Set Based probability Questions in English

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

(D) Let $A$ be the event that the first student qualifies and $B$ be the event that the second student qualifies.
Given $P(A) = \frac{1}{4}$ and $P(B) = \frac{2}{5}$.
Since the events are independent,the probability that neither qualifies is $P(A^c \cap B^c) = P(A^c) \times P(B^c)$.
$P(A^c) = 1 - P(A) = 1 - \frac{1}{4} = \frac{3}{4}$.
$P(B^c) = 1 - P(B) = 1 - \frac{2}{5} = \frac{3}{5}$.
Probability that neither qualifies = $\frac{3}{4} \times \frac{3}{5} = \frac{9}{20}$.
Probability that at least one qualifies = $1 - P(\text{neither qualifies}) = 1 - \frac{9}{20} = \frac{11}{20}$.

(A) Given,$P(A)=0.4, P(B)=0.3$ and $P(A \cap B)=0.2$.
We need to find the probability that neither $A$ nor $B$ occurs,which is $P(\overline{A} \cap \overline{B})$.
By De Morgan's Law,$P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.
First,calculate $P(A \cup B)$ using the formula:
$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0.2 = 0.5$.
Therefore,$P(\overline{A} \cap \overline{B}) = 1 - 0.5 = 0.5$.

(C) $A =$ Event that $A$ speaks the truth.
$B =$ Event that $B$ speaks the truth.
$P(A) = 4/5 \implies P(A^c) = 1/5$.
$P(B) = 3/4 \implies P(B^c) = 1/4$.
$A$ and $B$ contradict each other if one speaks the truth and the other lies.
Required probability $= P(A) \cdot P(B^c) + P(A^c) \cdot P(B)$.
$= (4/5 \times 1/4) + (1/5 \times 3/4)$.
$= 4/20 + 3/20 = 7/20$.

(A) Since $A$ and $B$ are mutually exclusive events,$A \cap B = \phi$,which implies $P(A \cap B) = 0$.
Given $P(A) = \frac{1}{4}$ and $P(B) = \frac{3}{7}$.
We need to find $P(A / A \cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)}$.
Since $A \subset (A \cup B)$,we have $A \cap (A \cup B) = A$,so $P(A \cap (A \cup B)) = P(A) = \frac{1}{4}$.
Also,$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{4} + \frac{3}{7} - 0 = \frac{7 + 12}{28} = \frac{19}{28}$.
Therefore,$P(A / A \cup B) = \frac{1/4}{19/28} = \frac{1}{4} \times \frac{28}{19} = \frac{7}{19}$.

(A) Since $A, B, C$ are mutually exclusive,we have $P(A \cap B) = P(B \cap C) = P(C \cap A) = P(A \cap B \cap C) = 0$.
For any event $E$,$0 \leq P(E) \leq 1$.
$1$. For $P(A) = \frac{3x+1}{3}$: $0 \leq \frac{3x+1}{3} \leq 1 \implies 0 \leq 3x+1 \leq 3 \implies -1 \leq 3x \leq 2 \implies -\frac{1}{3} \leq x \leq \frac{2}{3}$.
$2$. For $P(B) = \frac{1-x}{4}$: $0 \leq \frac{1-x}{4} \leq 1 \implies 0 \leq 1-x \leq 4 \implies -1 \leq -x \leq 3 \implies -3 \leq x \leq 1$.
$3$. For $P(C) = \frac{1-2x}{2}$: $0 \leq \frac{1-2x}{2} \leq 1 \implies 0 \leq 1-2x \leq 2 \implies -1 \leq -2x \leq 1 \implies -\frac{1}{2} \leq x \leq \frac{1}{2}$.
Intersection of these intervals: $[-\frac{1}{3}, \frac{2}{3}] \cap [-3, 1] \cap [-\frac{1}{2}, \frac{1}{2}] = [-\frac{1}{3}, \frac{1}{2}]$.
Also,for mutually exclusive events,$P(A \cup B \cup C) = P(A) + P(B) + P(C) \leq 1$.
$\frac{3x+1}{3} + \frac{1-x}{4} + \frac{1-2x}{2} \leq 1$
$\frac{4(3x+1) + 3(1-x) + 6(1-2x)}{12} \leq 1$
$\frac{12x+4+3-3x+6-12x}{12} \leq 1 \implies \frac{13-3x}{12} \leq 1 \implies 13-3x \leq 12 \implies -3x \leq -1 \implies x \geq \frac{1}{3}$.
Combining $x \geq \frac{1}{3}$ with the intersection $[-\frac{1}{3}, \frac{1}{2}]$,we get $x \in [\frac{1}{3}, \frac{1}{2}]$.

