Multiplication Theorem on Probability: Formulas and Proof

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).