If a card is drawn from a deck of 52 cards, then find the probability of getting a king or a heart or a red card.
$\because$ Number of possible event $=52$
and favourable events $=4$ king +13 heart +26 red $-13-2=28$
$\therefore \quad$ Required probability $=\frac{28}{52}=\frac{7}{13}$
A sample space consists of 9 elementary outcomes $E_1, E_2, \ldots, E_9$ whose probabilities are
$$\begin{aligned} & P\left(E_1\right)=P\left(E_2\right)=0.08, P\left(E_3\right)=P\left(E_4\right)=P\left(E_5\right)=0.1 \\ & P\left(E_6\right)=P\left(E_7\right)=0.2, P\left(E_8\right)=P\left(E_9\right)=0.07 \end{aligned}$$
Suppose $$A=\left\{E_1, E_5, E_8\right\}, B=\left\{E_2, E_5, E_8, E_9\right\}$$
(i) Calculate $P(A), P(B)$ and $P(A \cap B)$.
(ii) Using the addition law of probability, calculate $P(A \cup B)$.
(iii) List the composition of the event $A \cup B$ and calculate $P(A \cup B)$ by adding the probabilities of the elementary outcomes.
(iv) Calculate $P(\bar{B})$ from $P(B)$, also calculate $P(\bar{B})$ directly from the elementary outcomes of $\bar{B}$.
Given,
$$\begin{aligned} S & =\left\{E_1, E_2, E_3, E_4, E_5, E_6, E_7, E_8, E_9\right\} \\ A & =\left\{E_1, E_5, E_8\right\}, B=\left\{E_2, E_5, E_8, E_9\right\} \\ P\left(E_1\right) & =P\left(E_2\right)=0.08 \\ P\left(E_3\right) & =P\left(E_{4)}=P\left(E_5\right)=0.1\right. \\ P\left(E_6\right) & =P\left(E_7\right)=2, P\left(E_8\right)=P\left(E_9\right)=0.07 \end{aligned}$$
(i) $$\begin{aligned} P(A) & =P\left(E_1\right)+P\left(E_5\right)+P\left(E_8\right) \\ & =0.08+0.1+0.07=0.25 \end{aligned}$$
(ii)$$\begin{aligned} & P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ & \begin{aligned} \text { Now, } P(B) & =P\left(E_2\right)+P\left(E_5\right)+P\left(E_8\right)+P\left(E_9\right) \\ & =0.08+0.1+0.07+0.07=0.32 \\ A \cap B & =\left\{E_5, E_8\right\} \\ P(A \cap B) & =P\left(E_5\right)+P\left(E_8\right)=0.1+0.7=0.17 \end{aligned} \end{aligned}$$
On substituting these values in Eq.(i), we get
$$P(A \cup B)=0.25+0.32-0.17=0.40$$
(iii) $A \cup B=\left\{E_1, E_2, E_5, E_8, E_9\right\}$
$$\begin{aligned} P(A \cup B) & =P\left(E_1\right)+P\left(E_2\right)+P\left(E_5\right)+P\left(E_8\right)+P\left(E_9\right) \\ & =0.08+0.08+0.1+0.07+0.07=0.40 \end{aligned}$$
(iv) $$\begin{aligned} & \because P(\bar{B})=1-P(B)=1-0.32=0.68 \\ & \text { and } \quad \bar{B}=\left\{E_1, E_3, E_4, E_6, E_7\right\} \\ & \therefore \quad P(\bar{B})=P\left(E_1\right)+P\left(E_3\right)+P\left(E_4\right)+P\left(E_6\right)+P\left(E_7\right) \\ & =0.08+0.1+0.1+0.2+0.2=0.68 \end{aligned}$$
Determine the probability $p$, for each of the following events.
(i) An odd number appears in a single toss of a fair die.
(ii) Atleast one head appears in two tosses of a fair coin.
(iii) A king, 9 of hearts or 3 of spades appears in drawing a single card from a well shuffled ordinary deck of 52 cards.
(iv) The sum of 6 appears in a single toss of a pair of fair dice.
(i) When a die is throw the possible outcomes are
$$\begin{aligned} S & =\{1,2,3,4,5,6\} \text { out of which } 1,3,5 \text { are odd, } \\ \therefore \text { Required probability } & =\frac{3}{6}=\frac{1}{2} \end{aligned}$$
(ii) When a fair coin is tossed two times the sample space is
$$S=\{H H, H T, T H, T T\}$$
In at least one head favourable enonts are $H H, H T, T H$
$\therefore$ Required probability $=\frac{3}{4}$
(iii) Total cards $=52$
$$\begin{aligned} \text { Favourable } & =4 \text { king }+2 \text { of heart }+3 \text { of spade }=4+1+1=6 \\ \therefore \quad \text { Required probability } & =\frac{6}{52}=\frac{3}{26} \end{aligned}$$
(iv) When a pair of dice is rolled total sample parts are 36 . Out of which $(1,5),(5,1),(2,4)$, $(4,2)$ and $(3,3)$.
$\therefore$ Required probability $=\frac{5}{36}$
In a non-leap year, the probability of having 53 Tuesday or 53 Wednesday is
Three numbers are chosen from 1 to 20 . Find the probability that they are not consecutive