(B) Let the winning probability of $C$ be $P(C) = p$.
Since $A$ and $B$ are twice as likely to win as $C$,we have $P(A) = 2p$ and $P(B) = 2p$.
Since the sum of probabilities of all possible outcomes is $1$,we have $P(A) + P(B) + P(C) = 1$.
Substituting the values,we get $2p + 2p + p = 1$,which implies $5p = 1$,so $p = \frac{1}{5}$.
The probability that $B$ or $C$ wins is $P(B \cup C) = P(B) + P(C)$ (since the events are mutually exclusive).
$P(B \cup C) = 2p + p = 3p = 3 \times \frac{1}{5} = \frac{3}{5}$.

(A) 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,$P(\bar{A} \cap \bar{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.
Using the addition rule of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values:
$P(A \cup B) = 0.58 + 0.32 - 0.28 = 0.62$.
Therefore,$P(\bar{A} \cap \bar{B}) = 1 - 0.62 = 0.38$.

(D) Let the probability of getting a number $i$ be $P(i)$.
Since $P(i) \propto i$,we have $P(i) = Ki$ for some constant $K$.
The sum of all probabilities must be $1$:
$\sum_{i=1}^{6} P(i) = K(1+2+3+4+5+6) = 21K = 1$.
Thus,$K = \frac{1}{21}$.
The probability of getting an odd number is $P(1) + P(3) + P(5)$.
$= K(1 + 3 + 5) = 9K$.
$= 9 \times \frac{1}{21} = \frac{9}{21} = \frac{3}{7}$.

(D) Given $P(\bar{A}) = \frac{1}{4}$,so $P(A) = 1 - \frac{1}{4} = \frac{3}{4}$.
Given $P(\overline{A \cup B}) = \frac{1}{6}$,so $P(A \cup B) = 1 - \frac{1}{6} = \frac{5}{6}$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we have $\frac{5}{6} = \frac{3}{4} + P(B) - \frac{1}{4}$.
$\frac{5}{6} = \frac{2}{4} + P(B) = \frac{1}{2} + P(B)$.
$P(B) = \frac{5}{6} - \frac{1}{2} = \frac{5-3}{6} = \frac{2}{6} = \frac{1}{3}$.
Since $P(A) = \frac{3}{4}$ and $P(B) = \frac{1}{3}$,$P(A) \neq P(B)$,so they are not equally likely.
Check for independence: $P(A) \times P(B) = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4}$.
Since $P(A \cap B) = \frac{1}{4}$,we have $P(A \cap B) = P(A) \times P(B)$,which means $A$ and $B$ are independent.
Thus,the events are independent but not equally likely.

(C) Since $A, B, C$ are mutually exclusive and exhaustive events,$P(A) + P(B) + P(C) = 1$.
Given $P(B) = 2 P(C)$,we can write $P(C) = \frac{1}{2} P(B)$.
Also given $P(A) = \frac{2}{3} P(B)$.
Substituting these into the sum: $\frac{2}{3} P(B) + P(B) + \frac{1}{2} P(B) = 1$.
Finding a common denominator $(6)$: $\frac{4}{6} P(B) + \frac{6}{6} P(B) + \frac{3}{6} P(B) = 1$.
$\frac{13}{6} P(B) = 1 \implies P(B) = \frac{6}{13}$.
Then $P(A) = \frac{2}{3} \times \frac{6}{13} = \frac{4}{13}$ and $P(C) = \frac{1}{2} \times \frac{6}{13} = \frac{3}{13}$.
Since $A$ and $C$ are mutually exclusive,$P(A \cup C) = P(A) + P(C) = \frac{4}{13} + \frac{3}{13} = \frac{7}{13}$.

(A) Let $P, Q, R$ be the events that $P, Q, R$ hit the target respectively.
Given probabilities are $P(P) = \frac{2}{3}, P(Q) = \frac{3}{5}, P(R) = \frac{5}{7}$.
The probabilities of not hitting the target are $P(P') = 1 - \frac{2}{3} = \frac{1}{3}$,$P(Q') = 1 - \frac{3}{5} = \frac{2}{5}$,and $P(R') = 1 - \frac{5}{7} = \frac{2}{7}$.
We need the probability that the target is hit by $P$ or $Q$ but not by $R$. This can happen in three mutually exclusive ways:
$1$. $P$ hits,$Q$ misses,$R$ misses: $P(P) \times P(Q') \times P(R') = \frac{2}{3} \times \frac{2}{5} \times \frac{2}{7} = \frac{8}{105}$.
$2$. $P$ misses,$Q$ hits,$R$ misses: $P(P') \times P(Q) \times P(R') = \frac{1}{3} \times \frac{3}{5} \times \frac{2}{7} = \frac{6}{105}$.
$3$. $P$ hits,$Q$ hits,$R$ misses: $P(P) \times P(Q) \times P(R') = \frac{2}{3} \times \frac{3}{5} \times \frac{2}{7} = \frac{12}{105}$.
Summing these probabilities: $\frac{8}{105} + \frac{6}{105} + \frac{12}{105} = \frac{26}{105}$.

