Set Based probability Questions in English

(A) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
On a single die,the prime numbers are $\{2, 3, 5\}$ and the composite numbers are $\{4, 6\}$. Note that $1$ is neither prime nor composite.
We need the probability of getting a prime number on one die and a composite number on the other.
The possible outcomes are:
$(2, 4), (2, 6), (3, 4), (3, 6), (5, 4), (5, 6)$ (prime on first,composite on second)
$(4, 2), (6, 2), (4, 3), (6, 3), (4, 5), (6, 5)$ (composite on first,prime on second)
Total favorable outcomes = $6 + 6 = 12$.
Probability = $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{12}{36} = \frac{1}{3}$.

(A) Let $P(A)$ be the probability that student $A$ solves the problem and $P(B)$ be the probability that student $B$ solves the problem.
Given $P(A) = \frac{2}{5}$ and $P(B) = \frac{3}{4}$.
The problem is solved if at least one of them solves it.
$P(\text{problem is solved}) = 1 - P(\text{problem is not solved})$.
Since $A$ and $B$ try independently,the probability that neither solves it is $P(\bar{A}) \times P(\bar{B})$.
$P(\bar{A}) = 1 - P(A) = 1 - \frac{2}{5} = \frac{3}{5}$.
$P(\bar{B}) = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4}$.
$P(\text{problem is solved}) = 1 - (\frac{3}{5} \times \frac{1}{4}) = 1 - \frac{3}{20} = \frac{17}{20}$.

(A) Let $M$ be the set of students attending the Mathematics class and $P$ be the set of students attending the Physics class. Given: $P(M) = 40 \%$,$P(P) = 30 \%$,and $P(M \cap P) = 20 \%$.
The probability of students attending only the Mathematics class is $P(M) - P(M \cap P) = 40 \% - 20 \% = 20 \%$.
The probability of students attending only the Physics class is $P(P) - P(M \cap P) = 30 \% - 20 \% = 10 \%$.
The probability that the student attends only one class is the sum of the probabilities of attending only Mathematics and only Physics: $20 \% + 10 \% = 30 \%$.
Converting to a fraction,$30 \% = \frac{30}{100} = \frac{3}{10}$.

(B) For the equation $2x^2+x-1=0$,we have $2x^2+2x-x-1=0 \Rightarrow 2x(x+1)-1(x+1)=0$,so $x = \frac{1}{2}, -1$. Since probability $P \in [0, 1]$,we take $P_1 = \frac{1}{2}$.
For the equation $3x^2-10x+3=0$,we have $3x^2-9x-x+3=0 \Rightarrow 3x(x-3)-1(x-3)=0$,so $x = \frac{1}{3}, 3$. Since probability $P \in [0, 1]$,we take $P_2 = \frac{1}{3}$.
For the equation $6x^2+11x-2=0$,we have $6x^2+12x-x-2=0 \Rightarrow 6x(x+2)-1(x+2)=0$,so $x = \frac{1}{6}, -2$. Since probability $P \in [0, 1]$,we take $P_3 = \frac{1}{6}$.
Sum of probabilities $P_1 + P_2 + P_3 = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1$.
Since the sum of the probabilities of the events is $1$,the events are exhaustive.

(D) The total number of $4$-digit numbers that can be formed using the digits ${4, 5, 6, 7, 8, 9}$ with repetition is $6^4 = 1296$.
Let the number be $N = d_1 d_2 d_3 d_4$. The sum of digits is $S = d_1 + d_2 + d_3 + d_4$.
$A$ number is divisible by $3$ if and only if the sum of its digits is divisible by $3$.
Consider the digits modulo $3$:
$4 \equiv 1 \pmod{3}$,$5 \equiv 2 \pmod{3}$,$6 \equiv 0 \pmod{3}$,$7 \equiv 1 \pmod{3}$,$8 \equiv 2 \pmod{3}$,$9 \equiv 0 \pmod{3}$.
There are two digits for each remainder $0, 1, 2$ modulo $3$.
Let $S_n$ be the sum of the first $n$ digits. For any choice of $d_1, d_2, d_3$,the sum $S_3 = d_1 + d_2 + d_3$ will have some remainder $r \in \{0, 1, 2\}$ modulo $3$.
To make $S_4 = S_3 + d_4$ divisible by $3$,we need $d_4 \equiv -S_3 \pmod{3}$.
Since there are exactly two choices for $d_4$ for each remainder,and there are $6$ total choices for $d_4$,the probability that $d_4$ satisfies the condition is $\frac{2}{6} = \frac{1}{3}$.
Thus,the probability is $\frac{1}{3}$.

