On Conditional Probability, Multiplication, Independence, and Total Probability

Question 1: Course Participation and Conditional Probability

Question: A group of 240 students takes a short review course. The students are classified by study mode and whether they passed the final review test.

Study Mode	Passed	Did Not Pass	Total
Online	72	48	120
In Person	84	36	120
Total	156	84	240

One student is selected at random.

Let

\(O\) = the student studied online;
\(P\) = the student passed the review test.

Find:

\(P(P)\);
\(P(O\mid P)\);
\(P(P\mid O)\); and
explain why \(P(O\mid P)\) and \(P(P\mid O)\) are different.

Solution:

Part (a): Find \(P(P)\)

There are 240 students in total.

From the table, 156 students passed.

Therefore,

\[P(P) = \frac{156}{240} = \frac{13}{20} = 0.65.\]

Thus,

\[\boxed{P(P)=0.65}.\]

This is an ordinary probability. No condition has been given, so the denominator is the total number of students.

Part (b): Find \(P(O\mid P)\)

The notation

\[P(O\mid P)\]

means:

probability that the student studied online, given that the student passed.

The condition is \(P\).

Therefore, we no longer use all 240 students as the denominator.

We restrict attention to the 156 students who passed.

Among these 156 students, 72 studied online.

Thus,

\[P(O\mid P) = \frac{72}{156} = \frac{6}{13} \approx 0.462.\]

Therefore,

\[\boxed{P(O\mid P)=\frac{6}{13}\approx0.462}.\]

Using the conditional probability formula gives the same result:

\[P(O\mid P) = \frac{P(O\cap P)}{P(P)}.\]

Here,

\[P(O\cap P)=\frac{72}{240}\]

and

\[P(P)=\frac{156}{240}.\]

Therefore,

\[P(O\mid P) = \frac{72/240}{156/240} = \frac{72}{156} = \frac{6}{13}.\]

Part (c): Find \(P(P\mid O)\)

Now the condition changes.

The notation

\[P(P\mid O)\]

means:

probability that the student passed, given that the student studied online.

The condition is now \(O\).

There are 120 online students.

Among them, 72 passed.

Therefore,

\[P(P\mid O) = \frac{72}{120} = \frac{3}{5} = 0.60.\]

Thus,

\[\boxed{P(P\mid O)=0.60}.\]

Part (d): Why are the two conditional probabilities different?

We found

\[P(O\mid P) = \frac{72}{156}\]

but

\[P(P\mid O) = \frac{72}{120}.\]

The numerator is 72 in both cases because the same 72 students are both online students and students who passed.

However, the denominators are different.

For

\[P(O\mid P),\]

we already know that the student passed, so the new reference group contains 156 students.

For

\[P(P\mid O),\]

we already know that the student studied online, so the new reference group contains 120 students.

Therefore,

\[\boxed{P(O\mid P)\neq P(P\mid O)}.\]

The important idea is:

the event after the vertical bar determines the denominator.

Note

Conditional probability reduces the sample space.

In \(P(A\mid B)\), first restrict attention to event \(B\). Then ask how many of those outcomes also belong to \(A\).

Question 2: Two Special Cards Without Replacement

Question: A box contains 20 cards. Five cards are marked with a star and 15 cards are unmarked. Two cards are drawn one after another without replacement.

Let

\(A_1\) = the first card is a star;
\(A_2\) = the second card is a star.

Find:

\(P(A_1)\);
\(P(A_2\mid A_1)\);
the probability that both cards are stars; and
the probability that exactly one of the two cards is a star.

Solution:

Part (a): Probability that the first card is a star

There are 5 star cards among 20 cards.

Therefore,

\[P(A_1) = \frac{5}{20} = \frac{1}{4}.\]

Thus,

\[\boxed{P(A_1)=\frac{1}{4}}.\]

Part (b): Probability that the second card is a star, given that the first is a star

The first card is known to be a star.

Because the first card is not replaced:

19 cards remain;
only 4 star cards remain.

