Set Based probability Questions in English

(B) Given: $P(A) = \frac{2}{5}$ and $P(B) = \frac{4}{7}$.
We need to find the probability that exactly one of them is selected.
This is given by the formula: $P(\text{exactly one}) = P(A \cap B') + P(A' \cap B)$.
Since the events are independent,$P(A \cap B') = P(A) \times P(B')$ and $P(A' \cap B) = P(A') \times P(B)$.
Here,$P(A') = 1 - P(A) = 1 - \frac{2}{5} = \frac{3}{5}$ and $P(B') = 1 - P(B) = 1 - \frac{4}{7} = \frac{3}{7}$.
Substituting the values:
$P(\text{exactly one}) = \left(\frac{2}{5} \times \frac{3}{7}\right) + \left(\frac{3}{5} \times \frac{4}{7}\right)$
$= \frac{6}{35} + \frac{12}{35}$
$= \frac{18}{35}$.

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

(A) Let $P(A)$ be the probability of failing in Physics and $P(B)$ be the probability of failing in Mathematics.
Given $P(A) = \frac{20}{100} = 0.2$ and $P(B) = \frac{10}{100} = 0.1$.
Assuming the events are independent,the probability of failing in at least one subject is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since the events are independent,$P(A \cap B) = P(A) \cdot P(B) = 0.2 \times 0.1 = 0.02$.
Therefore,$P(A \cup B) = 0.2 + 0.1 - 0.02 = 0.28$.
Converting to percentage,$0.28 \times 100 = 28 \%$.

(D) When a pair of dice is rolled,the total number of possible outcomes is $6 \times 6 = 36$.
An even prime number is a number that is both even and prime. The only even prime number is $2$.
For each die,the probability of obtaining the number $2$ is $\frac{1}{6}$.
Since the two dice are rolled independently,the probability of obtaining an even prime number on each die is the product of their individual probabilities:
$P = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$.
Thus,the correct option is $D$.

(C) When a pair of dice is rolled,the total number of possible outcomes is $6 \times 6 = 36$.
An even number on a die can be $2, 4,$ or $6$.
So,there are $3$ favorable outcomes for each die.
The number of favorable outcomes for both dice to show an even number is $3 \times 3 = 9$.
The probability is given by the ratio of favorable outcomes to total outcomes:
$P = \frac{9}{36} = \frac{1}{4}$.

(B) When a pair of dice is rolled,the total number of possible outcomes is $6 \times 6 = 36$.
An even prime number is a number that is both even and prime. The only even prime number is $2$.
For each die,the probability of obtaining the number $2$ is $\frac{1}{6}$.
Since the rolls are independent,the probability of obtaining a $2$ on both dice is $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$.
Therefore,the correct option is $B$.

(C) When a pair of dice is thrown,the total number of possible outcomes is $6 \times 6 = 36$.
An even number on a die can be $2, 4,$ or $6$. Thus,there are $3$ favorable outcomes for each die.
The number of favorable outcomes for both dice to show an even number is $3 \times 3 = 9$.
The probability $P$ is given by the ratio of favorable outcomes to total outcomes:
$P = \frac{9}{36} = \frac{1}{4}$.

(B) Given,$P(A) = \frac{1}{2}, P(B) = \frac{1}{4}, P(C) = \frac{1}{3}$.
Therefore,the probabilities of not solving the problem are $P(\bar{A}) = 1 - \frac{1}{2} = \frac{1}{2}$,$P(\bar{B}) = 1 - \frac{1}{4} = \frac{3}{4}$,and $P(\bar{C}) = 1 - \frac{1}{3} = \frac{2}{3}$.
The problem is solved by exactly two persons if ($A$ and $B$ solve,$C$ fails) or ($A$ and $C$ solve,$B$ fails) or ($B$ and $C$ solve,$A$ fails).
Required Probability $= P(A)P(B)P(\bar{C}) + P(A)P(\bar{B})P(C) + P(\bar{A})P(B)P(C)$.
$= (\frac{1}{2} \times \frac{1}{4} \times \frac{2}{3}) + (\frac{1}{2} \times \frac{3}{4} \times \frac{1}{3}) + (\frac{1}{2} \times \frac{1}{4} \times \frac{1}{3})$.
$= \frac{2}{24} + \frac{3}{24} + \frac{1}{24} = \frac{6}{24} = \frac{1}{4}$.

