One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently the sample space consists of four elementary outcomes $S=\{$ John promoted, Rita promoted, Aslam promoted, Gurpreet promoted\}. You are told that the chances of John's promotion is same as that of Gurpreet Rita's chances of promotion are twice as likely as Johns. Aslam's chances are four times that of John.
(i) Determine
$P$ (John promoted), $\quad P$ (Rita promoted), $P$ (Aslam promoted), $\quad P$ (Gurpreet promoted).
(ii) If $A=\{$ John promoted or Gurpreet promoted $\}$, find $P(A)$
Let $E_1=$ John promoted
$E_2=$ Rita promoted
$E_3=$ Aslam promoted
$E_4=$ Gurpreet promoted
Given, sample space, S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}
i.e., $$S=\left\{E_1, E_2, E_3, E_4\right\}$$
It is given that, chances of John's promotion is same as that of Gurpreet.
$$P\left(E_1\right)=P\left(E_4\right)$$
Rita's chances of promotion are twice as likely as John.
$$P\left(E_2\right)=2 P\left(E_1\right)$$
$$\begin{aligned} &\text { And Aslam's chances of promotion are four times that of John. }\\ &\begin{array}{l} & P\left(E_3\right) =4 P\left(E_1\right) \\ \text { Now, } & P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)+P\left(E_4\right) =1 \\ \Rightarrow & P\left(E_1\right)+2 P\left(E_1\right)+4 P\left(E_1\right)+P\left(E_1\right) =1 \\ \Rightarrow & 8 P\left(E_1\right) =1 \\ \therefore & P\left(E_1\right)=\frac{1}{8} \end{array} \end{aligned}$$
$$\begin{aligned} & \text { (i) } P(\text { John promoted })=P\left(E_1\right)=\frac{1}{8} \\ & P(\text { Rita promoted })=P\left(E_2\right)=2 P\left(E_1\right)=2 \times \frac{1}{8}=\frac{2}{8}=\frac{1}{4} \\ & P(\text { Aslam promoted })=P\left(E_3\right)=4 P\left(E_1\right)=4 \times \frac{1}{8}=\frac{1}{2} \\ & P(\text { Gurpreet promoted })=P\left(E_4\right)=P\left(E_1\right)=\frac{1}{8} \end{aligned}$$
$$\begin{aligned} &\text { (ii) } A=\text { John promoted or Gurpreet promoted }\\ &\begin{aligned} & A=E_1 \cup E_4 \\ & P(A)=P\left(E_1 \cup E_4\right)=P\left(E_1\right)+P\left(E_4\right)-P\left(E_1 \cap E_4\right) \\ & =\frac{1}{8}+\frac{1}{8}-0 \quad {\left[\because P\left(E_1 \cap E_4\right)=0\right]} \\ & =\frac{2}{8}=\frac{1}{4} \end{aligned} \end{aligned}$$
The accompanying Venn diagram shows three events, $A, B$ and $C$ and also the probabilities of the various intersections [for instance, $P(A \cap B)=0.7$ ]. Determine
(i) $P$ (A)
(ii) $P(B \cap \bar{C})$
(iii) $P(A \cup B)$
(iv) $P(A \cap \bar{B})$
(v) $P(B \cap C)$
(vi) Probability of exactly one of the three occurs.
From the above Venn diagram,
(i) $P(A)=0.13+0.07=0.20$
(ii) $P(B \cap \bar{C})=P(B)-P(B \cap C)=0.07+0.10+0.15-0.15=0.07+0.10=0.17$
$$\begin{aligned} \text{(iii)}\quad P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\ & =0.13+0.07+0.07+0.10+0.15-0.07 \\ & =0.13+0.07+0.10+0.15=0.45 \end{aligned}$$
(iv) $P(A \cap \bar{B})=P(A)-P(A \cap B)=0.13+0.07-0.07=0.13$
(v) $P(B \cap C)=0.15$
(vi) $P$ (exactly one of the three occurs) $=0.13+0.10+0.28=0.51$
One urn contains two black balls (labelled $B_1$ and $B_2$ ) and one white ball. A second urn contains one black ball and two white balls (labelled $W_1$ and $W_2$ ). Suppose the following experiment is performed. One of the two urns is chosen at random. Next a ball is randomly chosen from the urn. Then, a second ball is chosen at random from the same urn without replacing the first ball.
(i) Write the sample space showing all possible outcomes.
(ii) What is the probability that two black balls are chosen?
(iii) What is the probability that two balls of opposite colour are chosen?
It is given that one of the two urn is chosen, then a ball is randomly chosen from the urn, then a second ball is chosen at random from the same urn without replacing the first ball.
(i) $\therefore$ Sample space $S=\left\{B_1 B_2, B_1 W, B_2 B_1, B_2 W, W B_1, W B_2, B W_1, B W_2, W_1 B, W_1 W_2, W_2 B, W_2 W_1\right\}$
$\therefore$ total sample point $=12$
(ii) If two black ball are chosen.
So, the favourable events are $B_1 B_2, B_2 B_1$ i.e., 2
$\therefore$ Required probability $=\frac{2}{12}=\frac{1}{6}$
(iii) If two balls of opposite colour are chosen.
