw the 3 on the first draw, and then there are 9 possibilities for the second draw. So $9$ (for 2 pulls). In total, I can draw 10 ways in the first and 9 on the second, so there are $90$ total possibilities. Assuming them all to be equally likely, this suggests that there is a $9 / 90 = 1/10$ probability of drawing the 3-ball first.

You could modify this for the second as well.

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There are $10!$ permutations of the set $\{1.2.3,\dots,10\}$, all equally likely. There are $9!$ permutations that have a $3$ in the second place.

Or else because all permutations are equally likely, the $3$ is in first, second, and so on up to tenth place with equal probabilities, which thus must be $1/10$.

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Other answers here use symmetry. This is very useful, but if the symmetry of the problem is not clear to you, this can always be calculated with conditional probability. Let's denote the event $A$ as "the first ball is #3", and $B$ "the second ball is #3". So conditional probability tells you that $$\begin{eqnarray} P(B)&=&P(B|A)\cdot P(A)+P(B|A^c)\cdot P(A^c)\\ &=&P(B|A)\cdot P(A)+P(B|A^c)\cdot(1- P(A)) \end{eqnarray} $$ ($A^c$ is the complement of $A$, i.e. the event "not A"). Now, the probability of $A$ is, as you probably know, $P(A)=\frac{1}{10}$. But given $A$, the probability of $B$ is $0$, because #3 was already taken out. The probability of $B$ given $A^c$ is, as you correctly stated, $P(B|A^c)=\frac{1}{9}$. So this gives

$$P(B)=0\cdot \frac{1}{10} + \frac{1}{9}\cdot\left(1-\frac{1}{10}\right)=\frac{1}{10}$$

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