(D) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
Let $S$ be the sum of the numbers on the two dice. We want to find the probability that $S > 5$.
It is easier to calculate the complement: the probability that $S \leq 5$.
The outcomes where $S \leq 5$ are:
$S=2: (1,1)$
$S=3: (1,2), (2,1)$
$S=4: (1,3), (2,2), (3,1)$
$S=5: (1,4), (2,3), (3,2), (4,1)$
Total outcomes where $S \leq 5$ is $1 + 2 + 3 + 4 = 10$.
Therefore,the number of outcomes where $S > 5$ is $36 - 10 = 26$.
The required probability is $P(S > 5) = \frac{26}{36} = \frac{13}{18}$.

(C) Since $A, B, C$ are mutually exclusive and exhaustive events,their sum of probabilities must be $1$:
$P(A) + P(B) + P(C) = 1$
Given that $P(A) = 2P(B) = 3P(C)$,we can express $P(A)$ and $P(C)$ in terms of $P(B)$:
$P(A) = 2P(B)$
$P(C) = \frac{2P(B)}{3}$
Substituting these into the sum equation:
$2P(B) + P(B) + \frac{2P(B)}{3} = 1$
Multiply the entire equation by $3$ to clear the denominator:
$6P(B) + 3P(B) + 2P(B) = 3$
$11P(B) = 3$
$P(B) = \frac{3}{11}$

(B) The total number of red marbles is $6$ (numbered $1, 2, 3, 4, 5, 6$).
The total number of white marbles is $4$ (numbered $12, 13, 14, 15$).
The total number of marbles in the box is $6 + 4 = 10$.
We are looking for a marble that is both white and odd-numbered.
The white marbles are ${12, 13, 14, 15}$.
Among these,the odd-numbered marbles are $13$ and $15$.
Thus,there are $2$ favorable outcomes.
The probability $P$ is given by the ratio of favorable outcomes to the total number of outcomes:
$P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{10} = \frac{1}{5}$.

(C) When two dice are thrown simultaneously,the total number of possible outcomes is $6 \times 6 = 36$.
The favorable outcomes for obtaining a total score of $5$ are the pairs $(1, 4), (4, 1), (2, 3), \text{ and } (3, 2)$.
The number of favorable outcomes is $4$.
The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of outcomes:
$P(E) = \frac{4}{36} = \frac{1}{9}$.

(A) Given that $A$ and $B$ are independent events.
We know that $P(A) = 1 - P(A^{\prime})$ and $P(B) = 1 - P(B^{\prime})$.
Given $P(A^{\prime}) = \frac{2}{3}$,so $P(A) = 1 - \frac{2}{3} = \frac{1}{3}$.
Given $P(B^{\prime}) = \frac{2}{7}$,so $P(B) = 1 - \frac{2}{7} = \frac{5}{7}$.
Since $A$ and $B$ are independent,$P(A \cap B) = P(A) \cdot P(B)$.
Therefore,$P(A \cap B) = \frac{1}{3} \cdot \frac{5}{7} = \frac{5}{21}$.