Therefore,

\[P(A_2\mid A_1) = \frac{4}{19}.\]

Thus,

\[\boxed{P(A_2\mid A_1)=\frac{4}{19}}.\]

The condition changes the probability because the composition of the box changes after the first draw.

Part (c): Probability that both cards are stars

The word both means an intersection:

\[A_1\cap A_2.\]

Use the multiplication rule:

\[P(A_1\cap A_2) = P(A_1)P(A_2\mid A_1).\]

Substitute the values:

\[P(A_1\cap A_2) = \frac{5}{20}\times\frac{4}{19}.\]

Therefore,

\[P(A_1\cap A_2) = \frac{20}{380} = \frac{1}{19}.\]

Thus,

\[\boxed{P(\text{both stars})=\frac{1}{19}}.\]

The calculation follows the sequence of the experiment:

first star \(\longrightarrow\) second star given the first star.

Part (d): Probability of exactly one star

Exactly one star can happen in two different ways:

first card is a star and second card is not a star;
first card is not a star and second card is a star.

These two possibilities cannot occur at the same time, so their probabilities can be added.

Path 1: Star, then not star

\[P(A_1\cap A_2^c) = \frac{5}{20}\times\frac{15}{19} = \frac{75}{380}.\]

Why is the second probability \(15/19\)?

After drawing a star first, all 15 unmarked cards are still present among the 19 remaining cards.

Path 2: Not star, then star

The probability that the first card is not a star is

\[\frac{15}{20}.\]

If an unmarked card is removed first, all 5 star cards remain among the 19 remaining cards.

Therefore,

\[P(A_1^c\cap A_2) = \frac{15}{20}\times\frac{5}{19} = \frac{75}{380}.\]

Add the two mutually exclusive paths:

\[P(\text{exactly one star}) = \frac{75}{380} + \frac{75}{380}.\]

Thus,

\[P(\text{exactly one star}) = \frac{150}{380} = \frac{15}{38}.\]

Therefore,

\[\boxed{P(\text{exactly one star})=\frac{15}{38}}.\]

Note

For a sequential problem:

multiply probabilities along one path;
add probabilities of different non-overlapping paths that produce the same final event.

Question 3: Checking Independence

Question: A survey of 200 students records whether each student attends a morning review session and whether the student brings a calculator.

Let

\(M\) = attends the morning review session;
\(C\) = brings a calculator.

The survey finds:

80 students attend the morning review session;
100 students bring a calculator;
40 students both attend the morning review session and bring a calculator.

One student is selected at random.

Find:

\(P(M)\), \(P(C)\), and \(P(M\cap C)\);
\(P(C\mid M)\);
determine whether \(M\) and \(C\) are independent, using two different checks; and
determine whether \(M\) and \(C\) are mutually exclusive.

Solution:

Part (a): Find the three probabilities

There are 200 students in total.

For the morning review session,

\[P(M) = \frac{80}{200} = 0.40.\]

For bringing a calculator,

\[P(C) = \frac{100}{200} = 0.50.\]

For doing both,

\[P(M\cap C) = \frac{40}{200} = 0.20.\]

Therefore,

\[\boxed{P(M)=0.40,\quad P(C)=0.50,\quad P(M\cap C)=0.20}.\]

Part (b): Find \(P(C\mid M)\)

The condition is \(M\).

There are 80 students in the morning review group.

Among these 80 students, 40 bring a calculator.

Therefore,

\[P(C\mid M) = \frac{40}{80} = 0.50.\]

Thus,

\[\boxed{P(C\mid M)=0.50}.\]

Part (c): Are the events independent?

We can check independence in two ways.

Check 1: Conditional probability

We found

\[P(C\mid M)=0.50.\]

We also know

\[P(C)=0.50.\]

Therefore,

\[P(C\mid M)=P(C).\]

Knowing that a student attends the morning session does not change the probability that the student brings a calculator.

This shows that \(M\) and \(C\) are independent.

