Theoretical Probability Questions in English

(N/A) No,we cannot say that the probability of getting a head is $1$.
When tossing a fair coin,the two possible outcomes (head and tail) are equally likely.
The probability of getting a head in a single toss is $P(\text{Head}) = \frac{1}{2}$.
Even if a head appears $6$ times in a row,each toss is an independent event.
The probability of getting a head remains $\frac{1}{2}$ for every individual toss,regardless of the previous outcomes.

(N/A) The outcome of the next toss may or may not be a tail. This is because tossing a coin is a random experiment where the occurrence of a head or a tail are equally likely events. Each toss is an independent event,meaning the result of the previous tosses does not influence the outcome of the next toss. Therefore,the probability of getting a tail in the next toss remains $1/2$.

(N/A) No,the outcome of each coin toss is an independent event. When we toss a fair coin,the probability of getting a head or a tail is equally likely,i.e.,$P(\text{Head}) = \frac{1}{2}$ and $P(\text{Tail}) = \frac{1}{2}$. Since the coin has no memory of previous outcomes,the probability of getting a tail in the $4^{th}$ toss remains $\frac{1}{2}$,which is the same as any other toss. Therefore,there is no reason to expect a higher chance for a tail.

(A) We know that,between $1$ to $100$,half the numbers are even and half are odd. Specifically,$50$ numbers $(2, 4, 6, 8, \dots, 96, 98, 100)$ are even and $50$ numbers $(1, 3, 5, 7, \dots, 97, 99)$ are odd.
Since the total number of slips is $100$,the number of favorable outcomes for an even number is $50$,and for an odd number is $50$.
Probability of getting an even number $= \frac{\text{Number of even slips}}{\text{Total number of slips}} = \frac{50}{100} = \frac{1}{2}$.
Probability of getting an odd number $= \frac{\text{Number of odd slips}}{\text{Total number of slips}} = \frac{50}{100} = \frac{1}{2}$.
Since both events have an equal number of favorable outcomes,they are equally likely,and the probability of each is indeed $\frac{1}{2}$.

(A-D) When two dice are thrown simultaneously,the total number of possible outcomes is $6 \times 6 = 36$.
$(i)$ Sum of the numbers appearing on the dice is $7$. The favorable outcomes are $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$.
Number of favorable outcomes $= 6$.
$\text{Probability} = \frac{6}{36} = \frac{1}{6}$.
$(ii)$ Sum of the numbers appearing on the dice is a prime number. The possible prime sums are $2, 3, 5, 7, 11$.
- Sum $= 2: (1, 1) \rightarrow 1$ outcome
- Sum $= 3: (1, 2), (2, 1) \rightarrow 2$ outcomes
- Sum $= 5: (1, 4), (2, 3), (3, 2), (4, 1) \rightarrow 4$ outcomes
- Sum $= 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) \rightarrow 6$ outcomes
- Sum $= 11: (5, 6), (6, 5) \rightarrow 2$ outcomes
Total favorable outcomes $= 1 + 2 + 4 + 6 + 2 = 15$.
$\text{Probability} = \frac{15}{36} = \frac{5}{12}$.
$(iii)$ Sum of the numbers appearing on the dice is $1$. Since the minimum sum of two dice is $1 + 1 = 2$,it is impossible to get a sum of $1$.
Number of favorable outcomes $= 0$.
$\text{Probability} = \frac{0}{36} = 0$.

(N/A) The total number of outcomes when two dice are thrown is $6 \times 6 = 36$.
$(i)$ For the product to be $6$,the possible outcomes are $(1, 6), (2, 3), (3, 2), (6, 1)$.
Number of favorable outcomes $= 4$.
Probability $= \frac{4}{36} = \frac{1}{9}$.
$(ii)$ For the product to be $12$,the possible outcomes are $(2, 6), (3, 4), (4, 3), (6, 2)$.
Number of favorable outcomes $= 4$.
Probability $= \frac{4}{36} = \frac{1}{9}$.
$(iii)$ For the product to be $7$,there are no possible outcomes since $7$ is a prime number and cannot be formed by the product of two numbers from $1$ to $6$ (except $1 \times 7$,but $7$ is not on a die).
Number of favorable outcomes $= 0$.
Probability $= \frac{0}{36} = 0$.