(C) The sum of probabilities of all outcomes in a random experiment is $1$.
$p(w_1) + p(w_2) + p(w_3) + p(w_4) + p(w_5) = 1$.
Given $p(w_1) = \frac{1}{6}$,$p(w_2) = a$,$p(w_3) = b$,$p(w_4) = c$,and $p(w_5) = \frac{1}{12}$.
So,$\frac{1}{6} + a + b + c + \frac{1}{12} = 1$.
$\frac{2+1}{12} + a + b + c = 1$ $\Rightarrow \frac{3}{12} + a + b + c = 1$ $\Rightarrow a + b + c = 1 - \frac{1}{4} = \frac{3}{4}$.
Given $12a + 12b - 1 = 0$,we have $a + b = \frac{1}{12}$.
Substituting this into the sum: $\frac{1}{12} + c = \frac{3}{4} \Rightarrow c = \frac{3}{4} - \frac{1}{12} = \frac{9-1}{12} = \frac{8}{12} = \frac{2}{3}$.
Assuming the outcomes $w_2, w_3$ are equally likely,$a = b$.
Since $a + b = \frac{1}{12}$,then $2b = \frac{1}{12} \Rightarrow b = \frac{1}{24}$.
However,based on the provided options and standard interpretation of such problems where $a=b=c$ is not implied but specific constraints are given,the intended answer is $p(w_3) = \frac{1}{24}$ which is not listed. Re-evaluating the prompt's constraint $12a+12b=1$,if $a=b$,then $b = \frac{1}{24}$. If the question implies $p(w_3) = b = \frac{1}{24}$,none match. Given the structure,if $a=b=c$,then $3b = 3/4 \Rightarrow b = 1/4$. If $12a+12b=1$ and $a=b$,$24b=1 \Rightarrow b=1/24$. Given the options,$b = 1/24$ is not present. Assuming a typo in the question where $12a+12b=1$ was meant to be $a+b=...$,we select the most logical fit.

(B) The total number of faces on the die is $2 + 3 + 1 = 6$.
The probability of getting a $1$ is $P(1) = \frac{2}{6}$.
The probability of getting a $3$ is $P(3) = \frac{1}{6}$.
Since the events are mutually exclusive,$P(1 \text{ or } 3) = P(1) + P(3) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.

(C) Given,$P(A)=0.59, P(B)=0.30$ and $P(A \cap B)=0.21$.
By De Morgan's Law,$P(A^{\prime} \cap B^{\prime}) = P((A \cup B)^{\prime}) = 1 - P(A \cup B)$.
Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values: $P(A \cup B) = 0.59 + 0.30 - 0.21 = 0.89 - 0.21 = 0.68$.
Therefore,$P(A^{\prime} \cap B^{\prime}) = 1 - 0.68 = 0.32$.

(C) Given: $P(A) = 0.5$ and $P(B) = 0.3$.
Since $A$ and $B$ are mutually exclusive events,$P(A \cap B) = 0$.
The probability of $A$ or $B$ occurring is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.3 - 0 = 0.8$.
The probability of neither $A$ nor $B$ is $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$.
Therefore,$P(A' \cap B') = 1 - 0.8 = 0.2$.

(C) When a pair of dice is thrown,the total number of possible outcomes is $6 \times 6 = 36$.
We need to find the probability of getting a sum greater than $7$.
The outcomes with a sum 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 $= 5 + 4 + 3 + 2 + 1 = 15$.
Therefore,the required probability is $P = \frac{15}{36} = \frac{5}{12}$.

(C) We are given the function $f(x) = x^2 - 5x + 4$. We need to find the probability that $f(x) > 10$ for $x \in \{1, 2, 3, \dots, 20\}$.
Solving the inequality $x^2 - 5x + 4 > 10$:
$x^2 - 5x - 6 > 0$
$(x - 6)(x + 1) > 0$
Since $x$ is a natural number,$x + 1$ is always positive. Thus,we need $x - 6 > 0$,which implies $x > 6$.
The natural numbers from $1$ to $20$ that satisfy $x > 6$ are $\{7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}$.
The number of such values is $20 - 7 + 1 = 14$.
The total number of natural numbers is $20$.
The probability is $\frac{14}{20} = \frac{7}{10}$.

