Set Based probability Questions in English

(A) Let $A$ be the event that the selected student has opted for $NCC$ and $B$ be the event that the selected student has opted for $NSS$.
Total number of students $= 60$.
Number of students who have opted for $NCC$,$n(A) = 30$.
Number of students who have opted for $NSS$,$n(B) = 32$.
Number of students who have opted for both $NCC$ and $NSS$,$n(A \cap B) = 24$.
We need to find the probability that the student has opted for $NSS$ but not $NCC$,which is $P(B - A)$.
The number of students who have opted for $NSS$ but not $NCC$ is given by $n(B - A) = n(B) - n(A \cap B)$.
$n(B - A) = 32 - 24 = 8$.
The probability is $P(B - A) = \frac{n(B - A)}{\text{Total students}} = \frac{8}{60}$.
Simplifying the fraction,we get $\frac{8}{60} = \frac{2}{15}$.

(C) Total number of faces on the die $= 6$.
Number of faces with the number $2 = 3$.
Probability $P(2) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2}$.

(A) The total number of faces on the die is $6$.
The number of faces with $1$ is $2$.
The number of faces with $3$ is $1$.
The probability of getting $1$ or $3$ is given by:
$P(1 \text{ or } 3) = P(1) + P(3) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.

(C) Total number of faces on the die $= 6$.
Number of faces with number $3 = 1$.
Therefore,the probability of getting $3$ is $P(3) = \frac{1}{6}$.
Thus,the probability of not getting $3$ is $P(\text{not } 3) = 1 - P(3) = 1 - \frac{1}{6} = \frac{5}{6}$.

(B) Total number of tickets sold $= 10,000$.
Number of prizes awarded $= 10$.
If you buy one ticket,the probability of winning a prize is $P(\text{winning}) = \frac{10}{10000} = \frac{1}{1000}$.
The probability of not getting a prize is $P(\text{not winning}) = 1 - P(\text{winning})$.
$P(\text{not winning}) = 1 - \frac{1}{1000} = \frac{999}{1000}$.

(A) Let the two sections be $S_1$ (size $40$) and $S_2$ (size $60$).
Total ways to place $2$ specific students into these sections is the total number of ways to choose $2$ spots out of $100$ for them,which is $100 \times 99$.
Alternatively,consider the probability that both are in the same section.
Probability both are in $S_1 = \frac{40}{100} \times \frac{39}{99} = \frac{2}{5} \times \frac{13}{33} = \frac{26}{165}$.
Probability both are in $S_2 = \frac{60}{100} \times \frac{59}{99} = \frac{3}{5} \times \frac{59}{99} = \frac{59}{165}$.
Probability both are in the same section $= \frac{26}{165} + \frac{59}{165} = \frac{85}{165} = \frac{17}{33}$.
Probability both are in different sections $= 1 - \frac{17}{33} = \frac{16}{33}$.

(A) Given that $P(A)=0.54$,$P(B)=0.69$,and $P(A \cap B)=0.35$.
We know that the probability of $A$ occurring and $B$ not occurring is given by the formula:
$P(A \cap B^{\prime}) = P(A) - P(A \cap B)$.
Substituting the given values:
$P(A \cap B^{\prime}) = 0.54 - 0.35$.
$P(A \cap B^{\prime}) = 0.19$.

(A) Given that $P(A)=0.54$,$P(B)=0.69$,and $P(A \cap B)=0.35$.
We know that the probability of the event $B$ occurring but not $A$ is given by $P(B \cap A^{\prime}) = P(B) - P(A \cap B)$.
Substituting the given values:
$P(B \cap A^{\prime}) = 0.69 - 0.35$.
Therefore,$P(B \cap A^{\prime}) = 0.34$.

(C) Let $S = \{1, 2, 3, 4, 5\}$. The total number of subsets of $S$ is $2^{5} = 32$.
Since two subsets $A$ and $B$ are selected randomly,the total number of pairs $(A, B)$ is $32 \times 32 = 2^{10}$.
For each element $x \in S$,there are $4$ possibilities for its membership in $A$ and $B$: $x \notin A, x \notin B$; $x \in A, x \notin B$; $x \notin A, x \in B$; or $x \in A, x \in B$.
We want the intersection $A \cap B$ to have exactly $2$ elements.
First,choose $2$ elements out of $5$ to be in the intersection: $\binom{5}{2} = 10$ ways.
For the remaining $3$ elements,each must not be in the intersection,meaning they can be in $(A \setminus B)$,$(B \setminus A)$,or in neither $A$ nor $B$. There are $3$ such choices for each of the $3$ remaining elements,giving $3^{3} = 27$ ways.
Thus,the number of favorable pairs is $10 \times 27 = 270$.
The probability is $\frac{270}{2^{10}} = \frac{270}{1024} = \frac{135}{512} = \frac{135}{2^{9}}$.