(C) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
We need to find the outcomes where the product of the two numbers is less than $9$.
The favorable outcomes are:
For $1$ on the first die: $(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)$ (Products: $1, 2, 3, 4, 5, 6$)
For $2$ on the first die: $(2,1), (2,2), (2,3), (2,4)$ (Products: $2, 4, 6, 8$)
For $3$ on the first die: $(3,1), (3,2)$ (Products: $3, 6$)
For $4$ on the first die: $(4,1), (4,2)$ (Products: $4, 8$)
For $5$ on the first die: $(5,1)$ (Product: $5$)
For $6$ on the first die: $(6,1)$ (Product: $6$)
Counting these,we have $6 + 4 + 2 + 2 + 1 + 1 = 16$ favorable outcomes.
The probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{16}{36} = \frac{4}{9}$.

(N/A) Total number of outcomes $= 6 \times 6 = 36$.
$(i)$ Event $E_1$ (sum $= 2$): Outcomes are $\{(1, 1), (1, 1)\}$. $n(E_1) = 2$. $P(E_1) = \frac{2}{36} = \frac{1}{18}$.
$(ii)$ Event $E_2$ (sum $= 3$): Outcomes are $\{(1, 2), (1, 2), (2, 1), (2, 1)\}$. $n(E_2) = 4$. $P(E_2) = \frac{4}{36} = \frac{1}{9}$.
$(iii)$ Event $E_3$ (sum $= 4$): Outcomes are $\{(1, 3), (1, 3), (2, 2), (2, 2), (3, 1), (3, 1)\}$. $n(E_3) = 6$. $P(E_3) = \frac{6}{36} = \frac{1}{6}$.
$(iv)$ Event $E_4$ (sum $= 5$): Outcomes are $\{(2, 3), (2, 3), (3, 2), (3, 2), (4, 1), (4, 1)\}$. $n(E_4) = 6$. $P(E_4) = \frac{6}{36} = \frac{1}{6}$.
$(v)$ Event $E_5$ (sum $= 6$): Outcomes are $\{(3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)\}$. $n(E_5) = 6$. $P(E_5) = \frac{6}{36} = \frac{1}{6}$.
$(vi)$ Event $E_6$ (sum $= 7$): Outcomes are $\{(4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)\}$. $n(E_6) = 6$. $P(E_6) = \frac{6}{36} = \frac{1}{6}$.
$(vii)$ Event $E_7$ (sum $= 8$): Outcomes are $\{(5, 3), (5, 3), (6, 2), (6, 2)\}$. $n(E_7) = 4$. $P(E_7) = \frac{4}{36} = \frac{1}{9}$.
$(viii)$ Event $E_8$ (sum $= 9$): Outcomes are $\{(6, 3), (6, 3)\}$. $n(E_8) = 2$. $P(E_8) = \frac{2}{36} = \frac{1}{18}$.

(A) The possible outcomes when a coin is tossed $2$ times are:
$S = \{(H, H), (H, T), (T, H), (T, T)\}$
Therefore,the total number of outcomes $n(S) = 4$.
Let $E$ be the event of getting at most one head.
This means getting zero heads or one head.
The outcomes favorable to event $E$ are $\{(T, T), (H, T), (T, H)\}$.
Therefore,the number of favorable outcomes $n(E) = 3$.
The probability of the event $E$ is given by $P(E) = \frac{n(E)}{n(S)}$.
Thus,$P(E) = \frac{3}{4}$.

(B) The sample space $S$ for tossing a coin $3$ times is:
$S = \{(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)\}$
Total number of outcomes $n(S) = 8$.
$(i)$ Let $E_1$ be the event of getting all heads.
$E_1 = \{(HHH)\}$
Number of favorable outcomes $n(E_1) = 1$.
Probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{1}{8}$.
$(ii)$ Let $E_2$ be the event of getting at least $2$ heads.
$E_2 = \{(HHH), (HHT), (HTH), (THH)\}$
Number of favorable outcomes $n(E_2) = 4$.
Probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{4}{8} = \frac{1}{2}$.

