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.

.. list-table::
   :header-rows: 1
   :widths: 30 25 25 20

   * - 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

- :math:`O` = the student studied online;
- :math:`P` = the student passed the review test.

Find:

(a) :math:`P(P)`;

(b) :math:`P(O\mid P)`;

(c) :math:`P(P\mid O)`; and

(d) explain why :math:`P(O\mid P)` and :math:`P(P\mid O)` are different.

**Solution:**
~~~~~~~~~~~~~

Part (a): Find :math:`P(P)`
^^^^^^^^^^^^^^^^^^^^^^^^^^^^

There are 240 students in total.

From the table, 156 students passed.

Therefore,

.. math::

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

Thus,

.. math::

   \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 :math:`P(O\mid P)`
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

The notation

.. math::

   P(O\mid P)

means:

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

The condition is :math:`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,

.. math::

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

Therefore,

.. math::

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

Using the conditional probability formula gives the same result:

.. math::

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

Here,

.. math::

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

and

.. math::

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

Therefore,

.. math::

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

Part (c): Find :math:`P(P\mid O)`
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Now the condition changes.

The notation

.. math::

   P(P\mid O)

means:

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

The condition is now :math:`O`.

There are 120 online students.

Among them, 72 passed.

Therefore,

.. math::

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

Thus,

.. math::

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

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

We found

.. math::

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

but

.. math::

   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

.. math::

   P(O\mid P),

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

For

.. math::

   P(P\mid O),

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

Therefore,

.. math::

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

- :math:`A_1` = the first card is a star;
- :math:`A_2` = the second card is a star.

Find:

(a) :math:`P(A_1)`;

(b) :math:`P(A_2\mid A_1)`;

(c) the probability that both cards are stars; and

(d) 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,

.. math::

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

Thus,

.. math::

   \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,

.. math::

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

Thus,

.. math::

   \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:

.. math::

   A_1\cap A_2.

Use the multiplication rule:

.. math::

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

Substitute the values:

.. math::

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

Therefore,

.. math::

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

Thus,

.. math::

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

The calculation follows the sequence of the experiment:

    first star
    :math:`\longrightarrow`
    second star given the first star.

Part (d): Probability of exactly one star
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

``Exactly one star`` can happen in two different ways:

1. first card is a star and second card is not a star;
2. 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**

.. math::

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

Why is the second probability :math:`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

.. math::

   \frac{15}{20}.

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

Therefore,

.. math::

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

Add the two mutually exclusive paths:

.. math::

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

Thus,

.. math::

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

Therefore,

.. math::

   \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

- :math:`M` = attends the morning review session;
- :math:`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:

(a) :math:`P(M)`, :math:`P(C)`, and :math:`P(M\cap C)`;

(b) :math:`P(C\mid M)`;

(c) determine whether :math:`M` and :math:`C` are independent, using two different checks; and

(d) determine whether :math:`M` and :math:`C` are mutually exclusive.

**Solution:**
~~~~~~~~~~~~~

Part (a): Find the three probabilities
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

There are 200 students in total.

For the morning review session,

.. math::

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

For bringing a calculator,

.. math::

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

For doing both,

.. math::

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

Therefore,

.. math::

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

Part (b): Find :math:`P(C\mid M)`
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

The condition is :math:`M`.

There are 80 students in the morning review group.

Among these 80 students, 40 bring a calculator.

Therefore,

.. math::

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

Thus,

.. math::

   \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

.. math::

   P(C\mid M)=0.50.

We also know

.. math::

   P(C)=0.50.

Therefore,

.. math::

   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 :math:`M` and :math:`C` are independent.

**Check 2: Product condition**

For independent events, we should have

.. math::

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

Calculate the product:

.. math::

   P(M)P(C)
   =
   (0.40)(0.50)
   =
   0.20.

But

.. math::

   P(M\cap C)=0.20.

Therefore,

.. math::

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

The second check also shows independence.

Thus,

.. math::

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

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

Mutually exclusive events cannot occur together.

If :math:`M` and :math:`C` were mutually exclusive, then

.. math::

   P(M\cap C)=0.

But here,

.. math::

   P(M\cap C)=0.20.

Forty students are in both events.

Therefore,

.. math::

   \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 :math:`B_1` is selected with probability :math:`0.40` and contains 3 red cards and 2 black cards.
- Box :math:`B_2` is selected with probability :math:`0.60` and contains 1 red card and 4 black cards.

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

Let :math:`R` be the event that the card is red.

Find:

(a) :math:`P(R\mid B_1)` and :math:`P(R\mid B_2)`;

(b) the probability of selecting :math:`B_1` and then drawing a red card;

(c) the probability of selecting :math:`B_2` and then drawing a red card; and

(d) the overall probability :math:`P(R)` of drawing a red card.

**Solution:**
~~~~~~~~~~~~~

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

Box :math:`B_1` contains 3 red cards among 5 cards.

Therefore,

.. math::

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

Box :math:`B_2` contains 1 red card among 5 cards.

Therefore,

.. math::

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

Thus,

.. math::

   \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 :math:`B_1\rightarrow R`
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

We want

.. math::

   P(B_1\cap R).

Use the multiplication rule:

.. math::

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

Therefore,

.. math::

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

Thus,

.. math::

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

Part (c): Probability of the path :math:`B_2\rightarrow R`
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Similarly,

.. math::

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

Therefore,

.. math::

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

Thus,

.. math::

   \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:

.. math::

   B_1\rightarrow R

or

.. math::

   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:

.. math::

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

Thus,

.. math::

   P(R)
   =
   0.24+0.12
   =
   0.36.

Therefore,

.. math::

   \boxed{P(R)=0.36}.

Using the total probability rule directly,

.. math::

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

Substituting the values,

.. math::

   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

   .. math::

      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.