(B) Let the pair of opposite faces be $(a, b)$ where $a+b=7$. The probability of getting $a$ is $P(a) = \frac{1}{6}-x$ and $P(b) = \frac{1}{6}+x$. For the other two pairs of opposite faces,the probability of each face is $\frac{1}{6}$.
The sum of two rolls is $7$ if the outcomes are $(1,6), (6,1), (2,5), (5,2), (3,4), (4,3)$.
Probability of sum $7 = 2[P(1)P(6) + P(2)P(5) + P(3)P(4)]$.
Assuming $(1,6)$ are the faces with probabilities $\frac{1}{6}-x$ and $\frac{1}{6}+x$,then $P(2)=P(5)=\frac{1}{6}$ and $P(3)=P(4)=\frac{1}{6}$.
Sum probability $= 2[(\frac{1}{6}-x)(\frac{1}{6}+x) + (\frac{1}{6})(\frac{1}{6}) + (\frac{1}{6})(\frac{1}{6})] = \frac{13}{96}$.
$2[(\frac{1}{36}-x^2) + \frac{1}{36} + \frac{1}{36}] = \frac{13}{96}$.
$2[\frac{3}{36}-x^2] = \frac{13}{96} \Rightarrow \frac{1}{6}-2x^2 = \frac{13}{96}$.
$2x^2 = \frac{1}{6}-\frac{13}{96} = \frac{16-13}{96} = \frac{3}{96} = \frac{1}{32}$.
$x^2 = \frac{1}{64} \Rightarrow x = \frac{1}{8}$.

(A) The total number of $2$-digit numbers is $90$ (from $10$ to $99$).
We need to check when $(2^{n}-2)$ is a multiple of $3$.
Consider the expression modulo $3$:
$2 \equiv -1 \pmod{3}$
So,$2^{n}-2 \equiv (-1)^{n}-2 \pmod{3}$.
If $n$ is even,$(-1)^{n}-2 = 1-2 = -1 \equiv 2 \pmod{3}$.
If $n$ is odd,$(-1)^{n}-2 = -1-2 = -3 \equiv 0 \pmod{3}$.
Thus,$(2^{n}-2)$ is a multiple of $3$ if and only if $n$ is an odd number.
In the set of $2$-digit numbers ${10, 11, 12, \dots, 99}$,the odd numbers are ${11, 13, 15, \dots, 99}$.
The number of odd integers in this range is $\frac{99-11}{2} + 1 = 44 + 1 = 45$.
The probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{45}{90} = \frac{1}{2}$.

(B) The region is bounded by the $y$-axis $(x=0)$,$2y+x=6$,and $5x-6y=30$. The vertices of the triangle are $B(0, 3)$,$C(0, -5)$,and $A(6, 0)$.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 6 = 24$.
The line $y=1$ intersects $2y+x=6$ at $D(4, 1)$ and the $y$-axis at $E(0, 1)$.
The region where $y < 1$ is the quadrilateral $ADEC$.
The area of $\triangle BDE$ (where $y \ge 1$) is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 2 = 4$.
Required probability $= 1 - \frac{\text{Area}(\triangle BDE)}{\text{Area}(\triangle ABC)} = 1 - \frac{4}{24} = 1 - \frac{1}{6} = \frac{5}{6}$.

(B) For any event $E_{i}$,$0 \leq P(E_{i}) \leq 1$.
For $P(E_{1}) = \frac{2+3p}{6}$,$0 \leq 2+3p \leq 6 \implies -2 \leq 3p \leq 4 \implies -\frac{2}{3} \leq p \leq \frac{4}{3}$.
For $P(E_{2}) = \frac{2-p}{8}$,$0 \leq 2-p \leq 8 \implies -2 \leq -p \leq 6 \implies -6 \leq p \leq 2$.
For $P(E_{3}) = \frac{1-p}{2}$,$0 \leq 1-p \leq 2 \implies -1 \leq -p \leq 1 \implies -1 \leq p \leq 1$.
Since $E_{1}, E_{2}, E_{3}$ are mutually exclusive,$P(E_{1}) + P(E_{2}) + P(E_{3}) \leq 1$.
$\frac{2+3p}{6} + \frac{2-p}{8} + \frac{1-p}{2} \leq 1$.
Multiplying by $24$: $4(2+3p) + 3(2-p) + 12(1-p) \leq 24$.
$8 + 12p + 6 - 3p + 12 - 12p \leq 24$.
$26 - 3p \leq 24 \implies 2 \leq 3p \implies p \geq \frac{2}{3}$.
Combining all constraints: $p \in [\frac{2}{3}, 1]$.
Thus,$p_{1} = 1$ and $p_{2} = \frac{2}{3}$.
$p_{1} + p_{2} = 1 + \frac{2}{3} = \frac{5}{3}$.

(C) The total number of $5$-digit numbers is $n(S) = 9 \times 10^4 = 90000$.
Let $A$ be the set of numbers that are multiples of $7$ but not divisible by $5$.
Smallest $5$-digit number divisible by $7$ is $10003$.
Largest $5$-digit number divisible by $7$ is $99995$.
Using the formula $a_n = a + (n-1)d$,we have $99995 = 10003 + (n-1)7$,which gives $n = 12857$.
Now,find numbers divisible by both $7$ and $5$,i.e.,divisible by $35$.
Smallest $5$-digit number divisible by $35$ is $10010$.
Largest $5$-digit number divisible by $35$ is $99995$.
$99995 = 10010 + (P-1)35$,which gives $P = 2572$.
The number of elements in $A$ is $12857 - 2572 = 10285$.
The probability $p = \frac{10285}{90000}$.
Therefore,$9p = 9 \times \frac{10285}{90000} = \frac{10285}{10000} = 1.0285$.