(C) The total number of outcomes in the sample space when two dice are thrown is $n(S) = 6 \times 6 = 36$.
Let $E$ be the event of getting numbers whose difference is $2$.
The favorable outcomes are:
$E = \{(1, 3), (2, 4), (3, 5), (4, 6), (3, 1), (4, 2), (5, 3), (6, 4)\}$.
Therefore,the number of favorable outcomes is $n(E) = 8$.
The probability of the event $E$ is given by $P(E) = \frac{n(E)}{n(S)}$.
$P(E) = \frac{8}{36} = \frac{2}{9}$.

(N/A) The total number of balls in the bag is $10 + 5 + 7 = 22$.
Thus,the total number of possible outcomes is $n(S) = 22$.
$(i)$ Let $E_1$ be the event of drawing a red ball. The number of red balls is $n(E_1) = 10$.
Therefore,the probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{10}{22} = \frac{5}{11}$.
$(ii)$ Let $E_2$ be the event of drawing a green ball. The number of green balls is $n(E_2) = 7$.
Therefore,the probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{7}{22}$.
$(iii)$ Let $E_3$ be the event of not drawing a blue ball. This means drawing either a red or a green ball.
The number of favorable outcomes is $n(E_3) = 10 + 7 = 17$.
Therefore,the probability $P(E_3) = \frac{n(E_3)}{n(S)} = \frac{17}{22}$.

(A) $(i)$ Let $E_1$ be the event of getting a club.
Total cards remaining $= 52 - 3 = 49$.
Number of clubs remaining $= 13 - 3 = 10$.
$\therefore$ Probability $= \frac{10}{49}$.
$(ii)$ Let $E_2$ be the event of getting $10$ of hearts.
There is only one $10$ of hearts in the deck.
$\therefore$ Probability $= \frac{1}{49}$.

(N/A) Total cards in a deck = $52$.
Number of jacks = $4$,queens = $4$,kings = $4$. Total removed = $4 + 4 + 4 = 12$.
Remaining cards $n(S) = 52 - 12 = 40$.
$(i)$ Let $E_1$ be the event of getting a card with value $7$. There are $4$ such cards (one of each suit).
$P(E_1) = \frac{n(E_1)}{n(S)} = \frac{4}{40} = \frac{1}{10}$.
$(ii)$ Let $E_2$ be the event of getting a card with a value greater than $7$. The possible values are $8, 9, 10$. There are $3$ values,each with $4$ suits,so $n(E_2) = 3 \times 4 = 12$.
$P(E_2) = \frac{n(E_2)}{n(S)} = \frac{12}{40} = \frac{3}{10}$.
$(iii)$ Let $E_3$ be the event of getting a card with a value less than $7$. The possible values are $1, 2, 3, 4, 5, 6$. There are $6$ values,each with $4$ suits,so $n(E_3) = 6 \times 4 = 24$.
$P(E_3) = \frac{n(E_3)}{n(S)} = \frac{24}{40} = \frac{3}{5}$.

(D) The total number of integers between $0$ and $100$ is $99$ (i.e.,${1, 2, 3, \dots, 99}$).
$n(S) = 99$
$(i)$ Let $E_1$ be the event of choosing an integer which is divisible by $7$.
The multiples of $7$ between $0$ and $100$ are ${7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98}$.
$n(E_1) = 14$
$P(E_1) = \frac{n(E_1)}{n(S)} = \frac{14}{99}$
$(ii)$ Let $E_2$ be the event of choosing an integer which is not divisible by $7$.
$n(E_2) = n(S) - n(E_1) = 99 - 14 = 85$
$P(E_2) = \frac{n(E_2)}{n(S)} = \frac{85}{99}$

(B) We know that in the English alphabet,there are $5$ vowels and $21$ consonants,making a total of $26$ letters.
The total number of possible outcomes is $n(S) = 26$.
Let $E$ be the event of choosing a consonant from the English alphabet.
The set of consonants is ${b, c, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, x, y, z}$.
Therefore,the number of favorable outcomes is $n(E) = 21$.
The probability of an event is given by $P(E) = \frac{n(E)}{n(S)}$.
Thus,the required probability is $\frac{21}{26}$.

(C) Total number of sealed envelopes in a box,$n(S) = 1000$.
Number of envelopes containing a cash prize $= 10 + 100 + 200 = 310$.
Number of envelopes containing no cash prize,$n(E) = 1000 - 310 = 690$.
The probability $P(E)$ that the envelope contains no cash prize is given by the ratio of the number of favorable outcomes to the total number of outcomes.
$P(E) = \frac{n(E)}{n(S)} = \frac{690}{1000} = 0.69$.

