# Thread: conditional probability?

1. ## conditional probability?

Alicia had planned to spend the morning studying for her afternoon statistics class. At the beginning of each class, her professor randomly selects three different students to answer questions about the material. There are 49 students in the class, so Alicia reasons that she is not likely to be selected. Rather than studying, she uses the time to browse the Web.

At the statistics class that afternoon, the 49 student names are written on slips of paper and put into a bag. One name is drawn (without replacement) for each of the three questions. Alicia twice breathes a sigh of relief. She is not picked to answer either of the first two questions. But probability is not in Alicia's favor on this day: She gets picked to answer the third question.

Define
C1, C2, and C3
to be the events that Alicia is called on to answer questions 1, 2, and 3, respectively.

can someone explain to me why all 3 of those events are 1/49? I thought it'd be 1/49 for c1 then 1/48 for c2, 1/47 for c3...

2. ## Re: conditional probability?

The explanation is a little more complicated than that, nilhery.

Your answer doesn't take into account that if Alicia is chosen for the first question she can not be chosen for the second or third. So as a result the probabilities should look like this:
$C_1 = \frac{1}{49}$
$C_2 = \frac{1}{48} * (1 - C_1) = \frac{1}{48} * \frac{48}{49} = \frac{1}{49}$
$C_2 = \frac{1}{48} * (1 - C_2) = \frac{1}{49}$

To shed some intuitive light on it: if the professor chose 49 different people without replacement, so everyone got chosen. Alicia will always be chosen and has a $\frac{1}{49}$ chance of being chosen in each of the positions.

Let me know if you need a little more explanation.

3. ## Re: conditional probability?

Originally Posted by Halty
The explanation is a little more complicated than that, nilhery.
You misunderstand. nilhery's only purpose in posting at all was to include that link to a comercial site!