(C) Total balls in box $P = 1 + 3 + 2 = 6$.
Total balls in box $Q = 2 + 3 + 4 = 9$.
Let $W_P, R_P, B_P$ be the events of drawing a white,red,and black ball from box $P$ respectively,and $W_Q, R_Q, B_Q$ be the corresponding events for box $Q$.
The probabilities are:
$P(W_P) = \frac{1}{6}, P(R_P) = \frac{3}{6}, P(B_P) = \frac{2}{6}$
$P(W_Q) = \frac{2}{9}, P(R_Q) = \frac{3}{9}, P(B_Q) = \frac{4}{9}$
The probability that the balls are of the same colour is:
$P(\text{same}) = P(W_P)P(W_Q) + P(R_P)P(R_Q) + P(B_P)P(B_Q)$
$P(\text{same}) = (\frac{1}{6} \times \frac{2}{9}) + (\frac{3}{6} \times \frac{3}{9}) + (\frac{2}{6} \times \frac{4}{9}) = \frac{2 + 9 + 8}{54} = \frac{19}{54}$
The probability that the balls are of different colours is:
$P(\text{different}) = 1 - P(\text{same}) = 1 - \frac{19}{54} = \frac{35}{54}$

(C) The total number of ways to select two numbers $x$ and $y$ from the set $\{1, 2, 3, \ldots, 10\}$ with replacement is $10 \times 10 = 100$.
We want to find the number of pairs $(x, y)$ such that $|x^2 - y^2|$ is divisible by $6$.
This condition is equivalent to $|(x - y)(x + y)|$ being divisible by $6$.
We analyze the pairs $(x, y)$ by iterating through $x$ from $1$ to $10$ and finding $y$ such that $x^2 \equiv y^2 \pmod{6}$.
The squares modulo $6$ are: $1^2 \equiv 1, 2^2 \equiv 4, 3^2 \equiv 3, 4^2 \equiv 4, 5^2 \equiv 1, 6^2 \equiv 0, 7^2 \equiv 1, 8^2 \equiv 4, 9^2 \equiv 3, 10^2 \equiv 4$.
Grouping by values: $0: \{6\}$,$1: \{1, 5, 7\}$,$3: \{3, 9\}$,$4: \{2, 4, 8, 10\}$.
Number of pairs $(x, y)$ such that $x^2 \equiv y^2 \pmod{6}$ is:
For $x^2 \equiv 0$: $1^2 = 1$ pair $(6, 6)$.
For $x^2 \equiv 1$: $3^2 = 9$ pairs (from $\{1, 5, 7\} \times \{1, 5, 7\}$).
For $x^2 \equiv 3$: $2^2 = 4$ pairs (from $\{3, 9\} \times \{3, 9\}$).
For $x^2 \equiv 4$: $4^2 = 16$ pairs (from $\{2, 4, 8, 10\} \times \{2, 4, 8, 10\}$).
Total favorable outcomes $= 1 + 9 + 4 + 16 = 30$.
Probability $= \frac{30}{100} = \frac{3}{10}$.

(C) Given that $A$ is the event that $X$ passes the exam and $B$ is the event that $Y$ passes the exam.
$P(A) = \frac{1}{7}$ and $P(B) = \frac{2}{9}$.
Since the events $A$ and $B$ are independent,the probability that both pass the exam is given by $P(A \cap B) = P(A) \times P(B)$.
Substituting the values,we get $P(A \cap B) = \frac{1}{7} \times \frac{2}{9} = \frac{2}{63}$.

(D) When three unbiased coins are tossed,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$
The total number of outcomes is $n(S) = 8$.
Let $E$ be the event of getting at most two heads.
It is easier to calculate the complement event $E'$,which is the event of getting three heads.
$E' = \{HHH\}$
The number of outcomes in $E'$ is $n(E') = 1$.
The probability of getting three heads is $P(E') = \frac{n(E')}{n(S)} = \frac{1}{8}$.
The probability of getting at most two heads is $P(E) = 1 - P(E') = 1 - \frac{1}{8} = \frac{7}{8}$.