(C) There are $32$ persons seated around a circular table. Fix the position of $X_1$. There are $31$ remaining seats where $X_3$ can be placed,each with equal probability of $\frac{1}{31}$.
$X_1$ and $X_3$ are within earshot if there are at most $3$ persons between them on the minor arc. Let $k$ be the number of persons between $X_1$ and $X_3$ on the minor arc.
If $k=0$,$X_3$ is adjacent to $X_1$. There are $2$ such positions (left or right).
If $k=1$,there is $1$ person between $X_1$ and $X_3$. There are $2$ such positions.
If $k=2$,there are $2$ persons between $X_1$ and $X_3$. There are $2$ such positions.
If $k=3$,there are $3$ persons between $X_1$ and $X_3$. There are $2$ such positions.
Total favorable positions for $X_3 = 2 + 2 + 2 + 2 = 8$.
Total possible positions for $X_3 = 31$.
Therefore,the probability is $\frac{8}{31}$.

(C) We have $S_n = \frac{n(n+1)}{2}$.
The sequence of sums is $S_1=1, S_2=3, S_3=6, S_4=10, S_5=15, S_6=21, S_7=28, S_8=36, \ldots$.
Observing the parity of $S_n$:
$S_1$ (odd),$S_2$ (odd),$S_3$ (even),$S_4$ (even),$S_5$ (odd),$S_6$ (odd),$S_7$ (even),$S_8$ (even),...
The pattern of parity repeats every $4$ terms: (odd,odd,even,even).
In the set ${S_1, S_2, \ldots, S_{99}}$,there are $99$ terms.
Since $99 = 4 \times 24 + 3$,we have $24$ full cycles of (odd,odd,even,even) and the first $3$ terms of the next cycle (odd,odd,even).
Number of even terms = $24 \times 2 + 1 = 49$.
Total number of cards = $99$.
Probability of drawing an even number = $\frac{\text{Number of even terms}}{\text{Total number of terms}} = \frac{49}{99}$.

(C) Player $A$ has cards $\{3, 5, 6\}$. The possible products of two cards are:
$(3 \times 5) = 15$,$(3 \times 6) = 18$,$(5 \times 6) = 30$.
Each product occurs with probability $\frac{1}{3}$.
Player $B$ has cards $\{8, 9, 10\}$. The possible sums of two cards are:
$(8 + 9) = 17$,$(8 + 10) = 18$,$(9 + 10) = 19$.
Each sum occurs with probability $\frac{1}{3}$.
$A$ wins if the product is greater than the sum.
Let $P_A$ be the product and $S_B$ be the sum.
If $P_A = 15$ (prob $\frac{1}{3}$): $A$ wins if $S_B < 15$. No outcomes satisfy this.
If $P_A = 18$ (prob $\frac{1}{3}$): $A$ wins if $S_B < 18$. Only $S_B = 17$ works (prob $\frac{1}{3}$). Probability = $\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$.
If $P_A = 30$ (prob $\frac{1}{3}$): $A$ wins if $S_B < 30$. All $S_B$ values $(17, 18, 19)$ satisfy this (prob $1$). Probability = $\frac{1}{3} \times 1 = \frac{1}{3}$.
Total probability = $\frac{1}{9} + \frac{1}{3} = \frac{1+3}{9} = \frac{4}{9}$.

(C) In probability theory,for a sample space $\Omega$ and an event $A \subseteq \Omega$:
$1$. If $P(A) = 0$,it implies that the event $A$ is an impossible event,which means $A = \phi$. Thus,$(S1)$ is true.
$2$. If $P(A) = 1$,it implies that the event $A$ is a sure event,which means $A = \Omega$. Thus,$(S2)$ is true.
Therefore,both statements $(S1)$ and $(S2)$ are true.

(B) The product of two integers $x$ and $66-x$ is $f(x) = x(66-x)$.
This is a downward parabola with maximum at $x = 33$.
Thus,$M = 33 \times 33 = 1089$.
We require $x(66-x) \geq \frac{5}{9} \times 1089 = 5 \times 121 = 605$.
$66x - x^2 \geq 605 \implies x^2 - 66x + 605 \leq 0$.
Solving $x^2 - 66x + 605 = 0$ using the quadratic formula: $x = \frac{66 \pm \sqrt{4356 - 2420}}{2} = \frac{66 \pm \sqrt{1936}}{2} = \frac{66 \pm 44}{2}$.
So,$x_1 = 11$ and $x_2 = 55$.
The set $S = \{11, 12, \ldots, 55\}$,so the number of elements $n(S) = 55 - 11 + 1 = 45$.
The event $A$ consists of multiples of $3$ in $S$: $A = \{12, 15, 18, \ldots, 54\}$.
This is an arithmetic progression with $a = 12$,$l = 54$,and $d = 3$.
$54 = 12 + (n-1)3 \implies 42 = (n-1)3 \implies n-1 = 14 \implies n = 15$.
Thus,$n(A) = 15$.
$P(A) = \frac{n(A)}{n(S)} = \frac{15}{45} = \frac{1}{3}$.

