Set Based probability Questions in English

(B) The sample space $S$ consists of tickets where the product of digits is $0$. $A$ product of digits is $0$ if at least one digit is $0$. The numbers are $00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40$.
Total number of such tickets $n(S) = 14$.
Let $A$ be the event that the sum of the digits is $8$.
Checking the numbers: $00 (0), 01 (1), 02 (2), 03 (3), 04 (4), 05 (5), 06 (6), 07 (7), 08 (8), 09 (9), 10 (1), 20 (2), 30 (3), 40 (4)$.
The only number with a sum of digits equal to $8$ is $08$.
So,$n(A) = 1$.
The probability $P(A) = \frac{n(A)}{n(S)} = \frac{1}{14}$.

(C) The sample space $S = \{1, 2, 3, 4, 5, 6\}$,so $n(S) = 6$.
Event $A$ is getting a number greater than $3$,so $A = \{4, 5, 6\}$ and $n(A) = 3$.
Event $B$ is getting a number less than $5$,so $B = \{1, 2, 3, 4\}$ and $n(B) = 4$.
The intersection $A \cap B$ is the set of numbers that are both greater than $3$ and less than $5$,so $A \cap B = \{4\}$ and $n(A \cap B) = 1$.
The union $A \cup B$ is the set of numbers that are either in $A$ or in $B$,so $A \cup B = \{1, 2, 3, 4, 5, 6\}$ and $n(A \cup B) = 6$.
Therefore,$P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{6}{6} = 1$.

(C) Total number of roses = $4 + 3 + 5 + 8 = 20$.
Number of red roses = $4$.
Number of white roses = $8$.
Number of favorable outcomes (red or white) = $4 + 8 = 12$.
Probability = $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{12}{20} = \frac{3}{5}$.

(C) The total number of outcomes when three fair dice are tossed is $6 \times 6 \times 6 = 216$. So,$n = 216$.
Let $A$ be the event that the integers on all three dice are the same.
The favorable outcomes are $A = \{(1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6)\}$.
Thus,the number of favorable outcomes is $r = 6$.
The probability $P(A)$ is given by $P(A) = \frac{r}{n} = \frac{6}{216} = \frac{1}{36}$.

(A) When a die is thrown twice,the total number of possible outcomes is $6 \times 6 = 36$.
Let $E$ be the event of getting $4$ at least once.
It is easier to find the probability of the complement event $E'$,which is the event of not getting $4$ in either throw.
For a single throw,the outcomes that are not $4$ are $\{1, 2, 3, 5, 6\}$,which are $5$ outcomes.
For two throws,the number of outcomes where $4$ does not appear is $5 \times 5 = 25$.
Therefore,the probability of not getting $4$ at all is $P(E') = \frac{25}{36}$.
The probability of getting $4$ at least once is $P(E) = 1 - P(E') = 1 - \frac{25}{36} = \frac{11}{36}$.

(C) Let $R_1$ be the event that the first ball is red and $W_1$ be the event that the first ball is white.
Let $R_2$ be the event that the second ball is red and $W_2$ be the event that the second ball is white.
We want to find the probability that the balls are of different colors,which means the outcomes are $(R_1, W_2)$ or $(W_1, R_2)$.
The total number of balls is $6$.
$P(R_1, W_2) = P(R_1) \times P(W_2 | R_1) = \frac{3}{6} \times \frac{3}{5} = \frac{9}{30} = \frac{3}{10}$.
$P(W_1, R_2) = P(W_1) \times P(R_2 | W_1) = \frac{3}{6} \times \frac{3}{5} = \frac{9}{30} = \frac{3}{10}$.
The total probability is $P(R_1, W_2) + P(W_1, R_2) = \frac{3}{10} + \frac{3}{10} = \frac{6}{10} = \frac{3}{5}$.

(A) leap year consists of $366$ days.
$366$ days = $52$ weeks and $2$ extra days.
These $2$ extra days can be: (Sunday,Monday),(Monday,Tuesday),(Tuesday,Wednesday),(Wednesday,Thursday),(Thursday,Friday),(Friday,Saturday),or (Saturday,Sunday).
There are $7$ possible outcomes for these $2$ days.
For the year to have $53$ Sundays,one of the $2$ extra days must be a Sunday.
This occurs in $2$ cases: (Saturday,Sunday) and (Sunday,Monday).
Therefore,the probability is $2/7$.
Since Statement-$II$ provides the basis for calculating the extra days in a leap year,it is the correct explanation for Statement-$I$.

(D) The sample space for rolling a fair die is $S = \{1, 2, 3, 4, 5, 6\}$.
Event $A$ is getting an integer greater than $3$,so $A = \{4, 5, 6\}$.
Event $B$ is getting an integer less than $5$,so $B = \{1, 2, 3, 4\}$.
The union of events $A$ and $B$ is $A \cup B = \{1, 2, 3, 4, 5, 6\}$.
Since $A \cup B$ is the entire sample space $S$,the probability $P(A \cup B) = P(S) = 1$.