(B) When two dice are thrown simultaneously,the total number of outcomes is $n(S) = 6 \times 6 = 36$.
Let $A$ be the event of getting a multiple of $2$ on the first die,i.e.,$A \in \{2, 4, 6\}$.
Let $B$ be the event of getting a multiple of $3$ on the second die,i.e.,$B \in \{3, 6\}$.
We want to find the probability of getting a multiple of $2$ on one die and a multiple of $3$ on the other.
Let $E$ be the set of favorable outcomes:
$E = \{(2,3), (4,3), (6,3), (2,6), (4,6), (6,6), (3,2), (3,4), (3,6), (6,2), (6,4)\}$.
Note that $(6,6)$ is included because it satisfies the condition of having a multiple of $2$ on one die and a multiple of $3$ on the other.
Counting the elements in $E$,we get $n(E) = 11$.
Therefore,the probability $P(E) = \frac{n(E)}{n(S)} = \frac{11}{36}$.

(B) Total number of cards $n(S) = 27$.
Let $E$ be the event that the number is even. The even numbers between $1$ and $27$ are $\{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26\}$.
Thus,$n(E) = 13$.
Let $F$ be the event that the number is divisible by $5$. The numbers divisible by $5$ are $\{5, 10, 15, 20, 25\}$.
Thus,$n(F) = 5$.
The intersection $E \cap F$ contains numbers that are both even and divisible by $5$,which are multiples of $10$. These are $\{10, 20\}$.
Thus,$n(E \cap F) = 2$.
Using the addition rule for probability,$n(E \cup F) = n(E) + n(F) - n(E \cap F) = 13 + 5 - 2 = 16$.
The probability is $P(E \cup F) = \frac{n(E \cup F)}{n(S)} = \frac{16}{27}$.

(B) Let $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$,and $P(C) = \frac{1}{4}$ be the probabilities of students $A, B$,and $C$ solving the problem respectively.
Then,the probabilities of them not solving the problem are $P(\bar{A}) = 1 - \frac{1}{2} = \frac{1}{2}$,$P(\bar{B}) = 1 - \frac{1}{3} = \frac{2}{3}$,and $P(\bar{C}) = 1 - \frac{1}{4} = \frac{3}{4}$.
The probability that exactly one of them solves the problem is given by:
$P(\text{Exactly one}) = P(A)P(\bar{B})P(\bar{C}) + P(\bar{A})P(B)P(\bar{C}) + P(\bar{A})P(\bar{B})P(C)$
$= (\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}) + (\frac{1}{2} \times \frac{1}{3} \times \frac{3}{4}) + (\frac{1}{2} \times \frac{2}{3} \times \frac{1}{4})$
$= \frac{6}{24} + \frac{3}{24} + \frac{2}{24} = \frac{11}{24}$.

(A) Let $W_i$ and $B_i$ denote the events of drawing a white ball and a black ball from box $B_i$ respectively. The probabilities are:
For $B_1$: $P(W_1) = \frac{1}{3}, P(B_1) = \frac{2}{3}$
For $B_2$: $P(W_2) = \frac{3}{4}, P(B_2) = \frac{1}{4}$
For $B_3$: $P(W_3) = \frac{2}{5}, P(B_3) = \frac{3}{5}$
We want the probability of drawing two black balls and one white ball. This can happen in three mutually exclusive ways:
$1$. $(W_1, B_2, B_3)$: $P(W_1) \times P(B_2) \times P(B_3) = \frac{1}{3} \times \frac{1}{4} \times \frac{3}{5} = \frac{3}{60}$
$2$. $(B_1, W_2, B_3)$: $P(B_1) \times P(W_2) \times P(B_3) = \frac{2}{3} \times \frac{3}{4} \times \frac{3}{5} = \frac{18}{60}$
$3$. $(B_1, B_2, W_3)$: $P(B_1) \times P(B_2) \times P(W_3) = \frac{2}{3} \times \frac{1}{4} \times \frac{2}{5} = \frac{4}{60}$
The total probability is the sum of these probabilities:
$P = \frac{3}{60} + \frac{18}{60} + \frac{4}{60} = \frac{25}{60} = \frac{5}{12}$

(C) Let $E$ be the event of getting a head from a coin.
Let $F$ be the event of getting an odd number $(1, 3, 5)$ from a die.
The probability of getting a head is $P(E) = \frac{1}{2}$.
The probability of getting an odd number on a six-faced die is $P(F) = \frac{3}{6} = \frac{1}{2}$.
Since the coin and the die are independent,the events $E$ and $F$ are independent.
Therefore,the probability of both events occurring is $P(E \cap F) = P(E) \times P(F)$.
$P(E \cap F) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