(N/A) Total number of pieces $= 8 \text{ triangles} + 10 \text{ squares} = 18$.
$(i)$ Probability that the lost piece is a triangle $= \frac{8}{18} = \frac{4}{9}$.
$(ii)$ Probability that the lost piece is a square $= \frac{10}{18} = \frac{5}{9}$.
$(iii)$ Number of blue squares $= 6$. Probability that the lost piece is a blue square $= \frac{6}{18} = \frac{1}{3}$.
$(iv)$ Number of red triangles $= 8 - 3 = 5$. Probability that the lost piece is a red triangle $= \frac{5}{18}$.

(A) Total possible outcomes of tossing a coin $3$ times:
$S = \{(HHH), (TTT), (HTT), (THT), (TTH), (THH), (HTH), (HHT)\}$
$\therefore n(S) = 8$
$(i)$ Let $E_1$ be the event that Sweta loses the entry fee.
This happens if she gets zero heads (i.e.,$TTT$).
$E_1 = \{(TTT)\}$,so $n(E_1) = 1$.
$P(E_1) = \frac{n(E_1)}{n(S)} = \frac{1}{8}$.
$(ii)$ Let $E_2$ be the event that Sweta gets double the entry fee.
This happens if she gets $3$ heads (i.e.,$HHH$).
$E_2 = \{(HHH)\}$,so $n(E_2) = 1$.
$P(E_2) = \frac{n(E_2)}{n(S)} = \frac{1}{8}$.
$(iii)$ Let $E_3$ be the event that Sweta gets her entry fee back.
This happens if she gets one or two heads.
$E_3 = \{(HTT), (THT), (TTH), (HHT), (HTH), (THH)\}$,so $n(E_3) = 6$.
$P(E_3) = \frac{n(E_3)}{n(S)} = \frac{6}{8} = \frac{3}{4}$.

(C) Given,a die has its six faces marked ${0, 1, 1, 1, 6, 6}$.
Total sample space,$n(S) = 6^2 = 36$.
$(i)$ The possible sums are obtained by adding the values on the two dice:
$0+0=0, 0+1=1, 0+6=6, 1+1=2, 1+6=7, 6+6=12$.
Thus,the different scores possible are ${0, 1, 2, 6, 7, 12}$. There are $6$ possible scores.
$(ii)$ Let $E$ be the event of getting a sum of $7$.
The pairs $(d_1, d_2)$ that result in a sum of $7$ are:
$(1, 6)$ occurs $3 \times 2 = 6$ times.
$(6, 1)$ occurs $2 \times 3 = 6$ times.
Total favorable outcomes $n(E) = 6 + 6 = 12$.
$P(E) = \frac{n(E)}{n(S)} = \frac{12}{36} = \frac{1}{3}$.

(A) Given that,the total number of balls in the bag $= 24$.
Number of red balls $= x$,number of white balls $= 2x$,and number of blue balls $= 3x$.
According to the condition,$x + 2x + 3x = 24$.
$6x = 24$,which gives $x = 4$.
Therefore,the number of red balls $= 4$,the number of white balls $= 2 \times 4 = 8$,and the number of blue balls $= 3 \times 4 = 12$.
The total number of outcomes $n(S) = 24$.
$(i)$ Let $E_1$ be the event of selecting a ball which is not red. This means the ball can be white or blue.
$n(E_1) = \text{Number of white balls} + \text{Number of blue balls} = 8 + 12 = 20$.
Required probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{20}{24} = \frac{5}{6}$.
$(ii)$ Let $E_2$ be the event of selecting a ball which is white.
$n(E_2) = \text{Number of white balls} = 8$.
Required probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{8}{24} = \frac{1}{3}$.

(A) Total number of cards in the box is $n(S) = 1000$.
$(i)$ Let $E_1$ be the event that the first player wins a prize. This happens if the player selects a perfect square greater than $500$.
The perfect squares between $1$ and $1000$ are $1^2, 2^2, \dots, 31^2 = 961$.
The perfect squares greater than $500$ are $23^2=529, 24^2=576, 25^2=625, 26^2=676, 27^2=729, 28^2=784, 29^2=841, 30^2=900, 31^2=961$.
There are $9$ such cards.
Thus, $n(E_1) = 9$.
The probability that the first player wins is $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{9}{1000} = 0.009$.
$(ii)$ If the first player has won, one card (which is a perfect square $> 500$) is removed from the box.
The remaining number of cards is $n(S') = 1000 - 1 = 999$.
The number of remaining perfect squares greater than $500$ is $n(E_2) = 9 - 1 = 8$.
The probability that the second player wins given the first has won is $P(E_2|E_1) = \frac{n(E_2)}{n(S')} = \frac{8}{999}$.