(A) When two dice are thrown,the sample space is $S = \{(1, 1), (1, 2), ..., (6, 6)\}$.
The total number of outcomes is $n(S) = 6 \times 6 = 36$.
$(1)$ Let $E_1$ be the event that the numbers on both dice are the same.
$E_1 = \{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\}$.
$n(E_1) = 6$.
$P(E_1) = \frac{n(E_1)}{n(S)} = \frac{6}{36} = \frac{1}{6}$.
$(2)$ Let $E_2$ be the event that the difference between the numbers is $1$.
$E_2 = \{(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4), (5, 6), (6, 5)\}$.
$n(E_2) = 10$.
$P(E_2) = \frac{n(E_2)}{n(S)} = \frac{10}{36} = \frac{5}{18}$.

(A) Let $P(A), P(B),$ and $P(C)$ be the probabilities of students $A, B,$ and $C$ solving the problem respectively.
$P(A) = 1/2, P(B) = 1/3, P(C) = 1/4$.
The probability that the problem is not solved by any of them is the probability that all three fail to solve it.
$P(\text{not } A) = 1 - 1/2 = 1/2$
$P(\text{not } B) = 1 - 1/3 = 2/3$
$P(\text{not } C) = 1 - 1/4 = 3/4$
Since the events are independent,the probability that none of them solve the problem is:
$P(\text{none solve}) = P(\text{not } A) \times P(\text{not } B) \times P(\text{not } C) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{1}{4}$.
The probability that the problem is solved is $1 - P(\text{none solve})$.
$P(\text{solved}) = 1 - \frac{1}{4} = \frac{3}{4}$.

(A) Given: $P(A) = 0.25$,$P(B) = 0.50$,and $P(A \cap B) = 0.14$.
We need to find the probability that neither $A$ nor $B$ occurs,which is $P(A^c \cap B^c)$.
By De Morgan's Law,$P(A^c \cap B^c) = P((A \cup B)^c) = 1 - P(A \cup B)$.
Using the Addition Theorem,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = 0.25 + 0.50 - 0.14 = 0.61$.
Therefore,$P(A^c \cap B^c) = 1 - 0.61 = 0.39$.

(B) Let $A$ and $B$ be the events that Krishna and Hari are alive in $10$ years,respectively.
Given $P(A) = \frac{7}{15}$ and $P(B) = \frac{7}{10}$.
The probability that Krishna dies is $P(A^c) = 1 - P(A) = 1 - \frac{7}{15} = \frac{8}{15}$.
The probability that Hari dies is $P(B^c) = 1 - P(B) = 1 - \frac{7}{10} = \frac{3}{10}$.
Since the events are independent,the probability that both die is $P(A^c \cap B^c) = P(A^c) \times P(B^c)$.
$P(A^c \cap B^c) = \frac{8}{15} \times \frac{3}{10} = \frac{24}{150}$.

(A) The total number of subsets of a set $X$ with $n$ elements is $2^n$.
Since $A$ and $B$ are chosen from these subsets,the total number of ways to choose the pair $(A, B)$ is $2^n \times 2^n = 2^{2n}$.
The number of subsets of $X$ having $r$ elements is given by ${}^nC_r$.
For $A$ and $B$ to have the same number of elements,say $r$,we must choose $A$ with $r$ elements and $B$ with $r$ elements.
The number of ways to choose $A$ and $B$ such that $|A| = |B| = r$ is ${}^nC_r \times {}^nC_r = ({}^nC_r)^2$.
Summing over all possible values of $r$ from $0$ to $n$,the total number of favorable outcomes is $\sum_{r=0}^{n} ({}^nC_r)^2$.
Using the identity $\sum_{r=0}^{n} ({}^nC_r)^2 = {}^{2n}C_n$,the total favorable outcomes is ${}^{2n}C_n$.
Thus,the probability is $\frac{{}^{2n}C_n}{2^{2n}}$.

(D) Let $b$ represent a boy and $g$ represent a girl.
Since there are $3$ families,the sample space $S$ is the Cartesian product of the choices from each family:
$S = \{b, g\} \times \{b, g\} \times \{b, g\}$
$S = \{bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg\}$
The event of selecting only girls means choosing $g$ from all $3$ families.
Therefore,the event is $\{ggg\}$.