(D) Let $P(C) = p$. Then $P(B) = 2p$ and $P(A) = 2(2p) = 4p$.
Since one of them must win,the sum of their probabilities is $1$:
$P(A) + P(B) + P(C) = 1$
$4p + 2p + p = 1$
$7p = 1$
$p = \frac{1}{7}$
Therefore,$P(A) = 4p = 4 \times \frac{1}{7} = \frac{4}{7}$.
The probability that $A$ loses is $P(\bar{A}) = 1 - P(A) = 1 - \frac{4}{7} = \frac{3}{7}$.

(A) Let $P(A)$ be the probability that student $A$ solves the problem,so $P(A) = \frac{2}{5}$.
Let $P(B)$ be the probability that student $B$ solves the problem,so $P(B) = \frac{3}{4}$.
The probability that student $A$ does not solve the problem is $P(A') = 1 - P(A) = 1 - \frac{2}{5} = \frac{3}{5}$.
The probability that student $B$ does not solve the problem is $P(B') = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4}$.
The problem is solved if at least one of them solves it. This is the complement of the event that neither of them solves the problem.
Probability that neither solves the problem = $P(A') \times P(B') = \frac{3}{5} \times \frac{1}{4} = \frac{3}{20}$.
Probability that the problem is solved = $1 - P(\text{neither solves}) = 1 - \frac{3}{20} = \frac{17}{20}$.

(A) Let $P(A)$ be the probability that person $A$ completes the work,$P(A) = \frac{2}{3}$.
Let $P(B)$ be the probability that person $B$ completes the work,$P(B) = \frac{3}{4}$.
The work is completed if at least one of them completes the work.
This is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since the events are independent,$P(A \cap B) = P(A) \times P(B) = \frac{2}{3} \times \frac{3}{4} = \frac{6}{12} = \frac{1}{2}$.
Therefore,$P(A \cup B) = \frac{2}{3} + \frac{3}{4} - \frac{1}{2}$.
Finding a common denominator of $12$:
$P(A \cup B) = \frac{8}{12} + \frac{9}{12} - \frac{6}{12} = \frac{11}{12}$.
Thus,the probability that the work is completed is $\frac{11}{12}$.

(C) Since $A$ and $B$ are independent events,we have $P(A \cap B) = P(A) \cdot P(B)$.
Given $P(A \cap B) = \frac{1}{6}$,let $P(A) = x$ and $P(B) = y$. Thus,$xy = \frac{1}{6}$.
The probability that neither occurs is $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) = \frac{1}{3}$.
Therefore,$P(A \cup B) = 1 - \frac{1}{3} = \frac{2}{3}$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we get $x + y - \frac{1}{6} = \frac{2}{3}$,which implies $x + y = \frac{5}{6}$.
Substituting $y = \frac{5}{6} - x$ into $xy = \frac{1}{6}$,we get $x(\frac{5}{6} - x) = \frac{1}{6}$.
This simplifies to $6x^2 - 5x + 1 = 0$.
Factoring the quadratic equation,we get $(2x - 1)(3x - 1) = 0$.
Thus,$x = \frac{1}{2}$ or $x = \frac{1}{3}$.
Hence,the probability of occurrence of $A$ is $\frac{1}{2}$ or $\frac{1}{3}$.

(B) The probability that none of the $n$ independent events $X_1, X_2, \ldots, X_n$ occur is given by $P(X_1' \cap X_2' \cap \ldots \cap X_n')$.
Since the events are independent,this is equal to $P(X_1') P(X_2') \ldots P(X_n')$.
Given $P(X_r) = \frac{1}{r+1}$,the probability of the complement event is $P(X_r') = 1 - P(X_r) = 1 - \frac{1}{r+1} = \frac{r}{r+1}$.
Thus,the required probability is:
$P(X_1') P(X_2') \ldots P(X_n') = \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{3}\right) \ldots \left(1 - \frac{1}{n+1}\right)$
$= \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \ldots \times \frac{n}{n+1}$
$= \frac{1}{n+1}$.