(D) The total number of natural numbers from $2$ to $1001$ is $1001 - 2 + 1 = 1000$.
We are looking for numbers $n$ such that $n \equiv 1 \pmod{7}$.
The sequence of such numbers starting from $2$ is $8, 15, 22, \dots, 995$.
This is an arithmetic progression where the first term $a = 8$,the common difference $d = 7$,and the last term $l = 995$.
Using the formula $l = a + (m - 1)d$,we have:
$995 = 8 + (m - 1)7$
$987 = (m - 1)7$
$m - 1 = 141$
$m = 142$.
The required probability is $\frac{m}{\text{Total numbers}} = \frac{142}{1000} = \frac{71}{500}$.

(D) Given,$P(A)=0.6$,$P(B)=0.4$,and $P(A \cap B)=0$.
We need to find the probability that neither $A$ nor $B$ occurs,which is $P(\bar{A} \cap \bar{B})$.
By De Morgan's Law,$P(\bar{A} \cap \bar{B}) = P(\overline{A \cup B})$.
We know that $P(\overline{A \cup B}) = 1 - P(A \cup B)$.
Using the addition rule of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $P(A \cup B) = 0.6 + 0.4 - 0 = 1.0$.
Therefore,$P(\overline{A \cup B}) = 1 - 1.0 = 0$.

(B) The total number of outcomes when two dice are thrown is $6 \times 6 = 36$.
The favorable outcomes where the sum is greater than $7$ are:
Sum $= 8$: $(2,6), (3,5), (4,4), (5,3), (6,2)$ ($5$ outcomes)
Sum $= 9$: $(3,6), (4,5), (5,4), (6,3)$ ($4$ outcomes)
Sum $= 10$: $(4,6), (5,5), (6,4)$ ($3$ outcomes)
Sum $= 11$: $(5,6), (6,5)$ ($2$ outcomes)
Sum $= 12$: $(6,6)$ ($1$ outcome)
Total favorable outcomes $= 5 + 4 + 3 + 2 + 1 = 15$.
The required probability is $\frac{15}{36} = \frac{5}{12}$.

(B) Let $S$ be the sample space of outcomes where the sum of the two dice is $7$. The possible outcomes are:
$S = \{(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)\}$
Total number of outcomes $n(S) = 6$.
Let $E$ be the event that the number $3$ appears at least once.
The favorable outcomes are:
$E = \{(3, 4), (4, 3)\}$
Number of favorable outcomes $n(E) = 2$.
The required probability is $P(E) = \frac{n(E)}{n(S)} = \frac{2}{6} = \frac{1}{3}$.

(D) Total sample points,$n(S) = 6 \times 6 = 36$.
Favourable outcomes for the sum to be $\ge 10$ are:
$E = \{(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)\}$.
Total number of favourable outcomes,$n(E) = 6$.
Required probability $P(E) = \frac{n(E)}{n(S)} = \frac{6}{36} = \frac{1}{6}$.