(D) Total number of horses = $5$.
Number of horses selected by Mr. $A$ = $2$.
Number of horses not selected by Mr. $A$ = $5 - 2 = 3$.
There is only $1$ winning horse.
The probability that the winning horse is $NOT$ among the $2$ horses selected by Mr. $A$ is the probability that the winning horse is among the $3$ horses not selected.
$P(\text{Winning horse not selected}) = \frac{3}{5}$.
The probability that the winning horse $IS$ among the $2$ horses selected by Mr. $A$ is:
$P(\text{Winning horse selected}) = 1 - P(\text{Winning horse not selected}) = 1 - \frac{3}{5} = \frac{2}{5}$.

(C) When two fair dice are thrown simultaneously,the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.
Let $A$ be the event of getting the same number on both dice.
The favorable outcomes are $A = \{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\}$.
The number of favorable outcomes is $n(A) = 6$.
The probability of event $A$ is $P(A) = \frac{n(A)}{n(S)} = \frac{6}{36} = \frac{1}{6}$.

(B) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
$A$ doublet occurs when both dice show the same number. The favorable outcomes are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$.
The number of favorable outcomes is $6$.
The probability of getting a doublet is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}$.

(C) The word $PROBABILITY$ contains $11$ letters in total.
The vowels in the word are $O, A, I, I$.
There are $4$ vowels in total.
The probability of choosing a vowel is given by the ratio of the number of favorable outcomes to the total number of outcomes.
Probability $= \frac{4}{11}$.

(B) The probability of hitting the bird in one attempt is $P(H) = 1/3$.
The probability of missing the bird in one attempt is $P(M) = 1 - 1/3 = 2/3$.
The probability of hitting the bird at least once in three attempts is $1 - P(\text{missing all three times})$.
The probability of missing all three times is $P(M)^3 = (2/3)^3 = 8/27$.
Therefore, the probability of hitting the bird is $1 - 8/27 = 19/27$.

(C) An event $A$ is independent of itself if $P(A \cap A) = P(A)P(A)$.
Since $A \cap A = A$,we have $P(A) = P(A)^2$.
This implies $P(A)^2 - P(A) = 0$.
$P(A)(P(A) - 1) = 0$.
Therefore,$P(A) = 0$ or $P(A) = 1$.

(D) The total number of elements in the Cartesian product $A \times B$ is $n(A \times B) = n(A) \times n(B) = 5 \times 4 = 20$.
The ordered pairs $(a, b)$ such that $a + b = 9$ are $\{(1, 8), (3, 6), (5, 4), (7, 2)\}$.
The number of favorable outcomes is $r = 4$.
The probability of the event is $P = \frac{r}{n} = \frac{4}{20} = \frac{1}{5}$.

(C) The total number of ways to draw $2$ cards from $52$ cards is $^{52}C_2$.
Let $A$ be the event that both cards are red.
Let $B$ be the event that both cards are kings.
We need to find $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
There are $26$ red cards,so the number of ways to choose $2$ red cards is $^{26}C_2$.
$P(A) = \frac{^{26}C_2}{^{52}C_2} = \frac{325}{1326}$.
There are $4$ kings,so the number of ways to choose $2$ kings is $^4C_2$.
$P(B) = \frac{^4C_2}{^{52}C_2} = \frac{6}{1326} = \frac{1}{221}$.
$A \cap B$ is the event that both cards are red kings. There are $2$ red kings (King of Hearts and King of Diamonds),so the number of ways to choose $2$ red kings is $^2C_2 = 1$.
$P(A \cap B) = \frac{^2C_2}{^{52}C_2} = \frac{1}{1326}$.
Using the formula: $P(A \cup B) = \frac{325}{1326} + \frac{6}{1326} - \frac{1}{1326} = \frac{330}{1326}$.
Simplifying the fraction: $\frac{330}{1326} = \frac{55}{221}$.

(B) The total number of possible outcomes when two dice are thrown is $6^2 = 36$.
Let $E$ be the event that at least one die shows a number greater than $3$.
It is easier to find the probability of the complement event $E'$,where neither die shows a number greater than $3$.
This means both dice must show a number from the set $\{1, 2, 3\}$.
The number of outcomes for $E'$ is $3 \times 3 = 9$.
The outcomes are: $(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)$.
Therefore,$P(E') = \frac{9}{36} = \frac{1}{4}$.
The probability of the required event is $P(E) = 1 - P(E') = 1 - \frac{1}{4} = \frac{3}{4}$.

(C) The total number of possible outcomes when two dice are rolled is $6^2 = 36$.
The sums that are multiples of $4$ are $4, 8,$ and $12$.
The pairs $(x, y)$ that result in these sums are:
For sum $= 4$: $(1, 3), (2, 2), (3, 1)$
For sum $= 8$: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$
For sum $= 12$: $(6, 6)$
Counting these,we have $3 + 5 + 1 = 9$ favorable outcomes.
The required probability is $P = \frac{9}{36} = \frac{1}{4}$.

