A group of 90 students is to be split at random into 3 classes of equal size. All partitions are equally likely. Joe and Jane are members of the 90-student group. Find the probability that Joe and Jane end up in the same class.
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
1/3 * 1/3 = ?
it is all narrowed down to:
29/89
To find the probability that Joe and Jane end up in the same class, we need to determine the total number of possible outcomes and the number of favorable outcomes.
First, we need to determine the total number of possible outcomes or the total number of ways to split the 90 students into 3 classes of equal size. For this, we use the concept of combinations.
The total number of ways to select students for the first class is given by the combination formula: C(90, 30), which represents choosing 30 students out of 90. Similarly, for the second class, we have C(60, 30) and for the third class, we have C(30, 30). However, since the order of the classes does not matter, we need to divide the total number of ways to select students for all three classes by 3!, which is the number of ways to arrange three classes.
Therefore, the total number of possible outcomes is:
C(90, 30) * C(60, 30) * C(30, 30) / (3!)
To determine the number of favorable outcomes or the number of ways Joe and Jane end up in the same class, we can treat Joe and Jane as a pair and consider the remaining 88 students. We can select 28 students to be in their class using the combination formula: C(88, 28). Then, the remaining students can be divided into two classes in the same way we did for the total number of possible outcomes.
Therefore, the number of favorable outcomes is:
C(88, 28) * C(60, 30) * C(30, 30) / (3!)
Finally, we can find the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = [C(88, 28) * C(60, 30) * C(30, 30) / (3!)] / [C(90, 30) * C(60, 30) * C(30, 30) / (3!)]
Simplifying this expression will give us the final answer for the probability that Joe and Jane end up in the same class.