(A) Given,$P(A)=0.5, P(B)=0.4$ and $P(A \cup B)=0.6$.
$(i) \ P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.5 + 0.4 - 0.6 = 0.3$. Thus,$(i) \rightarrow (3)$.
$(ii) \ P(A \cap \bar{B}) = P(A) - P(A \cap B) = 0.5 - 0.3 = 0.2$. Thus,$(ii) \rightarrow (2)$.
$(iii) \ P(\bar{A} \cap B) = P(B) - P(A \cap B) = 0.4 - 0.3 = 0.1$. Thus,$(iii) \rightarrow (4)$.
$(iv) \ P(\bar{A} \cap \bar{B}) = P((A \cup B)^c) = 1 - P(A \cup B) = 1 - 0.6 = 0.4$. Thus,$(iv) \rightarrow (1)$.
Therefore,the correct match is $(i)-3, (ii)-2, (iii)-4, (iv)-1$.

(C) Given,$P(\bar{A} \cup \bar{B}) = P(\overline{A \cap B}) = \frac{7}{10}$.
Since $P(A \cap B) + P(\overline{A \cap B}) = 1$,we have:
$P(A \cap B) = 1 - \frac{7}{10} = \frac{3}{10}$.
Also,the formula for the union of two events is:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values:
$\frac{4}{5} = P(A) + \frac{2}{5} - \frac{3}{10}$.
$P(A) = \frac{4}{5} - \frac{2}{5} + \frac{3}{10}$.
$P(A) = \frac{2}{5} + \frac{3}{10} = \frac{4+3}{10} = \frac{7}{10}$.

(A) The faces of the die are $\{2, 3, 5, 7, 11, 13\}$.
There is $1$ even number $(2)$ and $5$ odd numbers $(3, 5, 7, 11, 13)$.
The sum of two numbers is odd if one number is even and the other is odd.
Let $E$ be the event of getting an even number and $O$ be the event of getting an odd number.
$P(E) = \frac{1}{6}$ and $P(O) = \frac{5}{6}$.
The sum is odd in two cases: (First die is even,Second die is odd) or (First die is odd,Second die is even).
Required Probability $= P(E) \times P(O) + P(O) \times P(E)$
$= \left(\frac{1}{6} \times \frac{5}{6}\right) + \left(\frac{5}{6} \times \frac{1}{6}\right)$
$= \frac{5}{36} + \frac{5}{36} = \frac{10}{36} = \frac{5}{18}$.

(D) Total number of balls = $5 + 4 + 3 = 12$.
Number of black balls = $5$.
Number of red balls = $3$.
Number of favorable outcomes (black or red) = $5 + 3 = 8$.
Required probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{8}{12} = \frac{2}{3}$.

(C) We are given $P(A \cap \overline{B}) = 0.1$,$P(\overline{A} \cap B) = 0.2$,and $P(B) = 0.5$.
Since $B = (A \cap B) \cup (\overline{A} \cap B)$,and these are disjoint events,we have $P(B) = P(A \cap B) + P(\overline{A} \cap B)$.
Substituting the values,$0.5 = P(A \cap B) + 0.2$,which gives $P(A \cap B) = 0.3$.
Now,$P(A) = P(A \cap B) + P(A \cap \overline{B}) = 0.3 + 0.1 = 0.4$.
We want to find $P(\overline{A} \cap \overline{B})$. By De Morgan's Law,$P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we get $P(A \cup B) = 0.4 + 0.5 - 0.3 = 0.6$.
Therefore,$P(\overline{A} \cap \overline{B}) = 1 - 0.6 = 0.4$.

(C) non-leap year has $365$ days,which is equal to $52$ complete weeks and $1$ extra day.
This extra day can be any of the $7$ days of the week (Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday).
Let $E_1$ be the event of getting $53$ Sundays,$E_2$ be the event of getting $53$ Tuesdays,and $E_3$ be the event of getting $53$ Thursdays.
Since these events are mutually exclusive,the probability of getting $53$ Sundays $OR$ $53$ Tuesdays $OR$ $53$ Thursdays is $P(E_1 \cup E_2 \cup E_3) = P(E_1) + P(E_2) + P(E_3)$.
$P(E_1) = \frac{1}{7}$,$P(E_2) = \frac{1}{7}$,and $P(E_3) = \frac{1}{7}$.
Therefore,the required probability is $\frac{1}{7} + \frac{1}{7} + \frac{1}{7} = \frac{3}{7}$.

