The probability that atleast one of the two events $A$ and $B$ occurs is 0.6 . If $A$ and $B$ occur simultaneously with probability 0.3 , evaluate $P(\bar{A})+P(\bar{B})$
$$\begin{array}{ll} \text { Thus, } & P(A \cup B)=0.6 \text { and } P(A \cap B)=0.3 \\ \text { Also, } & P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ \Rightarrow & 0.6=P(A)+P(B)-0.3 \end{array}$$
$$\begin{array}{lrr} \Rightarrow & P(A)+P(B)=0.9 & \\ \Rightarrow & {[1-P(\bar{A})]+[1-P(\bar{B})]=0.9} & {[\because P(A)=1-P(\bar{A}) \text { and } P(B)=1-P(\bar{B})]} \\ \Rightarrow & P(\bar{A})+P(\bar{B})=2-0.9=1.1 & \end{array}$$
A bag contains 5 red marbles and 3 black marbles. Three marbles are drawn one by one without replacement. What is the probability that atleast one of the three marbles drawn be black, if the first marble is red?
Let $R=\{5$ red marbles $\}$ and $B=\{3$ black marbles $\}$
For atleast one of the three marbles drawn be black, if the first marble is red, then the following three conditions will be followed
(i) Second ball is black and third is red $\left(E_1\right)$.
(ii) Second ball is black and third is also black ( $E_2$ ).
(iii) Second ball is red and third is black $\left(E_3\right)$.
$$\begin{aligned} \therefore & P\left(E_1\right)=P\left(R_1\right) \cdot P\left(B_1 / R_1\right) \cdot P\left(R_2 / R_1 B_1\right)=\frac{5}{8} \cdot \frac{3}{7} \cdot \frac{4}{6}=\frac{60}{336}=\frac{5}{28} \\ & P\left(E_2\right)=P\left(R_1\right) \cdot P\left(B_1 / R_1\right) \cdot P\left(B_2 / R_1 B_1\right)=\frac{5}{8} \cdot \frac{3}{7} \cdot \frac{2}{6}=\frac{30}{336}=\frac{5}{56} \end{aligned}$$
$$ \begin{array}{ll} \text { and } & P\left(E_3\right)=P\left(R_1\right) \cdot P\left(R_2 / R_1\right) \cdot P\left(B_1 / R_1 R_2\right)=\frac{5}{8} \cdot \frac{4}{7} \cdot \frac{3}{6}=\frac{60}{336}=\frac{5}{28} \\ \therefore & P(E)=P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)=\frac{5}{28}+\frac{5}{56}+\frac{5}{28} \end{array}$$
$=\frac{10+5+10}{56}=\frac{25}{56}$
Two dice are thrown together and the total score is noted. The events $E$, $F$ and $G$ are 'a total of 4 ', 'a total of 9 or more' and 'a total divisible by 5 ', respectively. Calculate $P(E), P(F)$ and $P(G)$ and decide which pairs of events, if any are independent.
$$\begin{aligned} &\text { Two dice are thrown together i.e., sample space }(S)=36 \Rightarrow n(S)=36\\ &E=\text { A total of } 4=\{(2,2),(3,1),(1,3)\} \end{aligned}$$
$$\begin{array}{l} \Rightarrow & n(E) =3 \\ & F =\text { A total of } 9 \text { or more } \\ & =\{(3,6),(6,3),(4,5),(4,6),(5,4),(6,4),(5,5),(5,6),(6,5),(6,6)\} \\ \Rightarrow & n(F) =10 \\ \end{array}$$
$G=$ a total divisible by $5=\{(1,4),(4,1),(2,3),(3,2),(4,6),(6,4),(5,5)\}$
$\Rightarrow \quad n(G)=7$
$$\begin{array}{lc} \text { Here, } & (E \cap F)=\phi \text { and }(E \cap G)=\phi \\ \text { Also, } & (F \cap G)=\{(4,6),(6,4),(5,5)\} \\ \Rightarrow & n(F \cap G)=3 \text { and }(E \cap F \cap G)=\phi \\ \therefore & P(E)=\frac{n(E)}{n(S)}=\frac{3}{36}=\frac{1}{12} \end{array}$$
$$\begin{aligned} P(F) & =\frac{n(F)}{n(S)}=\frac{10}{36}=\frac{5}{18} \\ P(G) & =\frac{n(G)}{n(S)}=\frac{7}{36} \\ P(F \cap G) & =\frac{3}{36}=\frac{1}{12} \\ \text{and}\quad P(F) \cdot P(G) & =\frac{5}{18} \cdot \frac{7}{36}=\frac{35}{648} \end{aligned}$$
Here, we see that $P(F \cap G) \neq P(F) \cdot P(G)$
[since, only $F$ and $G$ have common events, so only $F$ and $G$ are used here]
Hence, there is no pair which is independent.
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) $P\left(A^{\prime} / B\right)=1-P(A / B)=1-\frac{3}{4}=\frac{1}{4}$
$$\begin{aligned} &\text { or } P\left(A^{\prime} / B\right)=\frac{P\left(A^{\prime} \cap B\right)}{P(B)}=\frac{P(B)-P(A \cap B)}{P(B)}=\frac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{3}}=\frac{\frac{1}{12}}{\frac{1}{3}}=\frac{1}{4}\\ &\begin{aligned} \text { (iv) } P\left(A^{\prime} / B^{\prime}\right)=\frac{P\left(A^{\prime} \cap B^{\prime}\right)}{P\left(B^{\prime}\right)} & =\frac{1-P(A \cup B)}{1-P(B)}=\frac{1-[P(A)+P(B)-P(A \cap B)]}{1-P(B)} \\ & =\frac{1-\left[\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right]}{1-\frac{1}{3}}=\frac{1-\left(\frac{5}{6}-\frac{1}{4}\right)}{\frac{2}{3}} \\ & =\frac{1-14 / 24}{2 / 3}=\frac{10 / 24}{2 / 3}=\frac{30}{48}=\frac{5}{8} \end{aligned} \end{aligned}$$