(B) The total number of outcomes when two dice are rolled is $n(S) = 6 \times 6 = 36$.
Given that $N - 2, \sqrt{3N}, N + 2$ are in geometric progression ($G$.$P$.),the square of the middle term must equal the product of the extremes:
$(\sqrt{3N})^2 = (N - 2)(N + 2)$
$3N = N^2 - 4$
$N^2 - 3N - 4 = 0$
Factoring the quadratic equation:
$(N - 4)(N + 1) = 0$
Since $N$ is the sum of two dice,$N \geq 2$,so $N = 4$ is the only valid solution.
The outcomes for which the sum $N = 4$ are: $(1, 3), (3, 1), (2, 2)$.
Thus,the number of favorable outcomes is $n(A) = 3$.
The probability is $P(A) = \frac{3}{36} = \frac{1}{12}$.
We are given $P(A) = \frac{k}{48}$,so:
$\frac{k}{48} = \frac{1}{12}$
$k = \frac{48}{12} = 4$.

(D) Let the two numbers be $x$ and $y$ where $0 \le x, y \le 60$. The total area of the sample space is $60 \times 60 = 3600$.
The condition is $|x - y| \le a$,which implies $-a \le x - y \le a$.
The area of the region where $|x - y| > a$ is the sum of the areas of two right-angled triangles with legs $(60 - a)$.
Area of region $|x - y| > a = \frac{1}{2}(60 - a)^2 + \frac{1}{2}(60 - a)^2 = (60 - a)^2$.
Thus,the area of the region where $|x - y| \le a$ is $3600 - (60 - a)^2$.
Given $P(A) = \frac{3600 - (60 - a)^2}{3600} = \frac{11}{36}$.
Multiplying by $3600$,we get $3600 - (60 - a)^2 = 1100$.
$(60 - a)^2 = 3600 - 1100 = 2500$.
$60 - a = 50 \Rightarrow a = 10$.

(A) Let the outcomes of the two dice be $(x, y)$ where $x, y \in \{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is $6 \times 6 = 36$.
Event $A$: $x < y$. The number of outcomes is $5 + 4 + 3 + 2 + 1 = 15$.
Event $B$: $x \in \{2, 4, 6\}$ and $y \in \{1, 3, 5\}$. The number of outcomes is $3 \times 3 = 9$.
Event $C$: $x \in \{1, 3, 5\}$ and $y \in \{2, 4, 6\}$. The number of outcomes is $3 \times 3 = 9$.
Now,$(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$.
$B \cap C$: $x$ is even and odd,which is impossible,so $B \cap C = \emptyset$.
$A \cap C$: $x < y$ and $x \in \{1, 3, 5\}, y \in \{2, 4, 6\}$.
If $x=1$,$y \in \{2, 4, 6\}$ ($3$ cases).
If $x=3$,$y \in \{4, 6\}$ ($2$ cases).
If $x=5$,$y \in \{6\}$ ($1$ case).
Total cases for $A \cap C = 3 + 2 + 1 = 6$.
Thus,the number of favourable cases for $(A \cup B) \cap C$ is $6 + 0 = 6$.

(A) Let $N$ be the sum of the numbers on two dice. The possible values for $N$ are $2, 3, 4, \dots, 12$.
We need to find the probability that $2^{N} < N!$.
Let us check the condition $2^{N} < N!$ for each $N$:
For $N=2: 2^2 = 4, 2! = 2$. $4 < 2$ is false.
For $N=3: 2^3 = 8, 3! = 6$. $8 < 6$ is false.
For $N=4: 2^4 = 16, 4! = 24$. $16 < 24$ is true.
For $N=5: 2^5 = 32, 5! = 120$. $32 < 120$ is true.
For $N \geq 4$,the condition $2^N < N!$ holds true.
Thus,we need to find the probability that $N \geq 4$.
$P(N \geq 4) = 1 - P(N < 4) = 1 - (P(N=2) + P(N=3))$.
The total number of outcomes is $6 \times 6 = 36$.
$P(N=2) = \frac{1}{36}$ (outcomes: $(1,1)$).
$P(N=3) = \frac{2}{36}$ (outcomes: $(1,2), (2,1)$).
$P(N < 4) = \frac{1}{36} + \frac{2}{36} = \frac{3}{36} = \frac{1}{12}$.
Therefore,$P(N \geq 4) = 1 - \frac{1}{12} = \frac{11}{12}$.
Here,$m = 11$ and $n = 12$. Since $11$ and $12$ are coprime,we calculate $4m - 3n = 4(11) - 3(12) = 44 - 36 = 8$.