(C) Given that the probability of at least one event occurring is $P(A \cup B) = 0.6$.
By De Morgan's Law,$P(A \cup B) = 1 - P(A' \cap B') = 0.6$,so $P(A' \cap B') = 0.4$.
We know that $P(A \cap B) = 0.3$.
Using the formula $P(A' \cup B') = P(A') + P(B') - P(A' \cap B')$,
where $P(A' \cup B') = 1 - P(A \cap B) = 1 - 0.3 = 0.7$.
Substituting the values: $0.7 = P(A') + P(B') - 0.4$.
Therefore,$P(A') + P(B') = 0.7 + 0.4 = 1.1$.

(D) Let the outcomes of the three coins be independent events $A, B,$ and $C$.
The probability of getting a head on the first coin is $P(A) = 1/2$.
The probability of getting a tail on the second coin is $P(B) = 1/2$.
The probability of getting a head on the third coin is $P(C) = 1/2$.
Since the events are independent,the joint probability is $P(A \cap B \cap C) = P(A) \times P(B) \times P(C)$.
$P(A \cap B \cap C) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$.

(B) Let $P(A) = p$ be the probability that $A$ dies,so $P(A') = 1 - p$ is the probability that $A$ lives.
Let $P(B) = q$ be the probability that $B$ dies,so $P(B') = 1 - q$ is the probability that $B$ lives.
The event that only one of them is alive at the end of the year means either ($A$ dies and $B$ lives) or ($B$ dies and $A$ lives).
Since the events are independent,the required probability is $P(A \cap B') + P(B \cap A') = P(A) \cdot P(B') + P(B) \cdot P(A')$.
Substituting the values: $p(1 - q) + q(1 - p) = p - pq + q - pq = p + q - 2pq$.

(A) $P(\overline{A \cup B}) = \frac{1}{6} \implies P(A \cup B) = 1 - \frac{1}{6} = \frac{5}{6}$.
Given $P(A \cap B) = \frac{1}{4}$ and $P(\bar{A}) = \frac{1}{4}$,we have $P(A) = 1 - \frac{1}{4} = \frac{3}{4}$.
Using the addition theorem,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values: $\frac{5}{6} = \frac{3}{4} + P(B) - \frac{1}{4}$.
$\frac{5}{6} = \frac{1}{2} + P(B) \implies P(B) = \frac{5}{6} - \frac{1}{2} = \frac{5-3}{6} = \frac{2}{6} = \frac{1}{3}$.
Now,check for independence: $P(A) \cdot P(B) = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4} = P(A \cap B)$.
Since $P(A \cap B) = P(A) \cdot P(B)$,the events are independent.
Since $P(A) = \frac{3}{4}$ and $P(B) = \frac{1}{3}$,$P(A) \neq P(B)$,so they are not equally likely.

(B) The sample space $S$ has $6 \times 6 = 36$ outcomes.
$E_1 = \{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\}$,so $P(E_1) = \frac{6}{36} = \frac{1}{6}$.
$E_2 = \{(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2)\}$,so $P(E_2) = \frac{6}{36} = \frac{1}{6}$.
$E_3$ is the event that the sum is odd,which occurs if one die is even and the other is odd. $P(E_3) = \frac{18}{36} = \frac{1}{2}$.
$E_1 \cap E_2 = \{(4, 2)\}$,so $P(E_1 \cap E_2) = \frac{1}{36} = P(E_1)P(E_2)$. Thus,$E_1$ and $E_2$ are independent.
$E_1 \cap E_3 = \{(4, 1), (4, 3), (4, 5)\}$,so $P(E_1 \cap E_3) = \frac{3}{36} = \frac{1}{12} = P(E_1)P(E_3)$. Thus,$E_1$ and $E_3$ are independent.
$E_2 \cap E_3 = \{(1, 2), (3, 2), (5, 2)\}$,so $P(E_2 \cap E_3) = \frac{3}{36} = \frac{1}{12} = P(E_2)P(E_3)$. Thus,$E_2$ and $E_3$ are independent.
For $E_1, E_2, E_3$ to be independent,we must have $P(E_1 \cap E_2 \cap E_3) = P(E_1)P(E_2)P(E_3)$.
$E_1 \cap E_2 \cap E_3 = \{(4, 2)\} \cap E_3 = \emptyset$ because the sum $4+2=6$ is even. So $P(E_1 \cap E_2 \cap E_3) = 0$.
Since $0 \neq \frac{1}{6} \times \frac{1}{6} \times \frac{1}{2} = \frac{1}{72}$,the events are not mutually independent.

