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The Multiplication Rule

The multiplication rule states that the probability that AAA and BBB both occur is equal to the probability that BBB occurs times the conditional probability that AAA occurs given that BBB occurs.

Learning Objective

  • Apply the multiplication rule to calculate the probability of both AAA and BBB occurring


Key Points

    • The multiplication rule can be written as: P(A∩B)=P(B)⋅P(A∣B)P(A \cap B) = P(B) \cdot P(A|B)P(A∩B)=P(B)⋅P(A∣B).
    • We obtain the general multiplication rule by multiplying both sides of the definition of conditional probability by the denominator.

Term

  • sample space

    The set of all possible outcomes of a game, experiment or other situation.


Full Text

The Multiplication Rule

In probability theory, the Multiplication Rule states that the probability that AAA and BBB occur is equal to the probability that AAA occurs times the conditional probability that BBB occurs, given that we know AAA has already occurred. This rule can be written:

P(A∩B)=P(B)⋅P(A∣B)\displaystyle P(A \cap B) = P(B) \cdot P(A|B)P(A∩B)=P(B)⋅P(A∣B)

Switching the role of AAA and BBB, we can also write the rule as:

P(A∩B)=P(A)⋅P(B∣A)\displaystyle P(A\cap B) = P(A) \cdot P(B|A)P(A∩B)=P(A)⋅P(B∣A)

We obtain the general multiplication rule by multiplying both sides of the definition of conditional probability by the denominator. That is, in the equation P(A∣B)=P(A∩B)P(B)\displaystyle P(A|B)=\frac{P(A\cap B)}{P(B)}P(A∣B)=​P(B)​​P(A∩B)​​, if we multiply both sides by P(B)P(B)P(B), we obtain the Multiplication Rule. 

The rule is useful when we know both P(B)P(B)P(B) and P(A∣B)P(A|B)P(A∣B), or both P(A)P(A)P(A) and P(B∣A).P(B|A).P(B∣A).

Example

Suppose that we draw two cards out of a deck of cards and let AAA be the event the the first card is an ace, and BBB be the event that the second card is an ace, then:

P(A)=452\displaystyle P(A)=\frac { 4 }{ 52 }P(A)=​52​​4​​

And:

P(B∣A)=351\displaystyle P\left( { B }|{ A } \right) =\frac { 3 }{ 51 }P(B∣A)=​51​​3​​

The denominator in the second equation is 515151 since we know a card has already been drawn. Therefore, there are 515151 left in total. We also know the first card was an ace, therefore:

P(A∩B)=P(A)⋅P(B∣A)=452⋅351=0.0045\displaystyle \begin{aligned} P(A \cap B) &= P(A) \cdot P(B|A)\\ &= \frac { 4 }{ 52 } \cdot \frac { 3 }{ 51 } \\ &=0.0045 \end{aligned}​P(A∩B)​​​​​=P(A)⋅P(B∣A)​=​52​​4​​⋅​51​​3​​​=0.0045​​

Independent Event

Note that when AAA and BBB are independent, we have that P(B∣A)=P(B)P(B|A)= P(B)P(B∣A)=P(B), so the formula becomes P(A∩B)=P(A)P(B)P(A \cap B)=P(A)P(B)P(A∩B)=P(A)P(B), which we encountered in a previous section. As an example, consider the experiment of rolling a die and flipping a coin. The probability that we get a 222 on the die and a tails on the coin is 16⋅12=112\frac{1}{6}\cdot \frac{1}{2} = \frac{1}{12}​6​​1​​⋅​2​​1​​=​12​​1​​, since the two events are independent.  

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