(C) The set of possible values for $p$ and $q$ is $S = \{1, 2, 3, 4, 5, 6\}$.
The total number of pairs $(p, q)$ is $6 \times 6 = 36$.
We need to find the number of distinct rational numbers $p/q$.
The distinct values are:
$1/1=1, 1/2, 1/3, 1/4, 1/5, 1/6$
$2/1=2, 2/2=1, 2/3, 2/4=1/2, 2/5, 2/6=1/3$
$3/1=3, 3/2, 3/3=1, 3/4, 3/5, 3/6=1/2$
$4/1=4, 4/2=2, 4/3, 4/4=1, 4/5, 4/6=2/3$
$5/1=5, 5/2, 5/3, 5/4, 5/5=1, 5/6$
$6/1=6, 6/2=3, 6/3=2, 6/4=3/2, 6/5, 6/6=1$
Listing the distinct values: $\{1, 2, 3, 4, 5, 6, 1/2, 1/3, 1/4, 1/5, 1/6, 2/3, 2/5, 3/2, 3/4, 3/5, 4/3, 4/5, 5/2, 5/3, 5/4, 5/6, 6/5\}$.
Counting these,we find there are $23$ distinct rational numbers.
$A$ proper fraction is a fraction where $p < q$.
The proper fractions are: $\{1/2, 1/3, 1/4, 1/5, 1/6, 2/3, 2/5, 3/4, 3/5, 4/5, 5/6\}$.
There are $11$ such distinct proper fractions.
Thus,the probability is $11/23$.

(B) When two dice are thrown,the total number of outcomes is $6 \times 6 = 36$.
Let $S$ be the sum of the numbers on the two dice. The possible values for $S$ are $\{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$.
Event $A$ is the event that the sum is a prime number: $A = \{2, 3, 5, 7, 11\}$.
Event $B$ is the event that the sum is greater than $8$: $B = \{9, 10, 11, 12\}$.
We need to find $P(A \cap \overline{B})$,which is the probability of the event $A$ occurring and $B$ not occurring.
This is equivalent to the event that the sum is a prime number $AND$ the sum is less than or equal to $8$.
$A \cap \overline{B} = \{2, 3, 5, 7\}$.
Now,we count the number of outcomes for each 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
Total outcomes for $A \cap \overline{B} = 1 + 2 + 4 + 6 = 13$.
Therefore,$P(A \cap \overline{B}) = \frac{13}{36}$.

(C) The total number of outcomes when throwing three dice is $6^3 = 216$.
First,we find the number of ways to get a sum $S$ with three dice,denoted by $n(S)$.
The number of ways to get a sum $S = 13$ is $n(13) = 21$.
The number of ways to get a sum $S > 13$ is $n(14) + n(15) + n(16) + n(17) + n(18)$.
Using the formula for the number of ways to get a sum $S$ with $n$ dice,each having $f$ faces:
$n(14) = 15, n(15) = 10, n(16) = 6, n(17) = 3, n(18) = 1$.
Summing these: $15 + 10 + 6 + 3 + 1 = 35$.
Thus,the probability that $B$ gets a sum higher than $13$ is $\frac{35}{216}$.

(B) The total number of outcomes when two dice are thrown is $6 \times 6 = 36$.
The outcomes where the sum is $10$ are $(4, 6), (5, 5), (6, 4)$.
The outcomes where the sum is $11$ are $(5, 6), (6, 5)$.
The total number of favorable outcomes is $3 + 2 = 5$.
Therefore,the required probability is $\frac{5}{36}$.

(A) Let $S$ be a set with $n$ elements. Each element $x \in S$ can belong to one of the following four disjoint sets:
$1$. $x \in A$ and $x \in B$
$2$. $x \in A$ and $x \notin B$
$3$. $x \notin A$ and $x \in B$
$4$. $x \notin A$ and $x \notin B$
Since there are $n$ elements,there are $4^n$ total ways to choose subsets $A$ and $B$.
We are given the conditions $A \cap B = \phi$ and $A \cup B = S$.
For each element $x \in S$,it must belong to either $A$ or $B$ (since $A \cup B = S$) but not both (since $A \cap B = \phi$).
Thus,for each element,there are only $2$ choices: either $x \in A$ (which implies $x \notin B$) or $x \in B$ (which implies $x \notin A$).
Therefore,the number of favorable outcomes is $2^n$.
The probability is $\frac{2^n}{4^n} = \frac{2^n}{(2^2)^n} = \frac{2^n}{2^{2n}} = \frac{1}{2^n}$.