(D) We are given that $P(A \cup B) = P(A \cap B)$.
Since $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we substitute the given condition:
$P(A \cap B) = P(A) + P(B) - P(A \cap B)$
$2P(A \cap B) = P(A) + P(B)$.
Also,since $A \cap B \subseteq A \subseteq A \cup B$ and $A \cap B \subseteq B \subseteq A \cup B$,the condition $P(A \cup B) = P(A \cap B)$ implies $P(A) = P(B) = P(A \cap B)$.
This means $P(A \cap B') = P(A) - P(A \cap B) = 0$ and $P(A' \cap B) = P(B) - P(A \cap B) = 0$.
Thus,options $A$,$B$,and $C$ are true.
However,$P(A) + P(B) = 2P(A \cap B)$,which is not necessarily equal to $1$. Therefore,option $D$ is not true.

(B) The set of integers is $S = \{x \in Z \mid -25 \leq x \leq 25, x \neq 0\}$. The total number of elements is $50$.
We need to solve the inequality $x + 6 \leq \frac{135}{x}$.
Rearranging the inequality,we get $\frac{x^2 + 6x - 135}{x} \leq 0$.
Factoring the numerator,we have $\frac{(x + 15)(x - 9)}{x} \leq 0$.
Using the sign scheme method,the solution set for the inequality is $x \in (-\infty, -15] \cup (0, 9]$.
Intersecting this with the given set $S$,we get $x \in [-25, -15] \cup \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
The number of integers in $[-25, -15]$ is $11$,and the number of integers in ${1, 2, 3, 4, 5, 6, 7, 8, 9}$ is $9$.
Total favorable outcomes $= 11 + 9 = 20$.
Probability $= \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{20}{50} = \frac{2}{5}$.

(D) Given,$P(A \cup B)=0.8$ and $P(A \cap B)=0.3$.
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Therefore,$P(A) + P(B) = P(A \cup B) + P(A \cap B) = 0.8 + 0.3 = 1.1$.
We need to find $P(A^C) + P(B^C)$.
Using the property $P(E^C) = 1 - P(E)$,we have:
$P(A^C) + P(B^C) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))$.
Substituting the value of $P(A) + P(B)$:
$P(A^C) + P(B^C) = 2 - 1.1 = 0.9$.

(D) Given,$P(B) = \frac{3}{2} P(A)$ and $P(C) = \frac{1}{2} P(B)$.
Since $A, B$ and $C$ are mutually exclusive and exhaustive events,their sum of probabilities is $1$:
$P(A) + P(B) + P(C) = 1$
Substituting the given relations in terms of $P(A)$:
$P(A) + \frac{3}{2} P(A) + \frac{1}{2} \left( \frac{3}{2} P(A) \right) = 1$
$P(A) \left( 1 + \frac{3}{2} + \frac{3}{4} \right) = 1$
$P(A) \left( \frac{4 + 6 + 3}{4} \right) = 1$
$\frac{13}{4} P(A) = 1 \implies P(A) = \frac{4}{13}$
Now,calculate $P(C)$:
$P(C) = \frac{3}{4} P(A) = \frac{3}{4} \times \frac{4}{13} = \frac{3}{13}$
Since $A$ and $C$ are mutually exclusive,$P(A \cup C) = P(A) + P(C)$:
$P(A \cup C) = \frac{4}{13} + \frac{3}{13} = \frac{7}{13}$

(A) Given that $P(A \cap B) = \frac{3}{25}$ and $P(B - A) = \frac{8}{25}$.
From the properties of sets and probability,the event $B$ can be expressed as the union of two disjoint sets: $(B - A)$ and $(A \cap B)$.
Therefore,$P(B) = P(B - A) + P(A \cap B)$.
Substituting the given values:
$P(B) = \frac{8}{25} + \frac{3}{25} = \frac{11}{25}$.

