A team of medical students doing their internship have to assist during surgeries at a city hospital. The probabilities of surgeries rated as very complex, complex, routine, simple or very simple are respectively, 0.15 , $0.20,0.31,0.26$ and 0.08 . Find the probabilities that a particular surgery will be rated
(i) complex or very complex.
(ii) neither very complex nor very simple.
(iii) routine or complex.
(iv) routine or simple.
Let $E_1, E_2, E_3, E_4$ and $E_5$ be the event that surgeries are rated as very complex, complex, routine, simple or very simple, respectively.
$$\therefore \quad P\left(E_1\right)=0.15, P\left(E_2\right)=0.20, P\left(E_3\right)=0.31, P\left(E_4\right)=0.26, P\left(E_5\right)=0.08$$
$$\begin{aligned} & \text { (i) } P \text { (complex or very complex })=P\left(E_1 \text { or } E_2\right) \\ & =P\left(E_1 \cup E_2\right)=P\left(E_1\right)+P\left(E_2\right)-P\left(E_1 \cap E_2\right) \\ & =0.15+0.20-0\left[P\left(E_1 \cap E_2\right)=0\right. \\ & \text { because all events are independent] } \\ & =0.35 \end{aligned}$$
(ii) $P$ (neither very complex nor very simple), $\left(P\left(E_1^{\prime} \cap E_5^{\prime}\right)=P\left(E_1 \cup E_5\right)^{\prime}\right.$
$$\begin{aligned} & =1-P\left(E_1 \cup E_5\right) \\ & =1-\left[P\left(E_1\right)+P\left(E_5\right)\right] \\ & =1-(0.15+0.08) \\ & =1-0.23 \\ & =0.77 \end{aligned}$$
(iii) $P$ (routine or complex) $=P\left(E_3 \cup E_2\right)=P\left(E_3\right)+P\left(E_2\right)$
$$=0.31+0.20=0.51$$
(iv) $P$ (routine or simple)
$$\begin{aligned} & =P\left(E_3 \cup E_4\right)=P\left(E_3\right)+P\left(E_4\right) \\ & =0.31+0.26=0.57 \end{aligned}$$
Four candidates $A, B, C$ and $D$ have applied for the assignment to coach a school cricket team. If $A$ is twice as likely to be selected as $B$ and $B$ and $C$ are given about the same chance of being selected, while $C$ is twice as likely to be selected as $D$, then what are the probabilities that
(i) $C$ will be selected?
(ii) A will not be selected?
It is given that $A$ is twice as likely to be selected as $D$.
$$\begin{array}{ll} & P(A)=2 P(B) \\ \Rightarrow \quad & \frac{P(A)}{2}=P(B) \end{array}$$
while $C$ is twice as likely to be selected as $D$.
$$\begin{aligned} & P(C)=2 P(D) \Rightarrow P(B)=2 P(D) \\ & \frac{P(A)}{2}=2 P(D) \Rightarrow P(D)=\frac{P(A)}{4} \end{aligned}$$
$B$ and $C$ are given about the same chance of being selected.
$$P(B)=P(C)$$
$$\begin{aligned} &\text { Now, } \quad \text { sum of probability }=1\\ &\begin{array}{l} P(A)+P(B)+P(C)+P(D) =1 \\ P(A)+\frac{P(A)}{2}+P \frac{(A)}{2}+\frac{P(A)}{4} =1 \\ \Rightarrow \quad \frac{4 P(A)+2 P(A)+2 P(A)+P(A)}{4} =1 \\ \Rightarrow \quad 9 P(A) =4 \Rightarrow P(A)=\frac{4}{9} \end{array} \end{aligned}$$
$$\begin{aligned} \text { (i) } \quad P(C \text { will be selected })=P(C)=P(B)= & \frac{P(A)}{2} \\ & =\frac{4}{9 \times 2} \quad\left[\because P(A)=\frac{9}{4}\right] \\ & =\frac{2}{9} \end{aligned}$$
(ii) $P(A$ will not be selected)$=P(A')=1-P(A)=1-\frac{4}{9}=\frac{5}{9}$
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 $=\frac{8}{12}=\frac{2}{3}$