(A) The total number of ways to choose two integers $x$ and $y$ with replacement from the set $\{0, 1, 2, \ldots, 10\}$ is $11 \times 11 = 121$.
We want to find the number of pairs $(x, y)$ such that $|x-y| > 5$.
Case $1$: $x - y > 5 \implies x - y \in \{6, 7, 8, 9, 10\}$.
If $x=6, y=0$; if $x=7, y=0, 1$; if $x=8, y=0, 1, 2$; if $x=9, y=0, 1, 2, 3$; if $x=10, y=0, 1, 2, 3, 4$.
The number of such pairs is $1 + 2 + 3 + 4 + 5 = 15$.
Case $2$: $y - x > 5 \implies y - x \in \{6, 7, 8, 9, 10\}$.
By symmetry,the number of such pairs is also $15$.
Total favorable outcomes $= 15 + 15 = 30$.
Therefore,the required probability is $\frac{30}{121}$.

(B) Let $A$ be the event that Ajay appears in the exam and $V$ be the event that Vijay appears in the exam.
Given: $P(\overline{A}) = \frac{2}{7}$,so $P(A) = 1 - P(\overline{A}) = 1 - \frac{2}{7} = \frac{5}{7}$.
Given: $P(A \cap V) = \frac{1}{5}$.
We need to find the probability that Ajay appears and Vijay does not appear,which is $P(A \cap \overline{V})$.
Using the property of sets,$P(A) = P(A \cap V) + P(A \cap \overline{V})$.
Substituting the values: $\frac{5}{7} = \frac{1}{5} + P(A \cap \overline{V})$.
$P(A \cap \overline{V}) = \frac{5}{7} - \frac{1}{5} = \frac{25 - 7}{35} = \frac{18}{35}$.

(D) Let the two positive integers be $x$ and $y$. Given $x+y=24$,where $x, y \in \mathbb{N}$.
The product $P = xy$. By the $AM-GM$ inequality,$\frac{x+y}{2} \geq \sqrt{xy}$,so $\sqrt{xy} \leq 12$,which implies $xy \leq 144$. The greatest positive product is $144$ (when $x=12, y=12$).
We need the probability that $xy \geq \frac{3}{4} \times 144$,which means $xy \geq 108$.
Since $y = 24-x$,we have $x(24-x) \geq 108$,or $24x - x^2 \geq 108$,which simplifies to $x^2 - 24x + 108 \leq 0$.
Solving $x^2 - 24x + 108 = 0$ using the quadratic formula: $x = \frac{24 \pm \sqrt{576 - 432}}{2} = \frac{24 \pm \sqrt{144}}{2} = \frac{24 \pm 12}{2}$.
Thus,$x = 6$ or $x = 18$. The inequality holds for $6 \leq x \leq 18$.
The possible values for $x$ are ${6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}$.
There are $18 - 6 + 1 = 13$ such values.
The total number of pairs $(x, y)$ such that $x+y=24$ for $x, y \in \mathbb{N}$ is $23$ (since $x$ can range from $1$ to $23$).
The probability is $\frac{13}{23} = \frac{m}{n}$.
Thus,$m=13$ and $n=23$. Then $n-m = 23-13 = 10$.

(B) The total number of non-negative integer solutions to $x+y+z=n$ is given by the formula $\binom{n+k-1}{k-1}$,where $k=3$ and $n=10$.
Total solutions $= \binom{10+3-1}{3-1} = \binom{12}{2} = \frac{12 \times 11}{2} = 66$.
For $z$ to be even,let $z \in \{0, 2, 4, 6, 8, 10\}$.
If $z=k$,then $x+y=10-k$. The number of solutions for $x+y=m$ is $m+1$.
For $z=0$,$x+y=10$,solutions $= 11$.
For $z=2$,$x+y=8$,solutions $= 9$.
For $z=4$,$x+y=6$,solutions $= 7$.
For $z=6$,$x+y=4$,solutions $= 5$.
For $z=8$,$x+y=2$,solutions $= 3$.
For $z=10$,$x+y=0$,solutions $= 1$.
Total favourable solutions $= 11+9+7+5+3+1 = 36$.
Probability $P = \frac{36}{66} = \frac{6}{11}$.

(A) Let $P(A) = \frac{1}{2}, P(B) = \frac{3}{4}, P(C) = \frac{1}{4}, P(D) = \frac{1}{8}$ be the probabilities of solving the problem correctly.
The probability that the problem is not solved by any of them is $P(\overline{A}) \times P(\overline{B}) \times P(\overline{C}) \times P(\overline{D})$.
$P(\overline{A}) = 1 - \frac{1}{2} = \frac{1}{2}$
$P(\overline{B}) = 1 - \frac{3}{4} = \frac{1}{4}$
$P(\overline{C}) = 1 - \frac{1}{4} = \frac{3}{4}$
$P(\overline{D}) = 1 - \frac{1}{8} = \frac{7}{8}$
$P(\text{none solve}) = \frac{1}{2} \times \frac{1}{4} \times \frac{3}{4} \times \frac{7}{8} = \frac{21}{256}$
The probability that the problem is solved by at least one person is $1 - P(\text{none solve}) = 1 - \frac{21}{256} = \frac{235}{256}$.

