MTH331 Lecture Notes 04 - Conditional Probability and Multiplication Rule

Conditional Probability Law

• What is it? Conditional probability tells us the chance of an event happening given that another event has already happened.

• We write it as $P(A\,||\,B)$, which means "the probability of event A happening, given that event B has already happened."

• The formula is: $P(A\,||\,B) = \frac{P(A \text{ and } B)}{P(B)}$.

  • This means: (Probability of both A and B happening) divided by (Probability of B happening).

• Think of it like this: When you know event B has happened, your entire possible world (sample space) shrinks from all outcomes ($\Omega$) to just the outcomes where B happened.

  • So, we only consider the outcomes within B, and event A inside this new world becomes "A and B".

  • We then calculate the probability of "A and B" relative to the size of "B".

• The rules still apply: Conditional probabilities still follow the basic rules of probability:

  • It's never negative: The chance of an event happening can't be less than zero. $P(A||B) \ge 0.$

  • The whole space is certain: If B has happened, then the probability of any outcome within B is 100%. $P(\text{all outcomes}||B) = 1.$

  • Adding probabilities for separate events: If two events, $A<em>1$ and $A</em>2$, can't happen at the same time (they are "disjoint"), then the probability of either $A<em>1$ or $A</em>2$ happening, given B, is simply the sum of their individual conditional probabilities.
    $P(A<em>1||B) + P(A</em>2||B) = P((A<em>1 \text{ or } A</em>2) || B).$

Multiplication Rule

• What does it do? This rule helps us find the probability that multiple events all happen one after another.

• The formula for events $A<em>1, A</em>2, \ldots, A<em>n$ all happening is: $P(A</em>1 \text{ and } A<em>2 \text{ and } \ldots \text{ and } A</em>n) = P(A<em>1) \cdot P(A</em>2 \mid A<em>1) \cdot P(A</em>3 \mid A<em>1 \text{ and } A</em>2) \cdot \ldots \cdot P(A<em>n \mid A</em>1 \text{ and } \ldots \text{ and } A_{n-1}).$

  • In simpler words: It's the probability of the first event, multiplied by the probability of the second event given the first one happened, multiplied by the probability of the third event given the first two happened, and so on.

• How it works (A quick look at why):

  • You can break down the probability of multiple events happening step-by-step using conditional probability.

  • For example, $P(A<em>1 \text{ and } A</em>2) = P(A<em>1) \cdot P(A</em>2 \mid A_1)$.

  • Then, $P(A<em>1 \text{ and } A</em>2 \text{ and } A<em>3) = P(A</em>1 \text{ and } A<em>2) \cdot P(A</em>3 \mid A<em>1 \text{ and } A</em>2)$.

  • If you keep substituting like this, you get the long multiplication chain shown in the rule.

• Example: Drawing 3 cards (without putting them back) and none are hearts.

  • Let's say:

    • $A_1$ = "The first card is not a heart."

    • $A_2$ = "The second card is not a heart."

    • $A_3$ = "The third card is not a heart."

  • Step 1: Probability the first card isn't a heart.

    • There are 39 non-heart cards out of 52 total. So, $P(A_1) = \frac{39}{52}$.

  • Step 2: Probability the second card isn't a heart, given the first wasn't a heart.

    • Now there are 51 cards left, and 38 of them are not hearts. So, $P(A<em>2 \mid A</em>1) = \frac{38}{51}$.

  • Step 3: Probability the third card isn't a heart, given the first two weren't hearts.

    • Now there are 50 cards left, and 37 of them are not hearts. So, $P(A<em>3 \mid A</em>1 \text{ and } A_2) = \frac{37}{50}$.

  • Final Result: To get the probability that none of the three cards are hearts, we multiply these probabilities:

    • $P(A<em>1 \text{ and } A</em>2 \text{ and } A_3) = \frac{39}{52} \cdot \frac{38}{51} \cdot \frac{37}{50}$.

Examples Using the Multiplication Rule

• Example 1: Drawing 3 cards (no hearts)

  • This is the same as the example in the "Multiplication Rule" section.

  • We want the probability that the 1st card isn't a heart, AND the 2nd isn't a heart, AND the 3rd isn't a heart.

  • The calculation is:

    • $P(\text{1st not heart}) = \frac{39}{52}$

    • $P(\text{2nd not heart} \mid \text{1st not heart}) = \frac{38}{51}$

    • $P(\text{3rd not heart} \mid \text{1st and 2nd not heart}) = \frac{37}{50}$

  • Total Probability: $P(\text{none are hearts}) = \frac{39}{52} \cdot \frac{38}{51} \cdot \frac{37}{50}.$

• Example 2: Grouping Graduates and Undergraduates

  • Imagine you have 4 graduates and 12 undergraduates (total 16 people). You divide them into 4 groups of 4 people each.

  • Question: What's the chance that each of the 4 groups ends up with exactly one graduate?

  • Let's call the graduates $G<em>1, G</em>2, G<em>3, G</em>4$.

  • We look at this step-by-step:

    1. Place the first graduate ($G<em>1$): It doesn't matter where $G</em>1$ goes.

    2. Place the second graduate ($G<em>2$): For $G</em>2$ to be in a different group from $G<em>1$, there are 15 spots left in total, and 12 of them lead to a different group (since $G</em>1$ took up one spot in their group, leaving 3 spots in that group, and 12 spots in the other 3 groups). So, $P(G<em>2 \text{ in different group from } G</em>1) = \frac{12}{15}$.

    3. Place the third graduate ($G<em>3$): Now, $G</em>1$ and $G<em>2$ are in different groups. There are 14 spots remaining. For $G</em>3$ to be in a different group from both $G<em>1$ and $G</em>2$, there are 8 valid spots left (since $G<em>1$'s group has 3 spots left, $G</em>2$'s group has 3 spots left, leaving $14 - 3 - 3 = 8$ spots in the remaining two empty groups). So, $P(G<em>3 \text{ in different group from } G</em>1 \text{ and } G_2) = \frac{8}{14}$.

    4. Place the fourth graduate ($G<em>4$): $G</em>1, G<em>2, G</em>3$ are all in different groups. There are 13 spots left. For $G<em>4$ to be in a different group from all three, it must go into the last remaining group that doesn't have a graduate yet. There are 4 valid spots in that last group. So, $P(G</em>4 \text{ in different group from } G<em>1, G</em>2 \text{ and } G_3) = \frac{4}{13}$.

  • Total Probability: Multiply these chances together:

    • $P(\text{each group has a graduate}) = \frac{12}{15} \cdot \frac{8}{14} \cdot \frac{4}{13}.$

• Example 3: The Prize Ticket Draw

  • You have 5 tickets, and only one is a prize winner. 5 people take turns drawing one ticket each, without putting them back.

  • Question: Is this game fair? Does the person who draws first have a better chance to win than someone who draws later?

  • Challenge for you: Try to use the conditional probability and multiplication rule ideas to figure out the probability of winning for each position (1st person, 2nd person, etc.). You'll find it's surprisingly fair!