Multiplication Theorem on Probability: Formulas and Proof

Probability is defined as the ratio of the number of favorable outcomes to the total number of outcomes. Multiplication Rule of probability is an important concept used for predicting the likelihood when two events occur. It is used to calculate joint probability. It is also known as the multiplication theorem of probability. In this probability of events is multiplied to get the likelihood of the events occurring together.

Multiplication Rule of Probability

The multiplication rule is closely linked to conditional probability. Conditional probability is a measure of the probability of one event occurring given that another event has already occurred. event $A$ given that $B$ has already occurred is written as $P(A \mid B), P(A / B)$ or $P\left(\frac{A}{B}\right)$.

The formula to calculate $P(A \mid B)$ is
$P(A \mid B)=\frac{P(A \cap B)}{P(B)}$ where $P(B)$ is greater than zero. written as $A B$.

The probability of event $A B$ or $A \cap B$ can be obtained by using the conditional probability.
The conditional probability of event $A$ given that $B$ has occurred is denoted by $P(A \mid B)$ and is given by

$
\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}, \mathrm{P}(\mathrm{B}) \neq 0
$
Using this result, we can write

$
\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{A} \mid \mathrm{B})
$
Also, we know that

$
\begin{aligned}
& P(B \mid A)=\frac{P(B \cap A)}{P(A)}, P(A) \neq 0 \\
\text { or } \quad & P(B \mid A)=\frac{P(A \cap B)}{P(A)}, P(A) \neq 0 \quad(\because A \cap B=B \cap A)
\end{aligned}
$

Thus, $\quad \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B} \mid \mathrm{A})$
Combining (1) and (2), we get

$
\begin{aligned}
\mathrm{P}(\mathrm{A} \cap \mathrm{B}) & =\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B} \mid \mathrm{A}) \\
& =\mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{A} \mid \mathrm{B}) \quad(\text { provided } \mathrm{P}(\mathrm{A}) \neq 0 \text { and } \mathrm{P}(\mathrm{B}) \neq 0)
\end{aligned}
$

The above result is known as the multiplication rule of probability.

Multiplication rule of probability for more than two events

If $A, B$, and $C$ are three events associated with sample space, then we have

$
\begin{aligned}
P(A \cap B \cap C) & =P(A) P(B \mid A) P(C \mid A \cap B) \\
& =P(A) P(B \mid A) P(C \mid A B)
\end{aligned}
$

Similarly, the multiplication rule of probability can be extended for four or more events.

Solved Example Based on Multiplication Rule:

Example 1: In a game, a man wins $Rs. 100$ if he gets $5$ or $6$ on a throw of a fair die and loses $Rs.50$ for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is:

1) $0$
2) $\frac{400}{3}$ gain
3) $\frac{400}{9}$ loss
4) $\frac{400}{3}$ loss

Solution
Let $A=$ getting $5$ or $6$
$B=$ not getting $5$ or $6$
So $P(A)=2 / 6=1 / 3$ and $P(B)=4 / 6=2 / 3$
Now as per the question, the following events are possible
$A$ or $BA$ or $BBA$ or $BBB$
Let us first find their probabilities

$
\begin{aligned}
& P(A)=1 / 3 \\
& P(B A)=P(B) \cdot P(A)=2 / 3 \cdot 1 / 3=2 / 9
\end{aligned}
$

Similarly $P(B B A)=4 / 27$ and $P(B B B)=8 / 27$
To get the expected value, let us first make a probability distribution

Expectation $=100 \times 1 / 3+50 \times 2 / 9+0 \times 4 / 27+(-150)^\times 8 / 27=0$
Hence, the answer is the option 1.

Example 2: Let $S=\left\{w_1, w_2, \ldots \ldots\right\}$ be the sample space associated to a random experiment. Let $P\left(w_n\right)=\frac{P\left(w_{n-1}\right)}{2}, n \geq 2$. Let $A=\{2 k+3 l: \boldsymbol{k}, l \in \mathbb{N}\}$ and $B=\left\{w_n: n \in A\right\}$. Then $P(B)$ is equal to
1) $\frac{3}{64}$
2) $\frac{1}{16}$
3) $\frac{1}{32}$
4) $\frac{3}{32}$