(C) non-leap year has $365$ days,which is $52$ weeks and $1$ extra day.
The extra day can be any one of the $7$ days of the week: {Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday}.
Let $A$ be the event that the year has $53$ Saturdays and $B$ be the event that the year has $53$ Sundays.
Since the year has $52$ weeks,it already contains $52$ Saturdays and $52$ Sundays.
For the year to have $53$ Saturdays,the extra day must be a Saturday. $P(A) = \frac{1}{7}$.
For the year to have $53$ Sundays,the extra day must be a Sunday. $P(B) = \frac{1}{7}$.
Since the extra day cannot be both a Saturday and a Sunday simultaneously,$A$ and $B$ are mutually exclusive events.
The probability that the year has $53$ Saturdays or $53$ Sundays is $P(A \cup B) = P(A) + P(B) = \frac{1}{7} + \frac{1}{7} = \frac{2}{7}$.
Note: The question asks for $52$ Saturdays $OR$ $53$ Sundays. Since every non-leap year has $52$ Saturdays,the event 'contains $52$ Saturdays' is a certain event with probability $1$.
However,interpreting the standard problem format for this specific question type,it asks for the probability of having $53$ Saturdays or $53$ Sundays. Given the options,the intended answer is $\frac{2}{7}$.

(A) Each of the $3$ persons can choose any of the $3$ houses independently.
Total number of ways for $3$ persons to choose houses $= 3 \times 3 \times 3 = 27$.
For all $3$ persons to apply for the same house,they must all choose house $1$,or all choose house $2$,or all choose house $3$.
Thus,the number of favourable cases $= 3$.
Therefore,the required probability $= \frac{3}{27} = \frac{1}{9}$.

(A) Given that,the probability that exactly one of them occurs is $P(A \cup B) - P(A \cap B) = \frac{2}{5} \dots (i)$
The probability that $A$ or $B$ occurs is $P(A \cup B) = \frac{1}{2} \dots (ii)$
We know that the probability of exactly one event occurring is given by $P(A \cup B) - P(A \cap B)$.
Substituting the values from $(i)$ and $(ii)$:
$\frac{1}{2} - P(A \cap B) = \frac{2}{5}$
$P(A \cap B) = \frac{1}{2} - \frac{2}{5}$
$P(A \cap B) = \frac{5 - 4}{10} = \frac{1}{10} = 0.1$

(C) When two fair dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
We want the sum of the numbers on the upper faces to be at least $9$,which means the sum can be $9, 10, 11,$ or $12$.
The favorable outcomes are:
Sum $= 9: (3, 6), (4, 5), (5, 4), (6, 3)$
Sum $= 10: (4, 6), (5, 5), (6, 4)$
Sum $= 11: (5, 6), (6, 5)$
Sum $= 12: (6, 6)$
The total number of favorable outcomes is $4 + 3 + 2 + 1 = 10$.
Therefore,the required probability is $\frac{10}{36} = \frac{5}{18}$.

(A) Let $X$ be the event that the number on the ticket is a perfect square.
$\therefore X = \{1, 4, 9, 16, 25, 36, 49, 64, 81, 100\}$
$\therefore n(X) = 10$
$\text{Also, } n(S) = 100$
$\therefore \text{Required probability} = \frac{n(X)}{n(S)} = \frac{10}{100} = \frac{1}{10}$

(D) Total number of cards $n(S) = 52$.
Let event $A$ be drawing a black card and event $B$ be drawing a face card.
Number of black cards $n(A) = 26$.
Number of face cards $n(B) = 12$.
Number of black face cards $n(A \cap B) = 6$ (since there are $3$ black face cards in each of the two black suits).
The probability of drawing a black card or a face card is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{26}{52} + \frac{12}{52} - \frac{6}{52} = \frac{32}{52} = \frac{8}{13}$.

(A) The sample space $S$ consists of all possible pairs $(x, y)$ where $x, y \in \{1, 2, 3, 5, 7, 11\}$.
Since there are $6$ faces on each die,the total number of outcomes is $n(S) = 6 \times 6 = 36$.
Let $A$ be the event that the sum of the numbers is a prime number.
The possible sums are:
$1+1=2$ (prime),$1+2=3$ (prime),$1+3=4$,$1+5=6$,$1+7=8$,$1+11=12$
$2+1=3$ (prime),$2+2=4$,$2+3=5$ (prime),$2+5=7$ (prime),$2+7=9$,$2+11=13$ (prime)
$3+1=4$,$3+2=5$ (prime),$3+3=6$,$3+5=8$,$3+7=10$,$3+11=14$
$5+1=6$,$5+2=7$ (prime),$5+3=8$,$5+5=10$,$5+7=12$,$5+11=16$
$7+1=8$,$7+2=9$,$7+3=10$,$7+5=12$,$7+7=14$,$7+11=18$
$11+1=12$,$11+2=13$ (prime),$11+3=14$,$11+5=16$,$11+7=18$,$11+11=22$
The favorable outcomes are: $(1,1), (1,2), (2,1), (2,3), (2,5), (2,11), (3,2), (5,2), (11,2)$.
Thus,$n(A) = 9$.
The probability $P(A) = \frac{n(A)}{n(S)} = \frac{9}{36} = \frac{1}{4}$.

