Explain why the experiment of tossing a coin three times is said to have Binomial distribution.
We know that, a random variable $X$ taking values $0,1,2, \ldots, n$ is said to have a binomial distribution with parameters $n$ and $P$, if its probability distribution is given by
$$\begin{aligned} P(X & =r)={ }^n C_r p^{\prime} q^{n-r} \\ \text{where,}\quad q & =1-p \\ \text{and}\quad r & =0,1,2, \ldots, n \end{aligned}$$
Similarly, in an experiment of tossing a coin three times, we have $n=3$ and random variable $X$ can take values $r=0,1,2$ and 3 with $p=\frac{1}{2}$ and $q=\frac{1}{2}$
| X | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P(X) | ${ }^3 C_0 q^3$ | ${ }^3 C_1 P q^2$ | ${ }^3 C_2 p^2 q$ | ${ }^3 C_3 P^3$ |
So, we see that in the experiment of tossing a coin three times, we have random variable $X$ which can take values $0,1,2$ and 3 with parameters $n=3$ and $P=\frac{1}{2}$. Therefore, it is said to have a Binomial distribution.
If $A$ and $B$ are two events such that $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}$ and $P(A \cap B)=\frac{1}{4}$, then find
(i) $P(A / B)$.
(ii) $P(B / A)$.
(iii) $P\left(A^{\prime} / B\right)$.
(iv) $P\left(A^{\prime} / B^{\prime}\right)$.
Here, $\quad P(A)=\frac{1}{2}, P(B)=\frac{1}{3}$ and $P(A \cap B)=\frac{1}{4}$
(i) $P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{1 / 4}{1 / 3}=\frac{3}{4}$
(ii) $P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{1 / 4}{1 / 2}=\frac{1}{2}$
(iii) $P\left(A^{\prime} / B\right)=1-P(A / B)=1-\frac{3}{4}=\frac{1}{4}$
$$\begin{aligned} &\text { or } P\left(A^{\prime} / B\right)=\frac{P\left(A^{\prime} \cap B\right)}{P(B)}=\frac{P(B)-P(A \cap B)}{P(B)}=\frac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{3}}=\frac{\frac{1}{12}}{\frac{1}{3}}=\frac{1}{4}\\ &\begin{aligned} \text { (iv) } P\left(A^{\prime} / B^{\prime}\right)=\frac{P\left(A^{\prime} \cap B^{\prime}\right)}{P\left(B^{\prime}\right)} & =\frac{1-P(A \cup B)}{1-P(B)}=\frac{1-[P(A)+P(B)-P(A \cap B)]}{1-P(B)} \\ & =\frac{1-\left[\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right]}{1-\frac{1}{3}}=\frac{1-\left(\frac{5}{6}-\frac{1}{4}\right)}{\frac{2}{3}} \\ & =\frac{1-14 / 24}{2 / 3}=\frac{10 / 24}{2 / 3}=\frac{30}{48}=\frac{5}{8} \end{aligned} \end{aligned}$$
Three events A, B and C have probabilities $\frac{2}{5}, \frac{1}{3}$ and $\frac{1}{2}$, respectively. If, $P(A \cap C)=\frac{1}{5}$ and $P(B \cap C)=\frac{1}{4}$, then find the values of $P(C / B)$ and $P\left(A^{\prime} \cap C^{\prime}\right)$.