(A) The sample space $S$ for tossing a coin three times contains $2^3 = 8$ outcomes: $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.
Event $A$ (getting three heads) is $\{HHH\}$,so $P(A) = \frac{1}{8}$.
Event $B$ (getting a head on the first toss) is $\{HHH, HHT, HTH, HTT\}$,so $P(B) = \frac{4}{8} = \frac{1}{2}$.
The intersection $A \cap B$ is $\{HHH\}$,so $P(A \cap B) = \frac{1}{8}$.
For events to be independent,we must have $P(A \cap B) = P(A) \times P(B)$.
Calculating $P(A) \times P(B) = \frac{1}{8} \times \frac{1}{2} = \frac{1}{16}$.
Since $P(A \cap B) = \frac{1}{8} \neq \frac{1}{16}$,the events are dependent.

(D) Total number of outcomes $= 30$.
Multiples of $3$ in the range $1$ to $30$ are: $3, 6, 9, 12, 15, 18, 21, 24, 27, 30$ (Total $10$).
Multiples of $5$ in the range $1$ to $30$ are: $5, 10, 15, 20, 25, 30$ (Total $6$).
Multiples of both $3$ and $5$ (i.e.,multiples of $15$) are: $15, 30$ (Total $2$).
Using the inclusion-exclusion principle,the number of favorable outcomes (multiples of $3$ or $5$) is: $10 + 6 - 2 = 14$.
Therefore,the required probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{14}{30} = \frac{7}{15}$.

(A) The total number of outcomes when tossing a coin $3$ times is $2^3 = 8$. The sample space is $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.
The boy wins if all tosses give the same outcome,which are the events $\{HHH, TTT\}$.
The number of winning outcomes is $2$.
The number of losing outcomes is $8 - 2 = 6$. These outcomes are $\{HHT, HTH, HTT, THH, THT, TTH\}$.
The probability that the boy loses the game is $P(\text{Lose}) = \frac{\text{Number of losing outcomes}}{\text{Total number of outcomes}} = \frac{6}{8} = \frac{3}{4}$.

(A) non-leap year contains $365$ days,which is equal to $52$ weeks and $1$ extra day.
Since there are $52$ weeks,there are definitely $52$ Sundays.
For the year to have $53$ Sundays,the extra day must be a Sunday.
The set of possible outcomes for the extra day is $\{ \text{Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday} \}$.
Total number of possible outcomes $= 7$.
Number of favorable outcomes (the day being a Sunday) $= 1$.
Therefore,the probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{7}$.

(D) Total number of tickets = $35$.
Number of prize-bearing tickets = $10$.
Number of tickets without a prize = $35 - 10 = 25$.
Probability of not getting a prize = $\frac{\text{Number of tickets without a prize}}{\text{Total number of tickets}} = \frac{25}{35} = \frac{5}{7}$.

(D) Total number of balls $= 7 + 5 = 12$.
Probability of drawing the first green ball $= \frac{7}{12}$.
Since the balls are not replaced,the number of green balls remaining is $6$ and the total number of balls remaining is $11$.
Probability of drawing the second green ball $= \frac{6}{11}$.
After drawing two green balls,the number of green balls remaining is $5$ and the total number of balls remaining is $10$.
Probability of drawing the third green ball $= \frac{5}{10} = \frac{1}{2}$.
The probability that all three balls are green is the product of these individual probabilities:
$P = \frac{7}{12} \times \frac{6}{11} \times \frac{5}{10} = \frac{7 \times 6 \times 5}{12 \times 11 \times 10} = \frac{210}{1320} = \frac{7}{44}$.

(D) The set for $x$ is $\{1, 2, 3, 4\}$ and the set for $y$ is $\{5, 6, 7\}$.
Total number of possible outcomes is $4 \times 3 = 12$.
The product $xy$ is even if at least one of $x$ or $y$ is even.
Alternatively,$xy$ is odd only if both $x$ and $y$ are odd.
Odd values in $x$ are $\{1, 3\}$ (count = $2$).
Odd values in $y$ are $\{5, 7\}$ (count = $2$).
Number of outcomes where $xy$ is odd = $2 \times 2 = 4$.
Number of outcomes where $xy$ is even = $\text{Total outcomes} - \text{Odd outcomes} = 12 - 4 = 8$.
Probability that $xy$ is even = $\frac{8}{12} = \frac{2}{3}$.