Check 2: Product condition

For independent events, we should have

\[P(M\cap C)=P(M)P(C).\]

Calculate the product:

\[P(M)P(C) = (0.40)(0.50) = 0.20.\]

But

\[P(M\cap C)=0.20.\]

Therefore,

\[P(M\cap C)=P(M)P(C).\]

The second check also shows independence.

Thus,

\[\boxed{M\text{ and }C\text{ are independent}}.\]

Part (d): Are the events mutually exclusive?

Mutually exclusive events cannot occur together.

If \(M\) and \(C\) were mutually exclusive, then

\[P(M\cap C)=0.\]

But here,

\[P(M\cap C)=0.20.\]

Forty students are in both events.

Therefore,

\[\boxed{M\text{ and }C\text{ are not mutually exclusive}}.\]

This example shows an important distinction:

independent events can occur together;
mutually exclusive events cannot occur together.

Note

Do not decide independence by asking whether the events overlap.

Independence asks whether knowing one event changes the probability of the other.

Question 4: Choosing One of Two Card Boxes

Question: One of two card boxes is selected.

Box \(B_1\) is selected with probability \(0.40\) and contains 3 red cards and 2 black cards.
Box \(B_2\) is selected with probability \(0.60\) and contains 1 red card and 4 black cards.

After a box is selected, one card is drawn from that box.

Let \(R\) be the event that the card is red.

Find:

\(P(R\mid B_1)\) and \(P(R\mid B_2)\);
the probability of selecting \(B_1\) and then drawing a red card;
the probability of selecting \(B_2\) and then drawing a red card; and
the overall probability \(P(R)\) of drawing a red card.

Solution:

Part (a): Conditional probability of red within each box

Box \(B_1\) contains 3 red cards among 5 cards.

Therefore,

\[P(R\mid B_1) = \frac{3}{5} = 0.60.\]

Box \(B_2\) contains 1 red card among 5 cards.

Therefore,

\[P(R\mid B_2) = \frac{1}{5} = 0.20.\]

Thus,

\[\boxed{P(R\mid B_1)=0.60,\qquad P(R\mid B_2)=0.20}.\]

These are conditional probabilities because the probability of red depends on which box was selected.

Part (b): Probability of the path \(B_1\rightarrow R\)

We want

\[P(B_1\cap R).\]

Use the multiplication rule:

\[P(B_1\cap R) = P(B_1)P(R\mid B_1).\]

Therefore,

\[P(B_1\cap R) = (0.40)(0.60) = 0.24.\]

Thus,

\[\boxed{P(B_1\cap R)=0.24}.\]

Part (c): Probability of the path \(B_2\rightarrow R\)

Similarly,

\[P(B_2\cap R) = P(B_2)P(R\mid B_2).\]

Therefore,

\[P(B_2\cap R) = (0.60)(0.20) = 0.12.\]

Thus,

\[\boxed{P(B_2\cap R)=0.12}.\]

Part (d): Find the overall probability of red

A red card can be obtained through two possible cases:

\[B_1\rightarrow R\]

or

\[B_2\rightarrow R.\]

The two boxes form a partition:

only one box is selected;
one of the two boxes must be selected.

Therefore, the two paths are mutually exclusive.

Add their probabilities:

\[P(R) = P(B_1\cap R) + P(B_2\cap R).\]

Thus,

\[P(R) = 0.24+0.12 = 0.36.\]

Therefore,

\[\boxed{P(R)=0.36}.\]

Using the total probability rule directly,

\[P(R) = P(B_1)P(R\mid B_1) + P(B_2)P(R\mid B_2).\]

Substituting the values,

\[P(R) = (0.40)(0.60) + (0.60)(0.20) = 0.36.\]

The structure is:

multiply within each case

then add all cases that lead to red.

Note

Do not simply calculate

\[P(R\mid B_1)+P(R\mid B_2).\]

The conditional probabilities must first be weighted by the probabilities of selecting their respective boxes.