(C) In the experiment of throwing a fair die,the possible equally likely outcomes are $1, 2, 3, 4, 5,$ and $6.$
Let event $A$ be the event of 'getting $2$ on the die.' Then,the number of outcomes favourable to $A$ is only $1.$
$\therefore P(A) = \text{P(getting } 2 \text{ on the die)}$
$= \frac{\text{Number of outcomes favourable to } A}{\text{Number of all possible outcomes}}$
$= \frac{1}{6}$
Thus,the probability of getting number $2$ on the die is $\frac{1}{6}$.

(D) There are six elementary outcomes in the experiment of throwing a balanced die once: $1, 2, 3, 4, 5, 6$.
Let event $A$ be: 'The number on the die is a prime number'.
The prime numbers among the outcomes are $2, 3, 5$.
Thus,the number of outcomes favourable to event $A$ is $3$.
The total number of possible outcomes is $6$.
Therefore,the probability of event $A$ is $P(A) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{3}{6} = \frac{1}{2}$.
Thus,the probability of getting a prime number on the die is $\frac{1}{2}$.

(A) There are six elementary outcomes in the experiment of throwing a balanced die once: $1, 2, 3, 4, 5, 6$.
Let event $A$ be: 'The number on the die is greater than $4$.'
The outcomes favourable to event $A$ are $5$ and $6$.
Hence,the number of outcomes favourable to event $A$ is $2$.
$\therefore P(A) = P(\text{The number on the die is greater than } 4) = \frac{\text{Number of outcomes favourable to } A}{\text{Number of all possible outcomes}}$.
$P(A) = \frac{2}{6} = \frac{1}{3}$.
Thus,the probability of getting a number greater than $4$ is $\frac{1}{3}$.

(B) There are six elementary outcomes in the experiment of throwing a balanced die once: $1, 2, 3, 4, 5, 6$.
Let event $B$ be: 'The number on the die is smaller than $4$.'
The outcomes favourable to event $B$ are $1, 2,$ and $3$.
Hence,the number of outcomes favourable to event $B$ is $3$.
Therefore,the probability $P(B)$ is given by:
$P(B) = \frac{\text{Number of outcomes favourable to } B}{\text{Total number of possible outcomes}}$
$P(B) = \frac{3}{6}$
$P(B) = \frac{1}{2}$
Thus,the probability of getting a number smaller than $4$ is $\frac{1}{2}$.

(C) In the experiment of selecting a card at random from a well-shuffled pack of $52$ cards,
the total number of elementary outcomes is $52$.
Let event $A$ be: "The selected card is a seven."
The number of outcomes favourable to event $A$ is $4$,as there are $4$ sevens in a standard pack of cards (one for each suit: Hearts,Diamonds,Clubs,and Spades).
Therefore,the probability $P(A)$ is given by the ratio of favourable outcomes to total outcomes:
$P(A) = \frac{4}{52} = \frac{1}{13}$.
Thus,the probability that the selected card is a seven is $\frac{1}{13}$.

(D) In the experiment of selecting a card at random from a well-shuffled pack of $52$ cards,the total number of elementary outcomes is $52$.
Let event $B$ be: "The selected card is a spade."
The number of outcomes favourable to event $B$ is $13$,as there are $13$ cards of spades in a standard pack of $52$ cards.
Therefore,the probability $P(B)$ is given by:
$P(B) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{13}{52} = \frac{1}{4}$.
Thus,the probability that the selected card is a spade is $\frac{1}{4}$.

(A) In the experiment of selecting a card at random from a well-shuffled pack of $52$ cards,the total number of elementary outcomes is $52$.
Let event $C$ be the event that the selected card is of a black suit.
$A$ pack of $52$ cards contains two black suits: Spades ($13$ cards) and Clubs ($13$ cards).
Therefore,the number of outcomes favourable to event $C$ is $13 + 13 = 26$.
The probability of event $C$ is given by $P(C) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$.
$P(C) = \frac{26}{52} = \frac{1}{2}$.
Thus,the probability that the selected card is of a black suit is $1/2$.