(B) Let the set be $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
Total number of ways to choose two distinct numbers is $n(S) = \binom{11}{2} = \frac{11 \times 10}{2} = 55$.
Let the two numbers be $x$ and $y$ where $x > y$.
We require $(x+y)$ to be a multiple of $4$ and $(x-y)$ to be a multiple of $4$.
If $(x+y)$ and $(x-y)$ are both multiples of $4$,then their sum $(x+y) + (x-y) = 2x$ must be a multiple of $4$,implying $x$ is an even number.
Similarly,their difference $(x+y) - (x-y) = 2y$ must be a multiple of $4$,implying $y$ is an even number.
Thus,both $x$ and $y$ must be even numbers from the set $\{0, 2, 4, 6, 8, 10\}$.
Let the chosen numbers be $x, y \in \{0, 2, 4, 6, 8, 10\}$.
For $(x+y)$ and $(x-y)$ to be multiples of $4$,$x+y \equiv 0 \pmod{4}$ and $x-y \equiv 0 \pmod{4}$.
This implies $x \equiv y \pmod{4}$.
Possible pairs $(x, y)$ with $x > y$ such that $x \equiv y \pmod{4}$:
If $x, y \equiv 0 \pmod{4}$: $(4, 0), (8, 0), (8, 4)$.
If $x, y \equiv 2 \pmod{4}$: $(6, 2), (10, 2), (10, 6)$.
Total favorable outcomes $n(E) = 3 + 3 = 6$.
Therefore,the probability $P(E) = \frac{n(E)}{n(S)} = \frac{6}{55}$.

(B) Since the events are mutually exclusive and exhaustive,the sum of their probabilities must be $1$.
$\frac{1 - 3p}{2} + \frac{1 + 4p}{3} + \frac{1 + p}{6} = 1$
Multiplying by $6$,we get $3(1 - 3p) + 2(1 + 4p) + (1 + p) = 6$.
$3 - 9p + 2 + 8p + 1 + p = 6$
$6 = 6$,which is always true for any $p$. However,each probability must lie in the interval $[0, 1]$.
$1) \ 0 \le \frac{1 - 3p}{2} \le 1 \Rightarrow 0 \le 1 - 3p \le 2 \Rightarrow -1 \le -3p \le 1 \Rightarrow -\frac{1}{3} \le p \le \frac{1}{3}$.
$2) \ 0 \le \frac{1 + 4p}{3} \le 1 \Rightarrow 0 \le 1 + 4p \le 3 \Rightarrow -1 \le 4p \le 2 \Rightarrow -\frac{1}{4} \le p \le \frac{1}{2}$.
$3) \ 0 \le \frac{1 + p}{6} \le 1 \Rightarrow 0 \le 1 + p \le 6 \Rightarrow -1 \le p \le 5$.
Taking the intersection of all these intervals: $[-\frac{1}{3}, \frac{1}{3}] \cap [-\frac{1}{4}, \frac{1}{2}] \cap [-1, 5] = [-\frac{1}{4}, \frac{1}{3}]$.

(A) Let $H$ be the event of getting a head in a coin toss,$P(H) = \frac{1}{2}$. The probability of not getting a head is $P(H') = 1 - \frac{1}{2} = \frac{1}{2}$.
Let $D$ be the event of getting a $5$ or $6$ in a dice throw,$P(D) = \frac{2}{6} = \frac{1}{3}$. The probability of not getting a $5$ or $6$ is $P(D') = 1 - \frac{1}{3} = \frac{2}{3}$.
The man starts with the coin and they alternate. He wins if he gets a head before getting a $5$ or $6$.
This can happen in the following ways:
$1$. He gets a head on the first toss: $P_1 = \frac{1}{2}$.
$2$. He fails on the first toss,fails on the first dice throw,and gets a head on the second toss: $P_2 = (\frac{1}{2} \times \frac{2}{3}) \times \frac{1}{2}$.
$3$. He fails on the first two tosses and first two dice throws,and gets a head on the third toss: $P_3 = (\frac{1}{2} \times \frac{2}{3})^2 \times \frac{1}{2}$.
The total probability is the sum of this infinite geometric series:
$P = \frac{1}{2} + (\frac{1}{2} \cdot \frac{2}{3}) \cdot \frac{1}{2} + (\frac{1}{2} \cdot \frac{2}{3})^2 \cdot \frac{1}{2} + \dots$
$P = \frac{1}{2} [1 + \frac{1}{3} + (\frac{1}{3})^2 + \dots]$
Using the sum formula for an infinite geometric series $S = \frac{a}{1-r}$ where $a=1$ and $r=\frac{1}{3}$:
$P = \frac{1}{2} \times \frac{1}{1 - 1/3} = \frac{1}{2} \times \frac{1}{2/3} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$.

