On Probability Rules

Exercise 1: A Product Identification Code

Question: A company creates a product identification code using 3 distinct English letters, followed by 4 distinct nonzero digits. For example, a code could look like KLM5832. Assume that every valid code is equally likely to be selected.

Find the probability that a randomly selected code:

begins with a vowel; and
ends with an even digit.

Solution:

We use

\[P(A) = \frac{\text{number of favorable codes}} {\text{number of possible codes}}.\]

The main work is therefore a counting problem.

Step 1: Count all possible codes

The code has 3 distinct letters.

For the first letter, there are 26 choices. For the second letter, there are 25 choices, and for the third letter, there are 24 choices.

Therefore, the number of possible letter parts is

\[26\times25\times24.\]

The code also has 4 distinct nonzero digits. The possible digits are

\[1,2,3,4,5,6,7,8,9.\]

There are 9 choices for the first digit, then 8, then 7, then 6.

Therefore, the number of possible digit parts is

\[9\times8\times7\times6.\]

Using the counting principle, the total number of valid codes is

\[N = (26\times25\times24) (9\times8\times7\times6).\]

Step 2: Count the favorable codes

The first letter must be a vowel. Using

\[A,E,I,O,U,\]

there are 5 choices for the first letter.

After choosing the first letter, the remaining two letters must still be different. Therefore, there are 25 choices for the second letter and 24 choices for the third letter.

So the number of favorable letter parts is

\[5\times25\times24.\]

The last digit must be even. Among the nonzero digits, the even digits are

\[2,4,6,8.\]

Therefore, there are 4 choices for the last digit.

Once the last digit is chosen, 8 nonzero digits remain. The first three digit positions can then be filled in

\[8\times7\times6\]

ways.

Thus, the number of favorable digit parts is

\[4\times8\times7\times6.\]

The total number of favorable codes is

\[n = (5\times25\times24) (4\times8\times7\times6).\]

Step 3: Form the probability

Therefore,

\[P(A) = \frac{ (5\times25\times24) (4\times8\times7\times6) }{ (26\times25\times24) (9\times8\times7\times6) }.\]

Cancel the common factors:

\[P(A) = \frac{5}{26}\times\frac{4}{9} = \frac{10}{117}.\]

Therefore,

\[\boxed{P(A)=\frac{10}{117}}.\]

Why This Method Works

The code is formed in stages, so the counting principle is appropriate.

The restrictions affect specific positions:

begins with a vowel affects the first letter;
ends with an even digit affects the last digit.

Because letters and digits cannot repeat within their respective parts, the number of choices decreases after each selection.

Note

A common mistake is to use \(26^3\) or \(9^4\).

Those counts allow repeated letters or digits, but the problem requires distinct letters and distinct digits.

Exercise 2: Two Review Workshops

Question: A group of 100 students is preparing for final examinations.

54 students attend a Statistics review workshop.
69 students attend a Programming review workshop.
35 students attend both workshops.

One student is selected at random.

Let \(S\) be the event that the student attends Statistics, and let \(P\) be the event that the student attends Programming.

Find the probability that the selected student:

attends Statistics or Programming, or both;
attends neither workshop; and
attends Programming but not Statistics.

Solution:

Part 1: Statistics or Programming, or both

The word or indicates a union:

\[S\cup P.\]

We know

\[P(S)=\frac{54}{100}, \qquad P(P)=\frac{69}{100}, \qquad P(S\cap P)=\frac{35}{100}.\]

The events overlap, so use the general addition rule:

\[P(S\cup P) = P(S)+P(P)-P(S\cap P).\]

Thus,

\[P(S\cup P) = \frac{54}{100} + \frac{69}{100} - \frac{35}{100} = \frac{88}{100} = 0.88.\]

Therefore,

\[\boxed{P(S\cup P)=0.88}.\]

Why subtract 35? If we add 54 and 69 directly, the 35 students who attend both workshops are counted twice. Subtracting the intersection once corrects the double counting.

Part 2: Neither workshop

Neither Statistics nor Programming is the complement of \(S\cup P\).

