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Multiplication
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Rule in Probability

**Multiplication**Rule in Probability

If $A$ and $B$ are two independent events in a probability experiment, then the probability that both events occur simultaneously is:

$P\left(A\text{and}B\right)=P\left(A\right)\cdot P\left(B\right)$

In case of dependent events , the probability that both events occur simultaneously is:

$P\left(A\text{and}B\right)=P\left(A\right)\cdot P\left(B\text{\hspace{0.17em}}|\text{\hspace{0.17em}}A\right)$

(The notation $P\left(B\text{\hspace{0.17em}}|\text{\hspace{0.17em}}A\right)$ means "the probability of $B$ , given that $A$ has happened.")

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Example 1:
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You have a cowboy hat, a top hat, and an Indonesian hat called a songkok. You also have four shirts: white, black, green, and pink. If you choose one hat and one shirt at random, what is the probability that you choose the songkok and the black shirt?

The two events are independent events; the choice of hat has no effect on the choice of shirt.

There are three different hats, so the probability of choosing the songkok is $\frac{1}{3}$ . There are four different shirts, so the probability of choosing the black shirt is $\frac{1}{4}$ .

So, by the Multiplication Rule:

$\begin{array}{l}P\left(\text{songokandblackshirt}\right)=\frac{1}{3}\cdot \frac{1}{4}\\ =\frac{1}{12}\end{array}$

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Example 2:
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Suppose you take out two cards from a standard pack of cards one after another, without replacing the first card. What is probability that the first card is the ace of spades, and the second card is a heart?

The two events are dependent events because the first card is not replaced.

There is only one ace of spades in a deck of $52$ cards. So:

$P\left({\text{1}}^{\text{st}}\text{cardistheaceofspades}\right)=\frac{1}{52}$

If the ace of spaces is drawn first, then there are $51$ cards left in the deck, of which $13$ are hearts:

$P\left({2}^{\text{nd}}{\text{cardisaheart|1}}^{\text{st}}\text{cardistheaceofspades}\right)=\frac{13}{51}$

So, by the multiplication rule of probability, we have:

$\begin{array}{l}P\left(\text{aceofspades,thenaheart}\right)=\frac{1}{52}\cdot \frac{13}{51}\\ =\frac{13}{4\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}13\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}51}\\ =\frac{1}{204}\end{array}$

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