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

(D) Let $A, B$ and $C$ be the events that the student passes tests $I, II$ and $III$ respectively. The student is successful if he passes in $(I \text{ and } II)$ or $(I \text{ and } III)$.
This is represented by the event $(A \cap B) \cup (A \cap C)$.
Using the inclusion-exclusion principle,$P((A \cap B) \cup (A \cap C)) = P(A \cap B) + P(A \cap C) - P(A \cap B \cap C)$.
Since the events are independent,$P(A \cap B) = P(A)P(B) = pq$,$P(A \cap C) = P(A)P(C) = p(\frac{1}{2})$,and $P(A \cap B \cap C) = P(A)P(B)P(C) = pq(\frac{1}{2})$.
Thus,the probability of success is $pq + \frac{p}{2} - \frac{pq}{2} = \frac{pq}{2} + \frac{p}{2} = \frac{p}{2}(q + 1)$.
Given the probability of success is $\frac{1}{2}$,we have $\frac{p}{2}(q + 1) = \frac{1}{2}$,which simplifies to $p(q + 1) = 1$.
If $p=1$,then $1+q=1 \Rightarrow q=0$. If $p=\frac{2}{3}$,then $\frac{2}{3}(q+1)=1 \Rightarrow q+1=\frac{3}{2} \Rightarrow q=\frac{1}{2}$.
Since there are multiple pairs $(p, q)$ satisfying $p(q+1)=1$,there are infinitely many values for $p$ and $q$. Thus,all options $A, B$ and $C$ are correct.

(A) Let $P(W_i)$ and $P(B_i)$ be the probabilities of drawing one white and one black ball from the $i$-th box where $i = 1, 2, 3$ respectively.
$P(W_1) = \frac{3}{4}, P(B_1) = \frac{1}{4}$
$P(W_2) = \frac{2}{4} = \frac{1}{2}, P(B_2) = \frac{2}{4} = \frac{1}{2}$
$P(W_3) = \frac{1}{4}, P(B_3) = \frac{3}{4}$
Two white and one black ball may be drawn from $3$ boxes in the following three ways:

$Way 1$	$W, W, B$
$Way 2$	$W, B, W$
$Way 3$	$B, W, W$

Required probability $= P(W_1)P(W_2)P(B_3) + P(W_1)P(B_2)P(W_3) + P(B_1)P(W_2)P(W_3)$
$= (\frac{3}{4} \times \frac{2}{4} \times \frac{3}{4}) + (\frac{3}{4} \times \frac{2}{4} \times \frac{1}{4}) + (\frac{1}{4} \times \frac{2}{4} \times \frac{1}{4})$
$= \frac{18}{64} + \frac{6}{64} + \frac{2}{64} = \frac{26}{64} = \frac{13}{32}$.

(D) The total number of ways to choose two distinct numbers $x$ and $y$ from the set $\{1, 2, \dots, 15\}$ is $15 \times 14 = 210$.
The equation of the line passing through $(0, 0)$ with a slope of $\frac{2}{3}$ is $y = \frac{2}{3}x$,which simplifies to $2x = 3y$.
We need to find pairs $(x, y)$ such that $x, y \in \{1, 2, \dots, 15\}$ and $2x = 3y$. This implies $x$ must be a multiple of $3$. Let $x = 3k$,then $2(3k) = 3y$,so $y = 2k$.
For $k=1, x=3, y=2$.
For $k=2, x=6, y=4$.
For $k=3, x=9, y=6$.
For $k=4, x=12, y=8$.
For $k=5, x=15, y=10$.
There are $5$ such favorable pairs.
The probability is $\frac{5}{210} = \frac{1}{42}$.

(A) Total number of balls $= 5 + 4 + 3 = 12$.
Let $R_1$ be the event that the first ball is not the particular red ball,$R_2$ be the event that the second ball is not the particular red ball,$R_3$ be the event that the third ball is not the particular red ball,and $E$ be the event that the fourth ball is the particular red ball.
The probability is given by $P(R_1 \cap R_2 \cap R_3 \cap E) = P(R_1) \cdot P(R_2|R_1) \cdot P(R_3|R_1 \cap R_2) \cdot P(E|R_1 \cap R_2 \cap R_3)$.
$P(R_1) = \frac{11}{12}$ (any ball except the particular red one).
$P(R_2|R_1) = \frac{10}{11}$ (any of the remaining $10$ balls except the particular red one).
$P(R_3|R_1 \cap R_2) = \frac{9}{10}$ (any of the remaining $9$ balls except the particular red one).
$P(E|R_1 \cap R_2 \cap R_3) = \frac{1}{9}$ (the particular red ball is now one of the $9$ remaining balls).
Required probability $= \frac{11}{12} \times \frac{10}{11} \times \frac{9}{10} \times \frac{1}{9} = \frac{1}{12}$.