(D) The given quadratic equation is $x^2 + 4x + c = 0$.
For the roots to be real,the discriminant $D$ must be greater than or equal to $0$.
$D = b^2 - 4ac \geq 0$
Substituting the values $a = 1, b = 4$,we get:
$4^2 - 4(1)(c) \geq 0$
$16 - 4c \geq 0$
$16 \geq 4c$
$c \leq 4$.
Since $c$ is chosen from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$,the possible values for $c$ are $\{1, 2, 3, 4\}$.
There are $4$ favorable outcomes out of $9$ total possible outcomes.
Therefore,the required probability is $\frac{4}{9}$.

(D) We know that for any two events $A$ and $B$,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Therefore,$P(A) + P(B) = P(A \cup B) + P(A \cap B)$.
Substituting the given values,$P(A) + P(B) = 0.8 + 0.3 = 1.1$.
We know that $P(\bar{A}) = 1 - P(A)$ and $P(\bar{B}) = 1 - P(B)$.
Thus,$P(\bar{A}) + P(\bar{B}) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))$.
Substituting $P(A) + P(B) = 1.1$,we get $P(\bar{A}) + P(\bar{B}) = 2 - 1.1 = 0.9$.

(A) Given that,$P(A) = 0.25$ and $P(B) = 0.50$.
The probability of both events occurring is $P(A \cap B) = 0.14$.
Using the addition rule for probability,the probability that at least one of the events occurs is:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) = 0.25 + 0.50 - 0.14 = 0.61$.
The probability that neither $A$ nor $B$ occurs is given by $P(\bar{A} \cap \bar{B})$,which is equal to $P(\overline{A \cup B})$.
$P(\overline{A \cup B}) = 1 - P(A \cup B) = 1 - 0.61 = 0.39$.

(C) Let $P(A), P(B), P(C)$ be the probabilities of $A, B, C$ hitting the balloon respectively.
$P(A) = \frac{4}{6} = \frac{2}{3}$,so $P(\bar{A}) = 1 - \frac{2}{3} = \frac{1}{3}$.
$P(B) = \frac{3}{5}$,so $P(\bar{B}) = 1 - \frac{3}{5} = \frac{2}{5}$.
$P(C) = \frac{2}{3}$,so $P(\bar{C}) = 1 - \frac{2}{3} = \frac{1}{3}$.
At least two hit the balloon means exactly two hit $OR$ all three hit.
Probability (exactly two hit) = $P(A)P(B)P(\bar{C}) + P(A)P(\bar{B})P(C) + P(\bar{A})P(B)P(C)$
$= (\frac{2}{3} \times \frac{3}{5} \times \frac{1}{3}) + (\frac{2}{3} \times \frac{2}{5} \times \frac{2}{3}) + (\frac{1}{3} \times \frac{3}{5} \times \frac{2}{3})$
$= \frac{6}{45} + \frac{8}{45} + \frac{6}{45} = \frac{20}{45} = \frac{4}{9}$.
Probability (all three hit) = $P(A)P(B)P(C) = \frac{2}{3} \times \frac{3}{5} \times \frac{2}{3} = \frac{12}{45} = \frac{4}{15}$.
Total probability = $\frac{20}{45} + \frac{12}{45} = \frac{32}{45}$.

(C) Let $G$ denote a girl and $B$ denote a boy. The probabilities for each group are as follows:
Group $A$: $P(G_A) = \frac{3}{4}$,$P(B_A) = \frac{1}{4}$
Group $B$: $P(G_B) = \frac{2}{4}$,$P(B_B) = \frac{2}{4}$
Group $C$: $P(G_C) = \frac{1}{4}$,$P(B_C) = \frac{3}{4}$
We need to select $1$ girl and $2$ boys. The possible cases are:

Case	Selection from $A, B, C$	Probability
$1$	$G, B, B$	$\frac{3}{4} \times \frac{2}{4} \times \frac{3}{4} = \frac{18}{64}$
$2$	$B, G, B$	$\frac{1}{4} \times \frac{2}{4} \times \frac{3}{4} = \frac{6}{64}$
$3$	$B, B, G$	$\frac{1}{4} \times \frac{2}{4} \times \frac{1}{4} = \frac{2}{64}$

Total probability $= \frac{18}{64} + \frac{6}{64} + \frac{2}{64} = \frac{26}{64} = \frac{13}{32}$.