(A) Let $P(X), P(Y), P(Z)$ be the probabilities of students $X, Y, Z$ passing the exam respectively.
Given:
$P(X) = \frac{1}{5} \implies P(X') = 1 - \frac{1}{5} = \frac{4}{5}$
$P(Y) = \frac{1}{4} \implies P(Y') = 1 - \frac{1}{4} = \frac{3}{4}$
$P(Z') = \frac{2}{3} \implies P(Z) = 1 - \frac{2}{3} = \frac{1}{3}$
We need the probability that at least two of them pass,which is the sum of probabilities of exactly two passing and all three passing.
$P(\text{at least 2 pass}) = P(X)P(Y)P(Z') + P(X)P(Y')P(Z) + P(X')P(Y)P(Z) + P(X)P(Y)P(Z)$
$= (\frac{1}{5} \times \frac{1}{4} \times \frac{2}{3}) + (\frac{1}{5} \times \frac{3}{4} \times \frac{1}{3}) + (\frac{4}{5} \times \frac{1}{4} \times \frac{1}{3}) + (\frac{1}{5} \times \frac{1}{4} \times \frac{1}{3})$
$= \frac{2}{60} + \frac{3}{60} + \frac{4}{60} + \frac{1}{60} = \frac{10}{60} = \frac{1}{6}$

(B) The total number of outcomes when two dice are thrown is $6 \times 6 = 36$.
The product of two numbers is odd only if both numbers are odd.
The odd numbers on a die are $\{1, 3, 5\}$.
The number of outcomes where both dice show an odd number is $3 \times 3 = 9$.
The product is even if it is not odd.
Therefore,the number of favorable outcomes where the product is even is $36 - 9 = 27$.
The probability of getting an even product is $\frac{27}{36} = \frac{3}{4}$.

(B) $A, B, C$ are pairwise independent events.
$\Rightarrow P(A \cap B) = P(A) \cdot P(B) = P^2$
$P(B \cap C) = P(B) \cdot P(C) = P^2$
$P(C \cap A) = P(C) \cdot P(A) = P^2$
Given that all of them cannot occur simultaneously,$P(A \cap B \cap C) = 0$.
Using the inclusion-exclusion principle:
$P(A \cup B \cup C) = P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)] + P(A \cap B \cap C)$
$P(A \cup B \cup C) = P + P + P - [P^2 + P^2 + P^2] + 0$
$P(A \cup B \cup C) = 3P - 3P^2 = 3P(1-P)$.

(A) Given that $E_k = \{(a, b) \in S : ab = k\}$ for $k \geq 1$ and $p_k = P(E_k)$.
Since there are $6 \times 6 = 36$ total outcomes in the sample space $S$,the probability $p_k$ is given by $\frac{|E_k|}{36}$.
For $k=1$: $E_1 = \{(1, 1)\}$,so $|E_1| = 1$ and $p_1 = \frac{1}{36}$.
For $k=2$: $E_2 = \{(1, 2), (2, 1)\}$,so $|E_2| = 2$ and $p_2 = \frac{2}{36}$.
For $k=4$: $E_4 = \{(1, 4), (4, 1), (2, 2)\}$,so $|E_4| = 3$ and $p_4 = \frac{3}{36}$.
For $k=6$: $E_6 = \{(1, 6), (6, 1), (2, 3), (3, 2)\}$,so $|E_6| = 4$ and $p_6 = \frac{4}{36}$.
For $k=30$: $E_{30} = \{(5, 6), (6, 5)\}$,so $|E_{30}| = 2$ and $p_{30} = \frac{2}{36}$.
Comparing the values: $p_1 = \frac{1}{36}$,$p_{30} = \frac{2}{36}$,$p_4 = \frac{3}{36}$,$p_6 = \frac{4}{36}$.
Thus,$p_1 < p_{30} < p_4 < p_6$.
Therefore,the correct option is $A$.

(A) The total number of elements in the set $S = \{1, 2, 3, \ldots, 100\}$ is $n(S) = 100$.
$A$ number is a perfect cube if it can be expressed as $k^3$ for some integer $k$.
The perfect cubes in the given set are $1^3 = 1$,$2^3 = 8$,$3^3 = 27$,and $4^3 = 64$. Note that $5^3 = 125$,which is greater than $100$.
Thus,the set of favorable outcomes is $A = \{1, 8, 27, 64\}$.
The number of favorable outcomes is $n(A) = 4$.
The required probability is $P(A) = \frac{n(A)}{n(S)} = \frac{4}{100} = \frac{1}{25}$.

(A) When two 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 prime numbers between $2$ and $12$ are $2, 3, 5, 7, 11$.
We list the outcomes for each prime sum:
- 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 $= 1 + 2 + 4 + 6 + 2 = 15$.
Probability $= \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{15}{36} = \frac{5}{12}$.