(C) To solve the inequality $\frac{(x - 10)(x - 50)}{(x - 30)} \geqslant 0$,we use the wavy curve method (sign scheme).
The critical points are $x = 10, 30, 50$.
Testing intervals:
For $x > 50$,the expression is positive.
For $30 < x < 50$,the expression is negative.
For $10 < x < 30$,the expression is positive.
For $x < 10$,the expression is negative.
Considering the inequality $\geqslant 0$:
- The expression is zero at $x = 10$ and $x = 50$.
- The expression is undefined at $x = 30$.
- Thus,the solution set is $x \in [10, 30) \cup [50, 100]$.
Counting the integers in this set:
- In $[10, 29]$,there are $29 - 10 + 1 = 20$ integers.
- In $[50, 100]$,there are $100 - 50 + 1 = 51$ integers.
- Total number of favorable integers $= 20 + 51 = 71$.
The total number of integers in the set $\{1, 2, ..., 100\}$ is $100$.
Therefore,the probability is $\frac{71}{100} = 0.71$.

(A) Let $P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4}$.
Then $P(\overline{A}) = \frac{1}{2}, P(\overline{B}) = \frac{2}{3}, P(\overline{C}) = \frac{3}{4}$.
$\lambda$ is the probability that exactly two hit the target:
$\lambda = P(A)P(B)P(\overline{C}) + P(A)P(\overline{B})P(C) + P(\overline{A})P(B)P(C)$
$\lambda = (\frac{1}{2} \times \frac{1}{3} \times \frac{3}{4}) + (\frac{1}{2} \times \frac{2}{3} \times \frac{1}{4}) + (\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4})$
$\lambda = \frac{3}{24} + \frac{2}{24} + \frac{1}{24} = \frac{6}{24} = \frac{1}{4}$.
$\mu$ is the probability that at least two hit the target:
$\mu = \lambda + P(A)P(B)P(C)$
$\mu = \frac{6}{24} + (\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}) = \frac{6}{24} + \frac{1}{24} = \frac{7}{24}$.
$\lambda + \mu = \frac{6}{24} + \frac{7}{24} = \frac{13}{24}$.

(B) The given inequation is $x^2 - 13x - 30 \leq 0$.
Factoring the quadratic expression,we get $(x - 15)(x + 2) \leq 0$.
For this product to be less than or equal to zero,$x$ must lie in the interval $[-2, 15]$.
Since $x$ is a natural number $(x \in \mathbb{N})$,the possible values for $x$ are ${1, 2, 3, \dots, 15}$.
The total number of natural numbers in the set ${1, 2, \dots, 100}$ is $100$.
The number of favorable outcomes is $15$.
Therefore,the required probability is $\frac{15}{100} = \frac{3}{20}$.

(D) Let $E_1, E_2, E_3, E_4$ be the events of hitting the plane in the first,second,third,and fourth shots respectively.
Given probabilities are $P(E_1) = 0.4, P(E_2) = 0.3, P(E_3) = 0.2, P(E_4) = 0.1$.
The probability of not hitting the plane in any shot is $P(E_i^c) = 1 - P(E_i)$.
$P(E_1^c) = 1 - 0.4 = 0.6$
$P(E_2^c) = 1 - 0.3 = 0.7$
$P(E_3^c) = 1 - 0.2 = 0.8$
$P(E_4^c) = 1 - 0.1 = 0.9$
The probability that the gun hits the plane at least once is $1 - P(\text{not hitting the plane in any shot})$.
Required probability $= 1 - (P(E_1^c) \times P(E_2^c) \times P(E_3^c) \times P(E_4^c))$
$= 1 - (0.6 \times 0.7 \times 0.8 \times 0.9)$
$= 1 - 0.3024$
$= 0.6976$

(A) Given the inequality $x + \frac{336}{x} \leq 50$.
Since $x$ is a positive integer,we can multiply by $x$ without changing the inequality sign: $x^2 + 336 \leq 50x$.
Rearranging the terms,we get $x^2 - 50x + 336 \leq 0$.
Factoring the quadratic expression,we find $(x - 8)(x - 42) \leq 0$.
This inequality holds for $x$ in the interval $[8, 42]$.
Since $x$ must be an integer,the possible values for $x$ are ${8, 9, 10, \dots, 42}$.
The number of such integers is $42 - 8 + 1 = 35$.
The total number of integers from $1$ to $50$ is $50$.
Therefore,the required probability is $\frac{35}{50} = \frac{7}{10}$.

(A) The total number of ways to have $6$ children is $2^6 = 64$.
Since at least one child is a girl,we exclude the case where all children are boys (which is $1$ case).
Thus,the total number of possible outcomes is $2^6 - 1 = 64 - 1 = 63$.
The number of ways to have $3$ boys and $3$ girls is given by the binomial coefficient $\binom{6}{3} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
Therefore,the required probability is $\frac{20}{63}$.

