Question

5.

The 2024 Olympics have just concluded! The local news has reported that the Arstotzkan Olympic team has

won exactly three medals, but has not said which ones.

Since we are unsure what the specific medal results are, we can create a sample space to represent the possible numbers of bronze, silver, and gold medals that are won. We can summarize the outcomes by abbreviating these medals by their first letter, so 2B0S1G represents the outcome where the team won two bronze medals, no silver medals, and one gold medal. Our sample space would be:

S = {3B0S0G, 2B0S1G, 2B1S0G, 1B0S2G, 1B1S1G, 1B2S0G, 0B0S3G, 0B1S2G, 0B2S1G, 0B3S0G}
(a) If we assume that this is an equiprobable sample space, determine the probability that at least one gold medal

is won by the Arstotzkan Olympic team.
(b) Let X be a random variable which counts the number of silver medals won by the Arstotzkan team, write

out the events, X ≤ 1, X > 2, and X = 4, as collections of sample points in S.
(c) Instead of assuming the sample space is equally likely, we assume the following:

1. all the outcomes where the Arstotzkan team wins at least one gold medal are equally likely

2. all the outcomes where the Arstotzkan team wins no gold medals are equally likely

3. outcomes where the Arstotzkan team wins at least one gold medal are twice as likely as outcomes where the Arstotzkan team wins no gold medals

Using this assumption, create a p.d.f for X, the random variable from (b).

Answer 1

(a) If the sample space is equiprobable, each event in the probability space has the same probability of occurrence. This means that if there are n events in the sample space, the probability of any event is
.

Here, there are n=10 events. Of this, the events containing gold are A = {2B0S1G, 1B0S2G, 1B1S1G, 0B0S3G, 0B1S2G, 0B2S1G}. So, 6 events contain gold. So, the probability of winning gold
.

Probability of winning at least one gold by the Arstotzkan Olympic team = 0.6.

(b) We can write the number of silver medals won and the corresponding probability like this:

X	Events E	Count n(E)
0	3B0S0G, 2B0S1G, 1B0S2G, 0B0S3G	4
1	2B1S0G, 1B1S1G, 0B1S2G	3
2	1B2S0G, 0B2S1G	2
3	0B3S0G	1

Using this table, we write the various events and their counts:

X ≤ 1, so X = 0 or 1; this corresponds to the first and second rows of the above table:
A = {3B0S0G, 2B0S1G, 1B0S2G, 0B0S3G, 2B1S0G, 1B1S1G, 0B1S2G}
n(A) = 7

X > 2, so X = 3; this corresponds to the last row of the above table:
B = (0B3S0G}
n(B) = 1

X = 4; this is not in the table, so we will get an empty set:
C = {}
n(C) = 0

(c) Let p be the probability of winning no gold. There are four such events. They are: {3B0S0G, 2B1S0G, 1B2S0G, 0B3S0G}.

Let q be the probability of winning at least one gold. There are six such events. They are: {2B0S1G, 1B1S1G, 1B0S2G, 0B2S1G, 0B1S2G, 0B0S3G}.

We have:

So we can write the PDF of X as: