SAT Probability

Correct Answer: C

Solution:

Tip: Assuming that all the possible outcomes of an event $A$ are equally likely, the probability that $A$ will occur is $P(A) = \frac{\text{ \# of favorable outcomes}}{\text{total \# of outcomes}}.$
Tip: The probability that two events, $A$ and $B,$ happen together is $P(A\ \text{and}\ B) = P(A)\cdot P(B\ \text{given}\ A).$

There are $n$ tickets in the hat. Since the numbers on the tickets do not repeat, the probability of first selecting the ticket numbered 1 is $P(\text{selecting \#1 first})=\frac{1}{n}.$

The first ticket is not replaced. That means that now in the hat there are $n-1$ tickets, and the probability of selecting 3 as the second ticket is $P(\text{selecting \#3, given we selected \#1})=\frac{1}{n-1}.$

The probability that the first two tickets are numbered 1 and 3 is

$\begin{aligned} &P(\text{selecting \#1 first and \#3 second, without replacement})\\ \\ &= P(\text{selecting \#1 first}) \cdot P(\text{selecting \#3, given we selected \#1})\\ &= \frac{1}{n} \cdot \frac{1}{n-1} \\ &= \frac{1}{n^2 -n}.\\ \end{aligned}$

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Incorrect Choices:

(A)
This is the probability that the first ticket is numbered 1.

(B)
This is the probability that the second ticket is numbered 3.

(D)
If you add instead of multiply the probabilities, you will get this wrong answer.

(E)
If you think that the first ticket is replaced, you will get this wrong answer.