(C) We are given the following probabilities:
$P(A \cap \bar{B} \cap \bar{C}) = 0.6$
$P(A) = 0.8$
$P(\bar{A} \cap B \cap C) = 0.1$
We know that $P(A) = P(A \cap \bar{B} \cap \bar{C}) + P(A \cap B \cap \bar{C}) + P(A \cap \bar{B} \cap C) + P(A \cap B \cap C)$.
Substituting the known values:
$0.8 = 0.6 + P(A \cap B \cap \bar{C}) + P(A \cap \bar{B} \cap C) + P(A \cap B \cap C)$
$P(A \cap B \cap \bar{C}) + P(A \cap \bar{B} \cap C) + P(A \cap B \cap C) = 0.8 - 0.6 = 0.2$.
The probability of at least two events occurring is given by:
$P(\text{at least two}) = P(A \cap B \cap \bar{C}) + P(A \cap \bar{B} \cap C) + P(\bar{A} \cap B \cap C) + P(A \cap B \cap C)$.
Substituting the sum calculated above and the given value for $P(\bar{A} \cap B \cap C)$:
$P(\text{at least two}) = 0.2 + 0.1 = 0.3$.

(A) The total number of outcomes when two dice are thrown is $6 \times 6 = 36$.
Let $A$ be the outcome of the first die and $B$ be the outcome of the second die. We want to find the number of outcomes where $A > B$.
If $A=2$,$B=1$ ($1$ outcome).
If $A=3$,$B=1, 2$ ($2$ outcomes).
If $A=4$,$B=1, 2, 3$ ($3$ outcomes).
If $A=5$,$B=1, 2, 3, 4$ ($4$ outcomes).
If $A=6$,$B=1, 2, 3, 4, 5$ ($5$ outcomes).
Total favorable outcomes $= 1 + 2 + 3 + 4 + 5 = 15$.
The probability is $\frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{15}{36} = \frac{5}{12}$.

(A) Let $P(A), P(B), P(C)$ be the probabilities that students $A, B, C$ solve the problem respectively.
$P(A) = \frac{1}{3}, P(B) = \frac{1}{4}, P(C) = \frac{1}{5}$.
The probability that the problem is not solved by any of them is:
$P(\text{not solved}) = (1 - P(A)) \times (1 - P(B)) \times (1 - P(C))$
$P(\text{not solved}) = (1 - \frac{1}{3}) \times (1 - \frac{1}{4}) \times (1 - \frac{1}{5})$
$P(\text{not solved}) = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{2}{5}$.
The probability that the problem is solved is:
$P(\text{solved}) = 1 - P(\text{not solved}) = 1 - \frac{2}{5} = \frac{3}{5}$.

(A) The total number of ways to have $6$ children is $2^6 = 64$.
Since at least one child is a girl,we exclude the case where all children are boys.
The number of favorable outcomes where at least one child is a girl is $64 - 1 = 63$.
The number of ways to have $3$ boys and $3$ girls is given by the binomial coefficient $\binom{6}{3} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
Therefore,the required probability is $\frac{20}{63}$.

(D) Total ways to choose two distinct numbers from $30$ is $^{30}C_{2} = \frac{30 \times 29}{2} = 435$.
For $a^2 - b^2$ to be divisible by $3$,$(a-b)(a+b)$ must be divisible by $3$.
This happens if $a \equiv b \pmod{3}$ or $a \equiv -b \pmod{3}$.
Let the sets based on remainders modulo $3$ be:
$D_{1} = \{1, 4, 7, \dots, 28\}$ (size $10$,remainder $1$)
$D_{2} = \{2, 5, 8, \dots, 29\}$ (size $10$,remainder $2$)
$D_{3} = \{3, 6, 9, \dots, 30\}$ (size $10$,remainder $0$)
Favorable cases:
$1$. Both $a, b$ from the same set: $^{10}C_{2} + ^{10}C_{2} + ^{10}C_{2} = 45 + 45 + 45 = 135$.
$2$. One from $D_{1}$ and one from $D_{2}$: $^{10}C_{1} \times ^{10}C_{1} = 100$.
Total favorable cases $= 135 + 100 = 235$.
Probability $P = \frac{235}{435} = \frac{47}{87}$.

(D) Let $P(i)$ be the probability of getting the number $i$ on the die.
Given that $P(i) \propto \frac{1}{i}$,we have $P(i) = \frac{K}{i}$ for $i = 1, 2, 3, 4, 5, 6$,where $K$ is a constant.
Since the sum of all probabilities must be $1$,we have $\sum_{i=1}^{6} P(i) = 1$.
$K \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} \right) = 1$.
Finding the common denominator for the sum: $\frac{60+30+20+15+12+10}{60} = \frac{147}{60}$.
So,$K \left( \frac{147}{60} \right) = 1 \Rightarrow K = \frac{60}{147} = \frac{20}{49}$.
The probability of getting $3$ is $P(3) = \frac{K}{3} = \frac{20}{49 \times 3} = \frac{20}{147}$.