So, the favourable events are $B_1 W_1, B_2 W_1, W B_1, W B_2, B W_1, B W_2, W_1 B_1, W_2$ Bi.e., 8 .
$\therefore$ Required probability $=\frac{8}{12}=\frac{2}{3}$
A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that
(i) all the three balls are white.
(ii) all the three balls are red.
(iii) one ball is red and two balls are white.
$\because$ Number of red balls $=8$
and number of white balls $=5$
$$\begin{aligned} &\begin{aligned} \text { (i) }\quad P(\text { all the three balls are white }) & =\frac{{ }^5 C_3}{{ }^{13} C_3}=\frac{\frac{5!}{3!2!}}{\frac{13!}{3!10!}}=\frac{5!}{3!2!} \times \frac{3!10!}{13!} \\ & =\frac{5 \times 4 \times 3 \times 2!}{2!} \times \frac{10!}{13 \times 12 \times 11 \times 10!}=\frac{5 \times 4 \times 3}{13 \times 12 \times 11} \\ & =\frac{5}{13 \times 11}=\frac{5}{143} \\ & =\frac{5 \times 4 \times 3}{13 \times 12 \times 11}=\frac{5}{13 \times 11}=\frac{5}{143} \end{aligned} \end{aligned}$$
(ii) P (all the three balls are red)
$$ \begin{aligned} & =\frac{{ }^8 C_3}{{ }^{13} C_3}=\frac{\frac{8!}{3!5!}}{\frac{13!}{3!10!}}=\frac{8!}{3!\times 5!} \times \frac{3!10!}{13!} \\ & =\frac{8 \times 7 \times 6 \times 5!}{5!} \times \frac{10!}{13 \times 12 \times 11 \times 10!} \\ & =\frac{8 \times 7 \times 6}{13 \times 12 \times 11}=\frac{28}{143} \end{aligned}$$
(iii) P(one ball is red and two balls are white)
$=\frac{{ }^8 C_1 \times{ }^5 C_2}{{ }^{13} C_3}=\frac{8 \times 10}{13 \times 6 \times 11}=\frac{40}{143}$
If the letters of the word 'ASSASSINATION' are arranged at random. Find the probability that
(i) four S's come consecutively in the word.
(ii) two I's and two N's come together.
(iii) all A's are not coming together.
(iv) no two A's are coming together.
Total number of letters in the word 'ASSASSINATION' are 13.
Out of which 3A's, 4S's, 2 I's, 2 N's, 1 T's and 10.
(i) If four S's come consecutively in the word, then we considers these 4 S's as 1 group. Now, the number of laters is 10 .
| S | S | S | S | A | A | A | I | I | N | N | T | O |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 | 9 | |||||||||||
Number of words when all S's are together $=\frac{10!}{3!2!2!}$
Total number of word using letters of the word 'ASSASSINATION'
$$\begin{aligned} \quad & =\frac{13!}{3!4!2!2!} \\ \therefore \quad \text { Required probability } & =\frac{10!}{\frac{3!2!2!\times 13!}{3!4!2!2!}} \\ & =\frac{10!\times 4!}{13!}=\frac{4!}{13 \times 12 \times 11}=\frac{24}{1716}=\frac{2}{143} \end{aligned}$$
(ii) If 2 I's and 2 N 's come together, then there as 10 alphabets.
Number of word when 2 l's and 2 N's are come together
$$\begin{aligned} & =\frac{10!}{3!4!} \times \frac{4!}{2!2!} \\ \therefore \quad \text { Required probability } & =\frac{\frac{10!4!}{3!4!2!2!}}{\frac{13!}{3!4!2!2!}}=\frac{4!10!}{2!2!3!4!} \times \frac{3!4!2!2!}{13!} \\ & =\frac{4!10!}{13!}=\frac{4!}{13 \times 12 \times 11}=\frac{24}{13 \times 12 \times 11}=\frac{2}{143} \end{aligned}$$
(iii) If all A's are coming together, then there are 11 alphabets.
Number of words when all A's come together
$$=\frac{11!}{4!2!2!}$$
Probability when all A's come together
$$=\frac{\frac{11!}{4!2!2!}}{\frac{13!}{4!3!2!2!}}=\frac{11!}{4!2!2!} \times \frac{4!3!2!2!}{13!}=\frac{11!\times 3!}{13!}=\frac{6}{13 \times 12}=\frac{1}{26}$$
Required probability when all A's does not come together
$$=1-\frac{1}{26}=\frac{25}{26}$$
(iv) If no two A's are together, then first we arrange the alphabets except A's.
| S | S | S | S | I | N | T | I | O | N |
All the alphabets except A's are arranged in $\frac{10!}{4!2!2!}$.
There are 11 vacant places between these alphabets.
So, 3 A's can be place in 11 places in ${ }^{11} C_3$ ways $=\frac{11!}{3!8!}$
$\therefore$ Total number of words when no two A's together
$$\begin{aligned} & =\frac{11!}{3!8!} \times \frac{10!}{4!2!2!} \\ \text { Required probability } & =\frac{11!\times 10!}{3!8!4!2!2!} \times \frac{4!3!2!2!}{13!}=\frac{10!}{8!\times 13 \times 12} \\ & =\frac{10 \times 9}{13 \times 12}=\frac{90}{156}=\frac{15}{26} \end{aligned}$$