Here, $P(A)=\frac{2}{5}, P(B)=\frac{1}{3}, P(C)=\frac{1}{2}, P(A \cap C)=\frac{1}{5}$ and $P(B \cap C)=\frac{1}{4}$
$$\begin{aligned} & \therefore \quad P(C / B)=\frac{P(B \cap C)}{P(B)}=\frac{1 / 4}{1 / 3}=\frac{3}{4} \\ & \text { and } \\ & P\left(A^{\prime} \cap C^{\prime}\right)=1-P(A \cup C)=1-[P(A)+P(C)-P(A \cap C)] \\ & =1-\left[\frac{2}{5}+\frac{1}{2}-\frac{1}{5}\right]=1-\left[\frac{4+5-2}{10}\right]=1-\frac{7}{10}=\frac{3}{10} \end{aligned}$$
Let $E_1$ and $E_2$ be two independent events such that $P\left(E_1\right)=P_1$ and $P\left(E_2\right)=P_2$. Describe in words of the events whose probabilities are
(i) $P_1 P_2$ (ii) $\left(1-P_1\right) P_2$ (iii) $1-\left(1-P_1\right)\left(1-P_2\right)$ (iv) $P_1+P_2-2 P_1 P_2$
$$P\left(E_1\right)=P_1 \text { and } P\left(E_2\right)=P_2$$
(i) $P_1 P_2 \Rightarrow P\left(E_1\right) \cdot P\left(E_2\right)=P\left(E_1 \cap E_2\right)$
So, $E_1$ and $E_2$ occur.
(ii) $\left(1-P_1\right) P_2=P\left(E_1\right)^{\prime} \cdot P\left(E_2\right)=P\left(E_1^{\prime} \cap E_2\right)$
So, $E_1$ does not occur but $E_2$ occurs.
(iii) $$\begin{aligned} 1-\left(1-P_1\right)\left(1-P_2\right) & =1-P\left(E_1\right)^{\prime} P\left(E_2\right)^{\prime}=1-P\left(E_1^{\prime} \cap E_2^{\prime}\right) \\ & =1-\left[1-P\left(E_1 \cup E_2\right)\right]=P\left(E_1 \cup E_2\right) \end{aligned}$$
So, either $E_1$ or $E_2$ or both $E_1$ and $E_2$ occurs.
(iv) $$\begin{aligned} P_1+P_2-2 P_1 P_2 & =P\left(E_1\right)+P\left(E_2\right)-2 P\left(E_1\right) \cdot P\left(E_2\right) \\ & =P\left(E_1\right)+P\left(E_2\right)-2 P\left(E_1 \cap E_2\right) \\ & =P\left(E_1 \cup E_2\right)-P\left(E_1 \cap E_2\right) \end{aligned}$$
So, either $E_1$ or $E_2$ occurs but not both.
A discrete random variable X has the probability distribution as given below
| X | 0.5 | 1 | 1.5 | 2 |
|---|---|---|---|---|
| P(X) | $k$ | $k^2$ | $2k^2$ | $k$ |
(i) Find the value of $k$.
(ii) Determine the mean of the distribution.
We have,
| X | 0.5 | 1 | 1.5 | 2 |
|---|---|---|---|---|
| P(X) | $k$ | $k^2$ | $2k^2$ | $k$ |
$$\begin{aligned} &\text { (i) We know that, } \quad \sum_{i=1}^n P_i=1 \text {, where } P_i \geq 0\\ &\begin{array}{lr} \Rightarrow & P_1+P_2+P_3+P_4=1 \\ \Rightarrow & k+k^2+2 k^2+k=1 \\ \Rightarrow & 3 k^2+2 k-1=0 \\ \Rightarrow & 3 k^2+3 k-k-1=0 \\ \Rightarrow & 3 k(k+1)-1(k+1)=0 \\ \Rightarrow & (3 k-1)(k+1)=0 \\ \Rightarrow & k=1 / 3 \Rightarrow k=-1 \\ \text { Since, } & k \text { is } \geq 0 \Rightarrow k=1 / 3 \end{array} \end{aligned}$$
(ii) Mean of the distribution $(\mu)=E(X)=\sum_{i=1}^n x_i P_i$
$$ \begin{aligned} & =0.5(k)+1\left(k^2\right)+1.5\left(2 k^2\right)+2(k)=4 k^2+2.5 k \\ & =4 \cdot \frac{1}{9}+2.5 \cdot \frac{1}{3} \quad \left[\because k=\frac{1}{3}\right]\\ & =\frac{4+7.5}{9}=\frac{23}{18} \end{aligned}$$