A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find $P(G)$, where $G$ is the event that a number greater than 3 occurs on a single roll of the die.
It is given that, $2 \times$ Probability of even number $=$ Probability of odd number
$\Rightarrow \quad P(O)=2 P(E)$
$\Rightarrow \quad P(O): P(E)=2: 1$
$\therefore$ Probability of occurring odd number, $P(O)=\frac{2}{2+1}=\frac{2}{3}$
and probability of occurring 5each number, $$P(E)=\frac{1}{2+1}=\frac{1}{3}$$
Now, $G$ be the even that a number greater than 3 occur in a single roll of die. So, the possible outcomes are 4,5 and 6 out of which two are even and one odd.
$$\begin{aligned} \therefore \quad \text { Required probability } & =P(G)=2 \times P(E) \times P(O) \\ & =2 \times \frac{1}{3} \times \frac{2}{3}=\frac{4}{9} \end{aligned}$$
In a large metropolitan area, the probabilities are $0.87,0.36,0.30$ that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set or both kinds of sets. What is the probability that a family owns either anyone or both kinds of sets?
Let $E_1$ be the event that family own colour television set and $E_2$ be the event that family owns a black and white television set.
$$\begin{aligned} \text{It is given that,}\quad P\left(E_1\right) & =0.87 \\ P\left(E_2\right) & =0.36 \\ \text{and}\quad P\left(E_1 \cap E_2\right) & =0.30 \end{aligned}$$
We have to find probability that a family owns either anyone or both kind of sets i.e., $P\left(E_1 \cup E_2\right)$.
$$\begin{aligned} \text{Now,}\quad P\left(E_1 \cup E_2\right) & =P\left(E_1\right)+P\left(E_2\right)-P\left(E_1 \cap E_2\right) \quad \text{by addition theorem]}\\ & =0.87+0.36-0.30 \\ & =0.93 \end{aligned}$$
If $A$ and $B$ are mutually exclusive events, $P(A)=0.35$ and $P(B)=0.45$, then find
(i) $P\left(A^{\prime}\right)$
(ii) $P\left(B^{\prime}\right)$
(iii) $P(A \cup B)$
(iv) $P(A \cap B)$
(v) $P\left(A \cap B^{\prime}\right)$
(vi) $P\left(A^{\prime} \cap B^{\prime}\right)$
$$\begin{aligned} &\text { Since, it is given that, } A \text { and } B \text { are mutually exclusive events. }\\ &\begin{aligned} & \therefore \quad P(A \cap B)=0 \quad {[\because A \cap B=\phi]} \\ & \text { and } \quad P(A)=0.35, P(B)=0.45 \end{aligned} \end{aligned}$$
(i) $P\left(A^{\prime}\right)=1-P(A)=1-0.35=0.65$
(ii) $P\left(B^{\prime}\right)=1-P(B)=1-0.45=0.55$
(iii) $P(A \cup B)=P(A)+P(B)-P(A \cap B)=0.35+0.45-0=0.80$
(iv) $P(A \cap B)=0$
(v) $P\left(A \cap B^{\prime}\right)=P(A)-P(A \cap B)=0.35-0=0.35$
(vi) $P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}=1-P(A \cup B)=1-0.8=0.2$
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)$=P(A')=1-P(A)=1-\frac{4}{9}=\frac{5}{9}$