(B) Let $B$ represent a boy and $G$ represent a girl. The sample space $S$ for a family with $3$ children is:
$S = \{BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG\}$
The total number of outcomes is $n(S) = 8$.
Let $E$ be the event that the oldest and youngest children are of the same gender.
The favorable outcomes are:
$E = \{BBB, BGB, GBG, GGG\}$
The number of favorable outcomes is $n(E) = 4$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{4}{8} = \frac{1}{2}$.

(B) The total number of outcomes when two dice are rolled is $n(S) = 6 \times 6 = 36$.
The possible sums range from $2$ to $12$. The prime numbers in this range are $2, 3, 5, 7, 11$.
The favorable outcomes for each sum are:
Sum $= 2: (1, 1) \rightarrow 1 \text{ outcome}$
Sum $= 3: (1, 2), (2, 1) \rightarrow 2 \text{ outcomes}$
Sum $= 5: (1, 4), (2, 3), (3, 2), (4, 1) \rightarrow 4 \text{ outcomes}$
Sum $= 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) \rightarrow 6 \text{ outcomes}$
Sum $= 11: (5, 6), (6, 5) \rightarrow 2 \text{ outcomes}$
Total favorable outcomes $n(E) = 1 + 2 + 4 + 6 + 2 = 15$.
The probability is $P(E) = \frac{n(E)}{n(S)} = \frac{15}{36} = \frac{5}{12}$.

(D) The total number of outcomes when two dice are thrown is $6 \times 6 = 36$.
Let $A$ be the event of getting a doublet. The outcomes are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$,so $n(A) = 6$.
Let $B$ be the event of getting a total of $10$. The outcomes are $(4,6), (5,5), (6,4)$,so $n(B) = 3$.
The intersection $A \cap B$ is the outcome $(5,5)$,so $n(A \cap B) = 1$.
The number of outcomes that are either a doublet or a total of $10$ is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 6 + 3 - 1 = 8$.
The number of outcomes that are neither a doublet nor a total of $10$ is $36 - 8 = 28$.
Therefore,the required probability is $\frac{28}{36} = \frac{7}{9}$.

(A) Let $H$ be the event of getting a head and $D$ be the event of getting a number greater than $4$ on a die.
$P(H) = \frac{1}{2}$.
The numbers greater than $4$ on a die are ${5, 6}$,so $P(D) = \frac{2}{6} = \frac{1}{3}$.
Since the events are independent,$P(H \cap D) = P(H) \times P(D) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$.
We need to find the probability of $H \cup D$,which is given by $P(H \cup D) = P(H) + P(D) - P(H \cap D)$.
$P(H \cup D) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{3+2-1}{6} = \frac{4}{6} = \frac{2}{3}$.

(C) We know that for any two events $A$ and $B$,the probability of their union is given by:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Substituting the given values:
$\frac{5}{6} = \frac{1}{6} + \frac{2}{3} - P(A \cap B)$
$\frac{5}{6} = \frac{1}{6} + \frac{4}{6} - P(A \cap B)$
$\frac{5}{6} = \frac{5}{6} - P(A \cap B)$
This implies $P(A \cap B) = 0$.
Since the probability of the intersection of events $A$ and $B$ is $0$,the events $A$ and $B$ are mutually exclusive.

(B) Total number of cards $= 52$.
Number of aces $= 4$.
Number of black kings (spade king and club king) $= 2$.
Number of queen of hearts $= 1$.
Since these events are mutually exclusive,the total number of favorable outcomes $= 4 + 2 + 1 = 7$.
$\therefore$ Required probability $= \frac{7}{52}$.

(C) When two dice are thrown,the total number of outcomes is $6 \times 6 = 36$.
The possible sums range from $2$ to $12$.
The prime numbers in this range are $2, 3, 5, 7, 11$.
Favourable outcomes for each sum:
Sum $= 2$: $(1, 1)$ - $1$ case
Sum $= 3$: $(1, 2), (2, 1)$ - $2$ cases
Sum $= 5$: $(1, 4), (2, 3), (3, 2), (4, 1)$ - $4$ cases
Sum $= 7$: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$ - $6$ cases
Sum $= 11$: $(5, 6), (6, 5)$ - $2$ cases
Total favourable cases $= 1 + 2 + 4 + 6 + 2 = 15$.
Required probability $= \frac{15}{36} = \frac{5}{12}$.

(A) The probability that exactly one of the events $A$ or $B$ occurs is given by $P(A \Delta B) = P(A \cup B) - P(A \cap B)$.
Given $P(A \cup B) = 0.6$ and $P(A \cap B) = 0.2$.
Therefore,the probability of exactly one event occurring is $0.6 - 0.2 = 0.4$.

(D) When two dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
Let the outcome of the first die be $x$ and the second die be $y$. We want the probability that $x < y$.
The possible pairs $(x, y)$ such that $x < y$ are:
For $x=1$,$y \in \{2, 3, 4, 5, 6\}$ ($5$ outcomes).
For $x=2$,$y \in \{3, 4, 5, 6\}$ ($4$ outcomes).
For $x=3$,$y \in \{4, 5, 6\}$ ($3$ outcomes).
For $x=4$,$y \in \{5, 6\}$ ($2$ outcomes).
For $x=5$,$y \in \{6\}$ ($1$ outcome).
For $x=6$,there are no possible values for $y$.
Total favorable outcomes $= 5 + 4 + 3 + 2 + 1 = 15$.
The probability is $\frac{15}{36} = \frac{5}{12}$.