Therefore,

\[P((S\cup P)^c) = 1-P(S\cup P) = 1-0.88 = 0.12.\]

Therefore,

\[\boxed{P(\text{neither})=0.12}.\]

A count check gives the same result:

\[100-88=12,\]

so

\[\frac{12}{100}=0.12.\]

Part 3: Programming but not Statistics

Programming but not Statistics means

\[P\cap S^c.\]

There are 69 students in Programming, but 35 of them also attend Statistics.

Therefore,

\[69-35=34\]

students attend Programming only.

Hence,

\[P(P\cap S^c) = \frac{34}{100} = 0.34.\]

Therefore,

\[\boxed{P(P\cap S^c)=0.34}.\]

Reasoning Summary

Words	Event notation	Main idea
Statistics or Programming	\(S\cup P\)	Add and subtract the overlap
Neither workshop	\((S\cup P)^c\)	Use the complement
Programming but not Statistics	\(P\cap S^c\)	Remove the overlap from \(P\)

Note

OR does not mean the events must be mutually exclusive.

Here, students can attend both workshops, so the overlap must be handled explicitly.

Exercise 3: Participation in Three Campus Activities

Question: A university surveys 500 students about participation in three campus activities:

\(A\): academic club;
\(S\): sports club;
\(V\): volunteer program.

The survey gives the following information:

Group	Number of students
Academic club	210
Sports club	258
Volunteer program	216
Academic and Sports	122
Sports and Volunteer	83
Academic and Volunteer	97
All three activities	52

One student is selected at random.

Find the probability that the student:

participates in Academic but not Sports;
participates in Sports and Volunteer, but not Academic;
participates in at least one of the three activities; and
participates in none of the three activities.

Solution:

Part 1: Academic but not Sports

We want

\[A\cap S^c.\]

There are 210 students in \(A\), and 122 of them are also in \(S\).

Therefore,

\[210-122=88.\]

Hence,

\[P(A\cap S^c) = \frac{88}{500} = 0.176.\]

Therefore,

\[\boxed{P(A\cap S^c)=0.176}.\]

Students who are in Academic and Volunteer but not Sports are still included. The question only excludes Sports.

Part 2: Sports and Volunteer, but not Academic

We want

\[S\cap V\cap A^c.\]

There are 83 students in \(S\cap V\). Of these, 52 are also in Academic.

Therefore,

\[83-52=31.\]

Thus,

\[P(S\cap V\cap A^c) = \frac{31}{500} = 0.062.\]

Therefore,

\[\boxed{P(S\cap V\cap A^c)=0.062}.\]

Part 3: At least one activity

At least one means

\[A\cup S\cup V.\]

Because the three events overlap, use the three-event addition rule:

\[\begin{split}\begin{aligned} P(A\cup S\cup V) ={}& P(A)+P(S)+P(V)\\ &-P(A\cap S)-P(A\cap V)-P(S\cap V)\\ &+P(A\cap S\cap V). \end{aligned}\end{split}\]

Using counts first,

\[\begin{split}\begin{aligned} n(A\cup S\cup V) ={}& 210+258+216\\ &-122-97-83\\ &+52. \end{aligned}\end{split}\]

Now,

\[210+258+216=684\]

and

\[122+97+83=302.\]

Therefore,

\[n(A\cup S\cup V) = 684-302+52 = 434.\]

Hence,

\[P(A\cup S\cup V) = \frac{434}{500} = 0.868.\]

Therefore,

\[\boxed{P(\text{at least one})=0.868}.\]

Why add the three-way intersection back?

Students in all three groups are first counted three times when the individual groups are added. They are then subtracted three times through the pairwise intersections. At that point, they have been counted zero times, so the triple intersection must be added back once.

Part 4: None of the three activities

None is the complement of at least one.

Therefore,

\[P((A\cup S\cup V)^c) = 1-P(A\cup S\cup V) = 1-0.868 = 0.132.\]

Therefore,

\[\boxed{P(\text{none})=0.132}.\]

Using counts gives the same result:

\[500-434=66,\]

so

\[\frac{66}{500}=0.132.\]

Reasoning Summary

Words	Event notation	Main idea
Academic but not Sports	\(A\cap S^c\)	Remove the overlap with Sports
Sports and Volunteer but not Academic	\(S\cap V\cap A^c\)	Remove the triple overlap
At least one	\(A\cup S\cup V\)	Use the three-event addition rule
None	\((A\cup S\cup V)^c\)	Use the complement