Apply the General Multiplication Rule

Lesson 1 of 2: When the Simple Product Fails

In this lesson:

• See why fails for dependent events
• State and apply the general rule

Advanced (+) standard — STEM-track scope.

What You Will Be Able to Do

By the end of this lesson, you should be able to:

1. State the general Multiplication Rule, both symmetric forms
2. Apply it to dependent events like draws without replacement
3. Explain why the second factor is a conditional probability

Two Kings Without Replacement: The Chance?

Draw two cards from a 52-card deck, without replacement.

What is ?

Try it before the next slide — what's your instinct for the calculation?

The Natural Move: Just Multiply Both Draws

The product rule from independence suggests:

This treats the two draws as if nothing changed between them. Did it?

After a King, the Deck Shrinks

One king is gone: 3 kings remain among 51 cards.

The Second Draw Depends on the First

The first king changed the second draw's probability:

• Before: kings
• After a king is drawn: kings

These events are dependent — the first outcome moves the second.

With Replacement Resets the Deck

The dependence comes from not replacing the card:

• Without replacement: deck shrinks → dependent
• With replacement: card returned, deck resets → independent

Is This Scenario Dependent or Independent?

You draw two names from a hat without putting the first back.

Does the first draw change the probability of the second?

Decide before advancing — this decides which rule you need.

The Corrected Factor Is Conditional

You fixed the second factor to — that's a conditional probability:

The rule we need is built from exactly this conditional factor.

The General Multiplication Rule, Both Forms

For any two events and :

• The second factor is conditional — after the first event
• Both symmetric forms give the same joint probability

Where the Rule Comes From

Start from the conditional-probability formula and rearrange:

Multiply both sides by — the rule is the conditional formula, solved for the joint.

Two Kings: (4/52)(3/51) = 1/221

Apply the rule to the without-replacement draw:

The conditional second factor, 3/51, makes it correct.

Update Both: 3 Kings Out of 51 Cards

A common slip: fixing the numerator but not the denominator.

• Kings drop: ✓
• Total also drops:

Say the full remaining deck — "3 kings out of 51" — before writing the fraction.

The Symmetric Form Gives the Same Answer

Condition the other way — second king first:

Both forms are two routes to one joint probability — use the easier conditional.

The Structure: First, Then Second Given First

Every application has the same shape:

• Probability of the first event
• Times the probability of the second, after the first

Your Turn: Apply From Given Probabilities

You're told and .

Find .

No cards this time — just apply the rule. Then sanity-check it's between 0 and 1.

Two Common Errors to Watch For

Using : applying the independent rule to dependent events
First ask: does the first event change the second?

Denominator stays 52: updating kings but not the total
Without replacement, both the favorable count and the total drop.

First, Times Second Given First

✓
✓ The second factor is conditional — after the first event
✓ Without replacement, update both the count and the total

Next: tree diagrams for sequences, and independence as the special case where the product rule returns.