(A) Since $A$,$B$,and $C$ are mutually exclusive and exhaustive events,their sum of probabilities is $1$:
$P(A) + P(B) + P(C) = 1$
Given $P(B) = \frac{3}{2} P(A)$ and $P(C) = \frac{1}{2} P(B) = \frac{1}{2} \times \frac{3}{2} P(A) = \frac{3}{4} P(A)$
Substituting these into the equation:
$P(A) + \frac{3}{2} P(A) + \frac{3}{4} P(A) = 1$
Taking the common denominator as $4$:
$\frac{4 P(A) + 6 P(A) + 3 P(A)}{4} = 1$
$\frac{13 P(A)}{4} = 1$
$P(A) = \frac{4}{13}$

(D) When two dice are thrown,the total number of outcomes is $n(S) = 6 \times 6 = 36$.
Let $E$ be the event that the sum of the numbers is divisible by $2$ or $3$.
The possible sums range from $2$ to $12$.
Sums divisible by $2$ are: $2, 4, 6, 8, 10, 12$.
Sums divisible by $3$ are: $3, 6, 9, 12$.
Combining these,the sums divisible by $2$ or $3$ are: $2, 3, 4, 6, 8, 9, 10, 12$.
Counting the outcomes for each sum:
Sum $2: (1,1) - 1$ outcome
Sum $3: (1,2), (2,1) - 2$ outcomes
Sum $4: (1,3), (2,2), (3,1) - 3$ outcomes
Sum $6: (1,5), (2,4), (3,3), (4,2), (5,1) - 5$ outcomes
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 $12: (6,6) - 1$ outcome
Total favorable outcomes $n(E) = 1 + 2 + 3 + 5 + 5 + 4 + 3 + 1 = 24$.
Therefore,the probability $P(E) = \frac{n(E)}{n(S)} = \frac{24}{36} = \frac{2}{3}$.

(A) Let $W$ denote a white ball and $B$ denote a black ball. The contents of the boxes are:
Box $I$: $3W, 1B$ (Total $4$ balls)
Box $II$: $2W, 2B$ (Total $4$ balls)
Box $III$: $1W, 3B$ (Total $4$ balls)
We need to draw $2$ white and $1$ black ball. This can happen in three mutually exclusive ways:
$1$. Box $I$ gives $B$,Box $II$ gives $W$,Box $III$ gives $W$
$2$. Box $I$ gives $W$,Box $II$ gives $B$,Box $III$ gives $W$
$3$. Box $I$ gives $W$,Box $II$ gives $W$,Box $III$ gives $B$
The required probability is:
$P = P(B_I)P(W_{II})P(W_{III}) + P(W_I)P(B_{II})P(W_{III}) + P(W_I)P(W_{II})P(B_{III})$
$P = (\frac{1}{4} \times \frac{2}{4} \times \frac{1}{4}) + (\frac{3}{4} \times \frac{2}{4} \times \frac{1}{4}) + (\frac{3}{4} \times \frac{2}{4} \times \frac{3}{4})$
$P = \frac{2}{64} + \frac{6}{64} + \frac{18}{64} = \frac{26}{64} = \frac{13}{32}$

(B) Let $H$ be the event that the husband is selected and $W$ be the event that the wife is selected.
Given $P(H) = \frac{1}{7}$ and $P(W) = \frac{1}{5}$.
The probability that the husband is not selected is $P(H') = 1 - \frac{1}{7} = \frac{6}{7}$.
The probability that the wife is not selected is $P(W') = 1 - \frac{1}{5} = \frac{4}{5}$.
The probability that only one of them is selected is given by $P(\text{only } H) + P(\text{only } W)$.
$P(\text{only } H) = P(H) \times P(W') = \frac{1}{7} \times \frac{4}{5} = \frac{4}{35}$.
$P(\text{only } W) = P(W) \times P(H') = \frac{1}{5} \times \frac{6}{7} = \frac{6}{35}$.
Therefore,the required probability is $\frac{4}{35} + \frac{6}{35} = \frac{10}{35} = \frac{2}{7}$.

(D) Let $A, B, C,$ and $D$ be the events that the four persons hit the target respectively.
Given probabilities are $P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4}, P(D) = \frac{1}{5}$.
The probability that the target is not hit by any of them is the probability that all four miss the target.
$P(A') = 1 - \frac{1}{2} = \frac{1}{2}$
$P(B') = 1 - \frac{1}{3} = \frac{2}{3}$
$P(C') = 1 - \frac{1}{4} = \frac{3}{4}$
$P(D') = 1 - \frac{1}{5} = \frac{4}{5}$
Since the events are independent,the probability that none hit the target is:
$P(\text{None hit}) = P(A') \times P(B') \times P(C') \times P(D') = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{1}{5}$.
The probability that the target is hit at least once is:
$P(\text{Target is hit}) = 1 - P(\text{None hit}) = 1 - \frac{1}{5} = \frac{4}{5}$.