(B) The set of numbers is $S = \{1, 2, 3, \ldots, 1000\}$,so the total number of outcomes $n(S) = 1000$.
We are looking for numbers $n$ such that $n \equiv 1 \pmod{7}$.
These numbers are of the form $n = 7k + 1$,where $k$ is an integer.
For $1 \le n \le 1000$,we have $1 \le 7k + 1 \le 1000$.
Subtracting $1$ from all parts: $0 \le 7k \le 999$.
Dividing by $7$: $0 \le k \le \frac{999}{7} \approx 142.71$.
Since $k$ must be an integer,$k$ can take values from $0, 1, 2, \ldots, 142$.
The number of such values is $142 - 0 + 1 = 143$.
Thus,the number of favorable outcomes $n(E) = 143$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{143}{1000}$.

(D) When two dice are thrown,the total number of outcomes is $6 \times 6 = 36$.
We need to find the number of pairs $(x, y)$ such that $x^2 + y^2 \leq 25$,where $x, y \in \{1, 2, 3, 4, 5, 6\}$.
Let us list the favorable outcomes for each value of $x$:
If $x = 1$,then $1^2 + y^2 \leq 25 \implies y^2 \leq 24$. Possible $y$ values are $\{1, 2, 3, 4\}$ ($4$ outcomes).
If $x = 2$,then $2^2 + y^2 \leq 25 \implies y^2 \leq 21$. Possible $y$ values are $\{1, 2, 3, 4\}$ ($4$ outcomes).
If $x = 3$,then $3^2 + y^2 \leq 25 \implies y^2 \leq 16$. Possible $y$ values are $\{1, 2, 3, 4\}$ ($4$ outcomes).
If $x = 4$,then $4^2 + y^2 \leq 25 \implies y^2 \leq 9$. Possible $y$ values are $\{1, 2, 3\}$ ($3$ outcomes).
If $x = 5$,then $5^2 + y^2 \leq 25 \implies y^2 \leq 0$. No possible $y$ values since $y \geq 1$.
If $x = 6$,then $6^2 + y^2 \leq 25$. No possible $y$ values.
Total favorable outcomes $= 4 + 4 + 4 + 3 = 15$.
The probability is $\frac{15}{36} = \frac{5}{12}$.
Thus,option $(d)$ is correct.

(C) Let $P(A)$ be the probability that the first person hits the target and $P(B)$ be the probability that the second person hits the target.
Given $P(A) = \frac{1}{4}$ and $P(B) = \frac{1}{5}$.
The target is hit if at least one of them hits the target.
It is easier to calculate the probability that the target is not hit at all.
The probability that the first person misses is $P(A') = 1 - P(A) = 1 - \frac{1}{4} = \frac{3}{4}$.
The probability that the second person misses is $P(B') = 1 - P(B) = 1 - \frac{1}{5} = \frac{4}{5}$.
Since they attempt independently,the probability that both miss is $P(A' \cap B') = P(A') \times P(B') = \frac{3}{4} \times \frac{4}{5} = \frac{3}{5}$.
The probability that the target is hit is $1 - P(A' \cap B') = 1 - \frac{3}{5} = \frac{2}{5}$.
Thus,the correct option is $C$.

