Conditional Probability: Solving Basic Problems

1. What is Conditional Probability?

At its heart, conditional probability is the chance of an event happening, given that another event has already occurred. This "given" part is crucial—it reduces our "universe" of possibilities, changing the calculation.

Imagine a bag of marbles. The probability of picking a red marble changes if you know that a blue marble has already been picked and not replaced.

The notation for conditional probability is $P(A|B)$, which we read as "the probability of event A happening, given that event B has happened."

• Event A is the event we are interested in.
• Event B is the condition that we know to be true.

When we know B has happened, we are no longer considering all possible outcomes. We are only looking at the outcomes where B is true. This is called reducing the sample space.

2. The Formula and Key Approaches

From our last lesson, we know the multiplication rule for dependent events:
$P(A\text{ and }B)=P(B)\times P(A|B)$
If we simply rearrange this formula to solve for $P(A|B)$, we get the formal definition of conditional probability:

$P(A|B)=\frac{P(A\text{ and }B)}{P(B)}$

Provided that $P(B)\ne 0$.

This formula is the algebraic foundation. However, for many CAT problems, you'll have two main ways to solve them:

1. Reduced Sample Space: Logically figuring out the new total outcomes.
2. Direct Formula Application: Using the formula above when probabilities are given.

Let's explore these.

Approach 1: Reduced Sample Space

This is the most intuitive method. You redefine your "total outcomes" based on the given condition.

Let's see this approach explained using a contingency table, which is a powerful tool for organizing information in probability questions.

As you read, notice how calculating P(mach learn is pred fashion | truth is fashion) was done by dividing the number of photos that are (pred fashion AND fashion) by the new total, which is just the total for the truth is fashion column.

Example:
A single fair die is rolled. What is the probability that the number is a 4, given that the outcome is an even number?

• Original Sample Space: {1, 2, 3, 4, 5, 6} (6 outcomes)
• Condition: The outcome is even.
• Reduced Sample Space: {2, 4, 6} (Now only 3 possible outcomes)
• Favorable Outcome: {4} (1 outcome)
• Conditional Probability: $P(\text{4 | even})=\frac{1}{3}$

This is often much faster than calculating $P(\text{4 and even})$ and $P(\text{even})$ separately and then dividing.

Approach 2: Direct Formula Application

Sometimes, the problem gives you probabilities directly, not counts. In these cases, using the formula is the only way to go.

Let's look at some CAT-specific examples that use both approaches.

Let's recap Example 3 from the reading:

• $P(\text{Passing both V and QA})=0.6$. This is $P(V\text{ and }QA)$.
• $P(\text{Passing V})=0.8$. This is $P(V)$.
• We need to find $P(\text{Passing QA | Passing V})$.

Here, we don't have counts, so we must use the formula:
$P(QA|V)=\frac{P(QA\text{ and }V)}{P(V)}=\frac{0.6}{0.8}=\frac{3}{4}=0.75$

3. Tree Diagrams for Sequential Events

For problems involving a sequence of events (like drawing two cards without replacement), a tree diagram is a fantastic way to organize all the conditional probabilities.

You start with branches for the first event. Then, from the end of each of those branches, you draw new branches for the second event. The probabilities on these second-stage branches are conditional.

Example: A bag has 3 red and 2 blue marbles. Two marbles are drawn without replacement. What is the probability the second marble is blue?

A tree diagram would look like this:

1. First Draw:

   • Path 1: Draw Red with $P(R_1)=3/5$.
   • Path 2: Draw Blue with $P(B_1)=2/5$.

2. Second Draw (Conditional):

   • If you drew Red first, the bag has 2R, 2B. So $P(B_2|R_1)=2/4$.
   • If you drew Blue first, the bag has 3R, 1B. So $P(B_2|B_1)=1/4$.

The question asks for $P(B_2)$. The second marble can be blue in two ways: (Red first, then Blue) OR (Blue first, then Blue).

• $P(R_1\text{ and }{B}_2)=P(R_1)\times P(B_2|R_1)=\frac{3}{5}\times \frac{2}{4}=\frac{6}{20}$
• $P(B_1\text{ and }{B}_2)=P(B_1)\times P(B_2|B_1)=\frac{2}{5}\times \frac{1}{4}=\frac{2}{20}$

Since these are "OR" events, we add the probabilities:
$P(B_2)=\frac{6}{20}+\frac{2}{20}=\frac{8}{20}=\frac{2}{5}$
Tree diagrams are extremely useful for organizing these sequential calculations and are foundational for more advanced topics like Bayes' theorem.

Conclusion

This lesson marks the end of our module on Combinatorics and Probability. You've built a strong foundation from basic counting principles all the way to handling conditional events.

Key Takeaways:

• Conditional Probability $P(A|B)$ is the probability of event A, given that event B has occurred.
• It operates on a reduced sample space, where the total possible outcomes are limited by the condition B.
• The governing formula is a rearrangement of the multiplication rule: $P(A|B)=\frac{P(A\text{ and }B)}{P(B)}$.
• For exam problems, you can often solve them faster by intuitively reducing the sample space rather than by mechanically applying the formula.
• Tree diagrams are an excellent tool for visualizing and solving problems that involve a sequence of dependent events.

Preview of the Next Lesson:
We will now transition to a new area of the CAT syllabus. In the next module, we'll begin our study of Plane Geometry. We will start with the building blocks: applying the properties of lines, angles, and triangles to find unknown values and solve geometric puzzles.