(C) In the experiment of selecting a card at random from a well-shuffled pack of $52$ cards,the total number of elementary outcomes is $52$.
Let event $E$ be: 'The selected card is the queen of hearts.'
The number of outcomes favourable to event $E$ is $1$,as there is only one queen of hearts in a standard pack of $52$ cards.
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$.
Therefore,$P(E) = \frac{1}{52}$.
Thus,the probability that the selected card is the queen of hearts is $\frac{1}{52}$.

(D) When two unbiased dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$. The sample space $S$ is given by:
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Let $A$ be the event that the sum of the numbers on the two dice is $8$.
The outcomes favourable to event $A$ are $(2,6), (3,5), (4,4), (5,3), (6,2)$.
The number of favourable outcomes is $n(A) = 5$.
The probability of event $A$ is given by $P(A) = \frac{n(A)}{n(S)} = \frac{5}{36}$.

(A) When two unbiased dice are thrown,the sample space $S$ consists of all possible ordered pairs $(x, y)$ where $x, y \in \{1, 2, 3, 4, 5, 6\}$.
The total number of elementary outcomes is $6 \times 6 = 36$.
The sample space is:
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Let $B$ be the event that the sum of the numbers on the two dice is $4$.
The outcomes favourable to event $B$ are $(1,3), (2,2), (3,1)$.
The number of favourable outcomes is $n(B) = 3$.
The probability of event $B$ is given by $P(B) = \frac{n(B)}{n(S)} = \frac{3}{36} = \frac{1}{12}$.

(B) When two unbiased dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$. The sample space $S$ is given by:
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Let event $C$ be the event that the sum of the numbers on the two dice is greater than $9$. This means the sum can be $10, 11,$ or $12$.
The outcomes favourable to event $C$ are:
Sum $= 10: (4,6), (5,5), (6,4)$
Sum $= 11: (5,6), (6,5)$
Sum $= 12: (6,6)$
Thus,the number of favourable outcomes is $n(C) = 3 + 2 + 1 = 6$.
The probability of event $C$ is given by $P(C) = \frac{n(C)}{n(S)} = \frac{6}{36} = \frac{1}{6}$.

(C) When two unbiased dice are thrown,the sample space $S$ consists of $6 \times 6 = 36$ possible outcomes:
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Total number of outcomes $n(S) = 36$.
Let $E$ be the event that the sum of the numbers on the two dice is smaller than $5$.
The possible sums smaller than $5$ are $2, 3,$ and $4$.
The outcomes favourable to event $E$ are:
Sum $= 2: (1,1)$
Sum $= 3: (1,2), (2,1)$
Sum $= 4: (1,3), (2,2), (3,1)$
Thus,the number of favourable outcomes $n(E) = 1 + 2 + 3 = 6$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{6}{36} = \frac{1}{6}$.

(A) When two unbiased dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$. The sample space $S$ is given by:
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
We need the sum of the numbers on the two dice to be a multiple of both $3$ and $4$. The least common multiple of $3$ and $4$ is $12$. Thus,we are looking for outcomes where the sum is $12$.
The only outcome where the sum is $12$ is $(6,6)$.
Number of favorable outcomes $= 1$.
Probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{36}$.

(B) When two unbiased dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$. The sample space $S$ is given by:
$S = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$
Let $E$ be the event that the sum of the numbers on the two dice is a prime number. The possible prime sums are $2, 3, 5, 7,$ and $11$.
The outcomes favorable to event $E$ are:
Sum $= 2: (1, 1)$
Sum $= 3: (1, 2), (2, 1)$
Sum $= 5: (1, 4), (2, 3), (3, 2), (4, 1)$
Sum $= 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$
Sum $= 11: (5, 6), (6, 5)$
Total number of favorable outcomes $= 1 + 2 + 4 + 6 + 2 = 15$.
Therefore,the probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{15}{36} = \frac{5}{12}$.