(C) It can be noted that if the ball is neither red nor green,then it must be blue.
Number of blue balls $= 7$.
Total number of balls $= 8 + 7 + 6 = 21$.
Thus,the probability of picking a blue ball is $= \frac{7}{21} = \frac{1}{3}$.

(D) Let the total amount be $S = 20000$. Let $X$ be the amount given to employee $X$ and $Y$ be the amount given to employee $Y$.
Given $X + Y = 20000$ and $X > Y$.
Since $X$ is the bigger part,$X$ must be in the range $(10000, 20000]$.
The total length of the sample space for $X$ is $20000 - 10000 = 10000$.
We want the probability that $X \geq 2Y$.
Substituting $Y = 20000 - X$,we get $X \geq 2(20000 - X)$.
$X \geq 40000 - 2X \Rightarrow 3X \geq 40000 \Rightarrow X \geq 40000/3$.
Since $X$ must be at most $20000$,the favorable range for $X$ is $[40000/3, 20000]$.
The length of this interval is $20000 - 40000/3 = 20000/3$.
The probability is $\frac{20000/3}{10000} = \frac{2}{3}$.

(C) When two dice are rolled together,the total number of outcomes is $6 \times 6 = 36$.
The pairs that result in a sum of $9$ are $(3,6), (4,5), (5,4), (6,3)$.
Thus,the number of favourable outcomes is $4$.
The probability of getting a sum of $9$ is given by the ratio of favourable outcomes to total outcomes:
$P(\text{sum of } 9) = \frac{4}{36} = \frac{1}{9}$.

(B) When two dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
The possible total scores range from $2$ to $12$.
The prime numbers in this range are $2, 3, 5, 7, 11$.
The outcomes resulting in these sums are:
Sum $= 2: (1,1)$
Sum $= 3: (1,2), (2,1)$
Sum $= 5: (1,4), (2,3), (3,2), (4,1)$
Sum $= 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$
Sum $= 11: (5,6), (6,5)$
Total number of favorable outcomes $= 1 + 2 + 4 + 6 + 2 = 15$.
The probability is $\frac{15}{36} = \frac{5}{12}$.

(A) The total number of outcomes when a die is rolled $4$ times is $6^4 = 1296$.
For each roll,the face value must be in the set $\{2, 3, 4, 5\}$ to satisfy the condition that the minimum is $\ge 2$ and the maximum is $\le 5$.
There are $4$ such favorable outcomes for each roll.
The number of favorable outcomes for $4$ rolls is $4^4 = 256$.
The required probability is $\frac{4^4}{6^4} = \left(\frac{4}{6}\right)^4 = \left(\frac{2}{3}\right)^4 = \frac{16}{81}$.

(C) Total cards $= 30$.
Number of white cards $= 17$.
Number of green cards $= 30 - 17 = 13$.
Number of $IMPORTANT$ cards $= 4 + 5 = 9$.
Number of cards that are both green and $IMPORTANT = 5$.
Let $G$ be the event of choosing a green card and $I$ be the event of choosing an $IMPORTANT$ card.
We need to find $P(G \cup I) = P(G) + P(I) - P(G \cap I)$.
$P(G) = \frac{13}{30}$,$P(I) = \frac{9}{30}$,$P(G \cap I) = \frac{5}{30}$.
$P(G \cup I) = \frac{13}{30} + \frac{9}{30} - \frac{5}{30} = \frac{17}{30}$.

(A) The total number of elements in the set is $n(S) = 100$.
The perfect cubes between $1$ and $100$ are $1^3 = 1$,$2^3 = 8$,$3^3 = 27$,and $4^3 = 64$.
Thus,the set of favorable outcomes is $E = \{1, 8, 27, 64\}$,so $n(E) = 4$.
The required probability is $P(E) = \frac{n(E)}{n(S)} = \frac{4}{100} = \frac{1}{25}$.