(A) Let the probabilities of horses $A$,$B$,and $C$ winning be $a$,$b$,and $c$ respectively.
Since one of them must win,the sum of their probabilities is $a + b + c = 1$ ... $(i)$
Given that $a = 2b$ and $b = 2c$.
Substituting $b = 2c$ into the expression for $a$,we get $a = 2(2c) = 4c$.
Now,substitute $a = 4c$ and $b = 2c$ into equation $(i)$:
$4c + 2c + c = 1$
$7c = 1$
$c = \frac{1}{7}$
Now,find $b$ and $a$:
$b = 2c = 2 \times \frac{1}{7} = \frac{2}{7}$
$a = 2b = 2 \times \frac{2}{7} = \frac{4}{7}$
Therefore,the probabilities of horses $A$,$B$,and $C$ winning are $\frac{4}{7}, \frac{2}{7}, \frac{1}{7}$ respectively.

(C) Let $A$ be the event that person $A$ solves the problem and $B$ be the event that person $B$ solves the problem.
Given,$P(A) = 0.90$ and $P(B) = 0.70$.
The probability that person $A$ fails to solve the problem is $P(\bar{A}) = 1 - 0.90 = 0.10$.
The probability that person $B$ fails to solve the problem is $P(\bar{B}) = 1 - 0.70 = 0.30$.
The probability that at least one of them solves the problem is given by $1 - P(\text{none solves the problem})$.
Since the events are independent,$P(\text{none solves}) = P(\bar{A}) \times P(\bar{B}) = 0.10 \times 0.30 = 0.03$.
Therefore,the probability that at least one of them solves the problem is $1 - 0.03 = 0.97$.

(B) Let $E_1$ be the event that $A$ speaks the truth.
$P(E_1) = \frac{20}{100} = \frac{1}{5}$.
Then,$P(\overline{E_1}) = 1 - \frac{1}{5} = \frac{4}{5}$.
Let $E_2$ be the event that $B$ speaks the truth.
$P(E_2) = \frac{80}{100} = \frac{4}{5}$.
Then,$P(\overline{E_2}) = 1 - \frac{4}{5} = \frac{1}{5}$.
Their statements do not match if ($A$ speaks the truth and $B$ lies) $OR$ ($A$ lies and $B$ speaks the truth).
Required Probability $= P(E_1) \cdot P(\overline{E_2}) + P(\overline{E_1}) \cdot P(E_2)$.
$= (\frac{1}{5} \times \frac{1}{5}) + (\frac{4}{5} \times \frac{4}{5})$.
$= \frac{1}{25} + \frac{16}{25} = \frac{17}{25}$.

(B) Let $X$ be the number drawn from box $A$ and $Y$ be the number drawn from box $B$. The total number of possible outcomes is $10 \times 10 = 100$.
We want to find the probability $P(X < Y)$.
The total outcomes can be divided into three cases: $X < Y$,$X > Y$,and $X = Y$.
The number of cases where $X = Y$ is $10$ (i.e.,$(1,1), (2,2), \ldots, (10,10)$).
Since the situation is symmetric,the number of cases where $X < Y$ is equal to the number of cases where $X > Y$.
Let $N$ be the number of cases where $X < Y$. Then $N + N + 10 = 100$.
$2N = 90 \implies N = 45$.
The required probability is $\frac{N}{100} = \frac{45}{100} = \frac{9}{20}$.
Hence,option $B$ is correct.

(D) Total number of slips for $III$ year students $= 3 \times 100 = 300$.
Total number of slips for $II$ year students $= 2 \times 150 = 300$.
Total number of slips for $I$ year students $= 1 \times 200 = 200$.
Total number of slips in the lottery $= 300 + 300 + 200 = 800$.
The probability that a $III$ year student gets the room is the ratio of the number of $III$ year student slips to the total number of slips.
Required probability $= \frac{300}{800} = \frac{3}{8}$.

(D) The set is given by $S = \{2k \mid -9 \leq k \leq 10\}$.
Since $k$ ranges from $-9$ to $10$,the number of values for $k$ is $10 - (-9) + 1 = 20$.
Thus,the total number of elements in set $S$ is $20$.
The elements are $\{-18, -16, -14, -12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20\}$.
An integer is divisible by both $4$ and $6$ if it is divisible by $\text{lcm}(4, 6) = 12$.
We look for elements in $S$ that are multiples of $12$.
These are $\{-12, 0, 12\}$.
There are $3$ such favourable outcomes.
Therefore,the required probability is $\frac{3}{20}$.