Solution
$
\begin{aligned}
&\begin{aligned}
& \mathrm{A}=\{5,7,8,9,10,11 \ldots \ldots\} . \\
& \mathrm{P}\left(\mathrm{W}_1\right)+\mathrm{P}\left(\mathrm{W}_2\right)+\mathrm{P}\left(\mathrm{W}_3\right)+\ldots \ldots \mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)=1 \\
& \mathrm{P}\left(\mathrm{W}_1\right)+\frac{\mathrm{P}\left(\mathrm{W}_1\right)}{2}+\frac{\mathrm{P}\left(\mathrm{W}_2\right)}{2^2}+\ldots .=1 \\
& \Rightarrow \mathrm{P}\left(\mathrm{W}_1\right) \cdot\left(\frac{1}{1-1 / 2}\right)=1 \\
& \mathrm{P}\left(\mathrm{W}_1\right)=\frac{1}{2} \quad \mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)=\frac{1}{2} \cdot\left(\frac{1}{2}\right)^{\mathrm{n}-1}=\frac{1}{2^{\mathrm{n}}} \\
& \because B=\left\{W_n: n \in A\right\} \\
& =\left\{\mathrm{W}_5, \mathrm{~W}_7, \mathrm{~W}_8, \ldots \ldots\right\} \\
& \mathrm{P}(\mathrm{B})=\mathrm{P}\left(\mathrm{W}_5\right)+\mathrm{P}\left(\mathrm{W}_7\right)+\mathrm{P}\left(\mathrm{W}_8\right)+\mathrm{P}\left(\mathrm{W}_9\right)+\mathrm{P}\left(\mathrm{W}_{10}\right)+\mathrm{P}\left(\mathrm{W}_{11}\right) \\
& =\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^8}+\ldots . \\
& =\frac{1}{32}+\frac{\frac{1}{2^7}}{1-\frac{1}{2}} \\
& =\frac{1}{32}+\frac{1}{2^7} \times 2 \\
& =\frac{1}{32}+\frac{1}{64}=\frac{2+1}{64}=\frac{3}{64}
\end{aligned}\\
&\mathrm{A}=\{5,7,8,9,10,11 \ldots \ldots .\\
&\mathrm{P}\left(\mathrm{W}_1\right)+\mathrm{P}\left(\mathrm{W}_2\right)+\mathrm{P}\left(\mathrm{W}_3\right)+\ldots \ldots \mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)=1\\
&\mathrm{P}\left(\mathrm{W}_1\right)+\frac{\mathrm{P}\left(\mathrm{W}_1\right)}{2}+\frac{\mathrm{P}\left(\mathrm{W}_2\right)}{2^2}+\ldots .=1\\
&\Rightarrow \mathrm{P}\left(\mathrm{W}_1\right) \cdot\left(\frac{1}{1-1 / 2}\right)=1\\
&\mathrm{P}\left(\mathrm{W}_1\right)=\frac{1}{2} \quad \mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)=\frac{1}{2} \cdot\left(\frac{1}{2}\right)^{\mathrm{n}-1}=\frac{1}{2^{\mathrm{n}}}\\
&\because B=\left\{W_n: n \in A\right\}\\
&=\left\{\mathrm{W}_5, \mathrm{~W}_7, \mathrm{~W}_8, \ldots \ldots\right\}\\
&\mathrm{P}(\mathrm{B})=\mathrm{P}\left(\mathrm{W}_5\right)+\mathrm{P}\left(\mathrm{W}_7\right)+\mathrm{P}\left(\mathrm{W}_8\right)+\mathrm{P}\left(\mathrm{W}_9\right)+\mathrm{P}\left(\mathrm{W}_{10}\right)+\mathrm{P}\left(\mathrm{W}_{11}\right)\\
&=\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^8}+\ldots .\\
&=\frac{1}{32}+\frac{\frac{1}{2^7}}{1-\frac{1}{2}}\\
&=\frac{1}{32}+\frac{1}{2^7} \times 2\\
&=\frac{1}{32}+\frac{1}{64}=\frac{2+1}{64}=\frac{3}{64}
\end{aligned}
$

Hence, the answer is option (1).

Example 3: Let $P(E)$ denote the probability of an event $E$. Given $P(A)=1, P(B)=\frac{1}{2}$ the values of $P(A / B)$ and $P(B / A)$ respectively are
1) $\frac{1}{4}, \frac{1}{2}$
2) $\frac{1}{2}, \frac{1}{4}$
3) $\frac{1}{2}, 1$
4) $1, \frac{1}{2}$

Solution
$\mathrm{P}(\mathrm{A})=1$
$\Rightarrow A$ is sure event
$\Rightarrow A$ will definitely occur
So $A$ is independent from $B$

Hence,

$
\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=1 \times \frac{1}{2}=\frac{1}{2}
$
So,

$
\left.\begin{array}{l}
P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=1 \\
P\left(\frac{B}{A}\right)=\frac{P(B \cap A)}{P(A)}=\frac{1}{2}
\end{array}\right\}
$

Hence, the answer is the option(4).

Example 4: An examination consists of two papers, Paper $1$ and Paper $2$. The probability of failing in Paper $1$ is $0.3$ and that in Paper $2$ is $0.2$ . Given that a student has failed in Paper $2$. the probability of failing in Paper $1$ is $0.6$ . The probability of a student failing in both the papers is
1) $0.5$
2) $0.18$
3) $ 0.12$
4) $0.06$

Solution
Let $A$ and $B$ be events of failing in paper $1 and paper 2 respectively.

$
\begin{aligned}
& \mathrm{P}(\mathrm{A})=0.3 \\
& \mathrm{P}(\mathrm{B})=0.2 \\
& \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=0.6
\end{aligned}
$
Required Probability:

$
\begin{aligned}
& =\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
& =\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \mathrm{P}(\mathrm{B})
\end{aligned}
$

Hence, the answer is the option (3).

Example 5: if $\mathrm{P}(\mathrm{X})=1 / 4, \mathrm{P}(\mathrm{Y})=1 / 3$ and $\mathrm{P}(\mathrm{X} \cap \mathrm{Y})=1 / 12$ the value of $\mathrm{P}(\mathrm{Y} / \mathrm{X})$ is
1) $\frac{1}{4}$
2) $\frac{4}{25}$
3) $\frac{1}{3}$
4) $\frac{29}{50}$

Solution

$
\begin{aligned}
& \mathrm{P}(\mathrm{X})=1 / 4 \\
& \mathrm{P}(\mathrm{Y})=1 / 3 \\
& \mathrm{P}(\mathrm{X} \cap \mathrm{Y})=1 / 12 \\
& \mathrm{P}(\mathrm{Y} / \mathrm{X})=\frac{\mathrm{P}(\mathrm{X} \cap \mathrm{Y})}{\mathrm{P}(\mathrm{X})} \\
& \quad=\frac{1 / 12}{1 / 4}=1 / 3
\end{aligned}
$
Hence, the answer is the option (3).