(A) Let $x$ be the number on the chit drawn from box $Q$ and $y$ be the number on the chit drawn from box $P$. Both $x$ and $y$ are integers such that $1 \leq x, y \leq 100$.
We are given the condition that $y = x^2$.
Since $y \leq 100$,we must have $x^2 \leq 100$,which implies $x \leq 10$.
Thus,the possible pairs $(x, y)$ are $(1, 1), (2, 4), (3, 9), (4, 16), (5, 25), (6, 36), (7, 49), (8, 64), (9, 81), (10, 100)$.
There are $10$ such favorable outcomes.
The total number of possible outcomes when drawing one chit from each box is $100 \times 100 = 10000$.
The probability is given by $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{10}{10000} = \frac{1}{1000}$.
Converting this to a percentage: $\frac{1}{1000} \times 100\% = 0.1\%$.

(C) Let $A$ be the event of qualifying for $IITJEE$ and $B$ be the event of qualifying for $EAMCET$.
Given: $P(A) = \frac{1}{5}$ and $P(B) = \frac{3}{5}$.
Assuming the events are independent,the probability of qualifying for at least one test is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since the events are independent,$P(A \cap B) = P(A) \times P(B) = \frac{1}{5} \times \frac{3}{5} = \frac{3}{25}$.
Therefore,$P(A \cup B) = \frac{1}{5} + \frac{3}{5} - \frac{3}{25}$.
$P(A \cup B) = \frac{5}{25} + \frac{15}{25} - \frac{3}{25} = \frac{17}{25}$.

(A) Let $P(B)=x$. Then,$P(A)=\frac{2}{3} x$ and $P(C)=\frac{1}{2} x$.
If $A, B, C$ are exhaustive and mutually exclusive,then $P(A \cup B \cup C)=1$.
Since they are mutually exclusive,$P(A \cup B \cup C) = P(A) + P(B) + P(C) = 1$.
Substituting the values: $\frac{2}{3} x + x + \frac{1}{2} x = 1$.
$\frac{4x + 6x + 3x}{6} = 1 \Rightarrow \frac{13x}{6} = 1 \Rightarrow x = \frac{6}{13}$.
Thus,$P(A) = \frac{2}{3} \times \frac{6}{13} = \frac{4}{13}$,$P(B) = \frac{6}{13}$,and $P(C) = \frac{1}{2} \times \frac{6}{13} = \frac{3}{13}$.
Now,$P(A \cup C) = P(A) + P(C) = \frac{4}{13} + \frac{3}{13} = \frac{7}{13}$.
Therefore,option $A$ is correct.

(B) The total number of outcomes when tossing two dice is $n(S) = 6 \times 6 = 36$.
$E$ is the event of getting a sum of $8$. The outcomes are $\{(2,6), (3,5), (4,4), (5,3), (6,2)\}$.
Thus,$n(E) = 5$,and $P(E) = \frac{n(E)}{n(S)} = \frac{5}{36}$. So,statement $I$ is false.
$F$ is the event of getting even numbers on both dice. The outcomes are $\{(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)\}$.
Thus,$n(F) = 9$,and $P(F) = \frac{n(F)}{n(S)} = \frac{9}{36} = \frac{1}{4}$. So,statement $II$ is false.
Therefore,neither $I$ nor $II$ is true.

(B) The probability of getting $5$ on a die is $P(A) = \frac{1}{6}$.
The probability of getting a tail on a coin is $P(B) = \frac{1}{2}$.
Since the events are independent,the required probability is $P(A \cap B) = P(A) \times P(B) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$.

(B) non-leap year has $365$ days,which is equal to $52$ weeks and $1$ extra day.
The sample space for this extra day is $S = \{ \text{Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday} \}$.
There are $7$ possible outcomes for this extra day.
Let $A$ be the event of having $53$ Sundays and $B$ be the event of having $53$ Saturdays.
The year will have $53$ Sundays if the extra day is a Sunday,so $P(A) = \frac{1}{7}$.
The year will have $53$ Saturdays if the extra day is a Saturday,so $P(B) = \frac{1}{7}$.
Since these events are mutually exclusive,the probability of having $53$ Sundays or $53$ Saturdays is $P(A \cup B) = P(A) + P(B) = \frac{1}{7} + \frac{1}{7} = \frac{2}{7}$.