(A) Given: $P(A \cup B) = \frac{3}{4}$,$P(A \cap B) = \frac{1}{4}$,and $P(\bar{A}) = \frac{2}{3}$.
First,find $P(A)$:
$P(A) = 1 - P(\bar{A}) = 1 - \frac{2}{3} = \frac{1}{3}$.
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values: $\frac{3}{4} = \frac{1}{3} + P(B) - \frac{1}{4}$.
$P(B) = \frac{3}{4} - \frac{1}{3} + \frac{1}{4} = 1 - \frac{1}{3} = \frac{2}{3}$.
We need to find $P(\bar{A} \cap B)$.
Using the property $P(\bar{A} \cap B) = P(B) - P(A \cap B)$:
$P(\bar{A} \cap B) = \frac{2}{3} - \frac{1}{4} = \frac{8 - 3}{12} = \frac{5}{12}$.

(D) The given inequation is $x + \frac{10}{x} \leq 11$.
Since $x \in \{1, 2, 3, \ldots, 50\}$,$x$ is always positive.
Multiplying by $x$,we get $x^2 + 10 \leq 11x$,which simplifies to $x^2 - 11x + 10 \leq 0$.
Factoring the quadratic,we get $(x - 1)(x - 10) \leq 0$.
This inequality holds for $1 \leq x \leq 10$.
The integers satisfying this condition are $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
There are $10$ such integers.
The total number of possible outcomes is $50$.
The probability is $\frac{10}{50} = \frac{1}{5}$.

(A) Let the $25$ consecutive integers be $n, n+1, n+2, \dots, n+24$. Since $n$ is odd,the sequence contains $13$ odd numbers and $12$ even numbers.
Total ways to choose $3$ integers from $25$ is $\binom{25}{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300$.
The sum of three integers is even if:
$1)$ All three are even: $\binom{12}{3} = \frac{12 \times 11 \times 10}{6} = 220$.
$2)$ One is even and two are odd: $\binom{12}{1} \times \binom{13}{2} = 12 \times \frac{13 \times 12}{2} = 12 \times 78 = 936$.
Total favorable outcomes = $220 + 936 = 1156$.
Probability = $\frac{1156}{2300} = \frac{289}{575}$.

(A) When two dice are thrown,the total number of possible outcomes is $6^2 = 36$.
Two numbers are coprime if their greatest common divisor $(GCD)$ is $1$.
It is easier to find the outcomes where the numbers are $NOT$ coprime (i.e.,$\text{GCD} > 1$).
The pairs $(x, y)$ where $\text{GCD}(x, y) > 1$ are:
$\{(2,2), (2,4), (2,6), (3,3), (3,6), (4,2), (4,4), (4,6), (5,5), (6,2), (6,3), (6,4), (6,6)\}$.
The number of such outcomes is $13$.
The number of favourable outcomes (where numbers are coprime) is $36 - 13 = 23$.
Therefore,the required probability is $\frac{23}{36}$.

(D) Let $p$ be the probability of drawing a spade,$p = \frac{13}{52} = \frac{1}{4}$.
Let $q$ be the probability of not drawing a spade,$q = 1 - p = \frac{3}{4}$.
$A$ wins if $A$ draws a spade on the $1^{st}, 5^{th}, 9^{th}, \dots$ turn.
The probability that $A$ wins is:
$P(A) = p + q^4 p + q^8 p + \dots$
This is an infinite geometric series with first term $a = p$ and common ratio $r = q^4$.
The sum is $S = \frac{a}{1-r} = \frac{p}{1-q^4}$.
Substituting the values:
$P(A) = \frac{1/4}{1 - (3/4)^4} = \frac{1/4}{1 - 81/256} = \frac{1/4}{175/256} = \frac{1}{4} \times \frac{256}{175} = \frac{64}{175}$.

(C) When a pair of dice is thrown,the total number of outcomes is $n(S) = 6 \times 6 = 36$.
Prime numbers on a die are $\{2, 3, 5\}$. The outcomes where both dice show prime numbers are $\{(2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5)\}$. There are $9$ such outcomes.
Probability of getting prime numbers on both dice in the first throw is $P(A) = \frac{9}{36} = \frac{1}{4}$.
Composite numbers on a die are $\{4, 6\}$ (Note: $1$ is neither prime nor composite). The outcomes where both dice show composite numbers are $\{(4,4), (4,6), (6,4), (6,6)\}$. There are $4$ such outcomes.
Probability of getting composite numbers on both dice in the second throw is $P(B) = \frac{4}{36} = \frac{1}{9}$.
Since the two throws are independent,the required probability is $P(A) \times P(B) = \frac{1}{4} \times \frac{1}{9} = \frac{1}{36}$.