(D) The sample space $S$ for throwing two unbiased dice is given by the set of all possible ordered pairs $(x, y)$ where $x, y \in \{1, 2, 3, 4, 5, 6\}$.
There are $6 \times 6 = 36$ possible outcomes:
$S = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$
Let $E$ be the event that the sum of the numbers on the two dice is smaller than $13$.
The minimum sum is $1 + 1 = 2$ and the maximum sum is $6 + 6 = 12$.
Since the maximum possible sum is $12$,which is always less than $13$,every outcome in the sample space satisfies the condition.
Thus,$E = S$ and the number of favorable outcomes is $36$.
The probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{36}{36} = 1$.
Therefore,the probability is $1$.

(A) All the possible outcomes in the experiment of throwing two unbiased dice are as follows:
$(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$
$(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)$
$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)$
$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)$
$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)$
$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$
Hence,the total number of elementary outcomes is $36$.
Let event $J$ be: 'The sum of numbers on two dice is smaller than $2$.'
The minimum sum of numbers on two dice is $1 + 1 = 2$.
Since the sum of numbers on two dice is always from $2$ to $12$,it is never smaller than $2$.
Therefore,event $J$ is an impossible event.
The probability of an impossible event is $0$.
Thus,$P(J) = 0$.

(B) The total number of elementary outcomes in the experiment of selecting a bulb at random from a carton carrying $20$ bulbs is $20$.
Total number of bulbs in the carton $= 20$.
Number of defective bulbs in the carton $= 4$.
Therefore,the number of non-defective bulbs in the carton $= 20 - 4 = 16$.
Let event $A$ be the event that the bulb selected at random is non-defective.
The number of outcomes favourable to event $A = 16$.
The probability of event $A$ is given by $P(A) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$.
$P(A) = \frac{16}{20} = \frac{4}{5}$.
Thus,the probability that the bulb selected at random is non-defective is $\frac{4}{5}$.

(C) Total number of balls in the box $= 5 + 8 + 4 = 17$.
Therefore,the total number of possible outcomes in the experiment of drawing a ball from the box $= 17$.
Let event $A$ be the event that the ball drawn is red.
There are $5$ red balls in the box.
Therefore,the number of outcomes favourable to event $A = 5$.
The probability of an event is given by the formula: $P(A) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$.
Thus,$P(A) = \frac{5}{17}$.

(D) Total number of balls in the box $= 5 + 8 + 4 = 17$.
Therefore,the total number of elementary outcomes in the experiment of drawing a ball from the box $= 17$.
Let event $B$ be the event that the ball drawn is white.
There are $8$ white balls in the box.
Therefore,the number of outcomes favourable to event $B = 8$.
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$.
Thus,$P(B) = \frac{8}{17}$.

(B) The month of July in any year has $31$ days.
These $31$ days consist of $4$ complete weeks and $3$ additional days.
Therefore,every day of the week will occur at least $4$ times.
The remaining $3$ days can be any of the following $7$ possible triplets:
$1$. (Sunday,Monday,Tuesday)
$2$. (Monday,Tuesday,Wednesday)
$3$. (Tuesday,Wednesday,Thursday)
$4$. (Wednesday,Thursday,Friday)
$5$. (Thursday,Friday,Saturday)
$6$. (Friday,Saturday,Sunday)
$7$. (Saturday,Sunday,Monday)
Total number of elementary outcomes $= 7$.
The outcomes that contain a Sunday are:
$1$. (Sunday,Monday,Tuesday)
$6$. (Friday,Saturday,Sunday)
$7$. (Saturday,Sunday,Monday)
Number of favourable outcomes $= 3$.
Therefore,the required probability $= \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{3}{7}$.

(C) leap year has $366$ days,which is equivalent to $52$ complete weeks and $2$ extra days.
Since $52$ weeks contain $52$ Thursdays,the occurrence of a $53$rd Thursday depends on the remaining $2$ days.
The possible pairs for these $2$ days are:
(Sunday,Monday),(Monday,Tuesday),(Tuesday,Wednesday),(Wednesday,Thursday),(Thursday,Friday),(Friday,Saturday),(Saturday,Sunday).
There are $7$ total possible outcomes.
The outcomes that include a Thursday are (Wednesday,Thursday) and (Thursday,Friday).
Thus,there are $2$ favorable outcomes.
The probability is calculated as $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{7}$.

