An experiment consists of rolling a die until a 2 appears.
(i) How many elements of the sample space correspond to the event that the 2 appears on the $k$ th roll of the die?
(ii) How many elements of the sample space correspond to the event that the 2 appears not later than the $k$ th roll of the die?
In a through of a die there is 6 sample points.
(i) If 2 appears on the $k$ th roll of the die.
So, first $(k-1)$ roll have 5 outcomes each and $k$ th roll results 2 i.e., 1 outcome.
$\therefore$ Number of element of sample space correspond to the event that 2 appears on the $k$ th roll of the die $=5^{k-1}$
(ii) If we consider that 2 appears not later than $k$ th roll of the die, then it is possible that 2 comes in first throw i.e., 1 outcome.
If 2 does not appear in first throw, then outcomes will be 5 and 2 comes in second throw i.e., 1 outcome, possible outcome $=5 \times 1=5$
$$\begin{aligned} &\text { Similarly, if } 2 \text { does not appear in second throw and appears in third throw. }\\ &\begin{aligned} \therefore \quad \text { Possible outcomes } & =5 \times 5 \times 1 \\ \text { Given, } \quad \text { series } & =1+5+5 \times 5+5 \times 5 \times 5+\ldots+5^{k-1} \\ & =1+5+5^2+5^3+\ldots+5^{k-1} \\ & =\frac{1\left(5^k-1\right)}{5-1}=\frac{5^k-1}{4} \end{aligned} \end{aligned}$$
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)
$$\begin{aligned} & =P\left(E_3 \cup E_4\right)=P\left(E_3\right)+P\left(E_4\right) \\ & =0.31+0.26=0.57 \end{aligned}$$