(A) In the experiment of tossing three balanced coins simultaneously,the sample space $S$ consists of the following elementary outcomes:
$S = \{(HHH), (HHT), (HTH), (THH), (HTT), (THT), (TTH), (TTT)\}$
where $H$ represents head and $T$ represents tail.
Total number of elementary outcomes $= 8$.
Let event $A$ be the event of receiving at least two heads (i.e.,$2$ or $3$ heads).
The outcomes favourable to event $A$ are: $(HHH), (HHT), (HTH), (THH)$.
Number of favourable outcomes $= 4$.
The probability of event $A$ is given by $P(A) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$.
$P(A) = \frac{4}{8} = \frac{1}{2}$.
Thus,the probability of receiving at least two heads is $\frac{1}{2}$.

(B) The total number of possible outcomes when a balanced die is thrown is $S = \{1, 2, 3, 4, 5, 6\}$,so the total number of outcomes $n(S) = 6$.
Let $E$ be the event of getting an even number on the die.
The even numbers on a die are $\{2, 4, 6\}$,so the number of favorable outcomes $n(E) = 3$.
The probability of the event $E$ is given by $P(E) = \frac{n(E)}{n(S)} = \frac{3}{6} = \frac{1}{2}$.

(C) When a balanced die is thrown,the sample space $S$ is given by $S = \{1, 2, 3, 4, 5, 6\}$.
The total number of possible outcomes is $n(S) = 6$.
Let $E$ be the event that the number on the die is a multiple of $3$.
The multiples of $3$ in the sample space are $3$ and $6$.
Thus,the favorable outcomes are $E = \{3, 6\}$.
The number of favorable outcomes is $n(E) = 2$.
The probability $P(E)$ is given by the formula $P(E) = \frac{n(E)}{n(S)}$.
Substituting the values,$P(E) = \frac{2}{6} = \frac{1}{3}$.

(D) When a balanced die is thrown,the total number of possible outcomes is $S = \{1, 2, 3, 4, 5, 6\}$.
Thus,the total number of outcomes $n(S) = 6$.
Let $E$ be the event that the number on the die is smaller than $3$.
The favorable outcomes are $E = \{1, 2\}$.
Thus,the number of favorable outcomes $n(E) = 2$.
The probability $P(E)$ is given by the formula $P(E) = \frac{n(E)}{n(S)}$.
$P(E) = \frac{2}{6} = \frac{1}{3}$.

(A) When two dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
Let the outcomes be $(x, y)$ where $x$ and $y$ are the numbers on the first and second die respectively.
The sum $(x + y)$ is even if:
$1.$ Both $x$ and $y$ are odd: There are $3$ choices for $x$ $(1, 3, 5)$ and $3$ choices for $y$ $(1, 3, 5)$. Total outcomes = $3 \times 3 = 9$.
$2.$ Both $x$ and $y$ are even: There are $3$ choices for $x$ $(2, 4, 6)$ and $3$ choices for $y$ $(2, 4, 6)$. Total outcomes = $3 \times 3 = 9$.
Total favorable outcomes = $9 + 9 = 18$.
Probability = $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{18}{36} = \frac{1}{2}$.

(B) When two balanced dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
The sum of the numbers on the two dice is odd if one die shows an even number and the other shows an odd number.
Let $E$ be the event that the sum is odd.
Possible outcomes for (Odd,Even) are: $(1,2), (1,4), (1,6), (3,2), (3,4), (3,6), (5,2), (5,4), (5,6)$ (Total $9$ outcomes).
Possible outcomes for (Even,Odd) are: $(2,1), (2,3), (2,5), (4,1), (4,3), (4,5), (6,1), (6,3), (6,5)$ (Total $9$ outcomes).
Total favorable outcomes = $9 + 9 = 18$.
Probability $P(E) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{18}{36} = \frac{1}{2}$.

(C) When two balanced dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
Let $S$ be the sum of the numbers on the two dice. The possible values for $S$ range from $2$ to $12$.
The multiples of $5$ in this range are $5$ and $10$.
We identify the pairs $(d_1, d_2)$ that result in these sums:
For $S = 5$: $(1, 4), (2, 3), (3, 2), (4, 1)$ (Total $4$ outcomes).
For $S = 10$: $(4, 6), (5, 5), (6, 4)$ (Total $3$ outcomes).
The total number of favorable outcomes is $4 + 3 = 7$.
Therefore,the probability $P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{7}{36}$.