(C) Total number of balls = $3 + 2 + 5 = 10$.
Number of balls that are not red = $3$ (white) + $2$ (blue) = $5$.
Probability of drawing a ball that is not red = $\frac{\text{Number of non-red balls}}{\text{Total number of balls}} = \frac{5}{10} = \frac{1}{2}$.

(B) The total number of cards in a pack is $52$.
There are $4$ Jacks,$4$ Queens,and $4$ Kings in a pack.
Total number of face cards = $4 + 4 + 4 = 12$.
The probability of drawing a face card is given by the ratio of the number of face cards to the total number of cards.
$P(\text{Face Card}) = \frac{12}{52} = \frac{3}{13}$.

(C) Let $A$ be the event of getting a road contract and $B$ be the event of getting a building contract.
Given: $P(A) = \frac{2}{9}$,$P(B) = \frac{5}{9}$,and $P(A \cap B) = \frac{1}{6}$.
We need to find the probability of getting neither contract,which is $P(A' \cap B') = P((A \cup B)')$.
First,calculate the probability of getting at least one contract:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) = \frac{2}{9} + \frac{5}{9} - \frac{1}{6} = \frac{7}{9} - \frac{1}{6}$
Finding a common denominator $(18)$:
$P(A \cup B) = \frac{14}{18} - \frac{3}{18} = \frac{11}{18}$.
Now,the probability of getting neither contract is:
$P((A \cup B)') = 1 - P(A \cup B) = 1 - \frac{11}{18} = \frac{7}{18}$.
Thus,the probability is $\frac{7}{18}$.

(C) Given,$P(A)=\frac{1}{3}, P(B)=\frac{1}{4}, P(A \cup B)=\frac{1}{2}$.
First,we find $P(A \cap B)$ using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:
$P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{1}{3} + \frac{1}{4} - \frac{1}{2} = \frac{4+3-6}{12} = \frac{1}{12}$.
Check for independence: $P(A) \cdot P(B) = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$. Since $P(A \cap B) = P(A) \cdot P(B)$,$A$ and $B$ are independent. (Option $A$ is correct).
Calculate $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/12}{1/4} = \frac{1}{3}$. (Option $B$ is correct).
Calculate $P(A^C \cap B) = P(B) - P(A \cap B) = \frac{1}{4} - \frac{1}{12} = \frac{3-1}{12} = \frac{2}{12} = \frac{1}{6}$. (Option $C$ is incorrect,as it states $1/3$).
Calculate $P(A \cap B^C) = P(A) - P(A \cap B) = \frac{1}{3} - \frac{1}{12} = \frac{4-1}{12} = \frac{3}{12} = \frac{1}{4}$. (Option $D$ is correct).
Thus,the incorrect statement is $P(A^C \cap B) = \frac{1}{3}$.

(C) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Rearranging the terms,we get $P(A) + P(B) = P(A \cup B) + P(A \cap B)$.
Substituting the given values,$P(A) + P(B) = 0.65 + 0.15 = 0.8$.
We need to find $P(A^C) + P(B^C)$.
Using the complement rule,$P(A^C) = 1 - P(A)$ and $P(B^C) = 1 - P(B)$.
Therefore,$P(A^C) + P(B^C) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))$.
Substituting the sum $P(A) + P(B) = 0.8$,we get $2 - 0.8 = 1.2$.

(C) The set $S$ consists of equations $x^2+bx+c=0$ where $b, c \in \{1, 2, 3, 4, 5, 6\}$.
The total number of such equations is $n(S) = 6 \times 6 = 36$.
$A$ quadratic equation $x^2+bx+c=0$ has real roots if the discriminant $D = b^2-4ac \geq 0$.
Since $a=1$,the condition becomes $b^2 \geq 4c$.
We test values of $b$ from $1$ to $6$:
If $b=1$,$1 \geq 4c$ (No values for $c$)
If $b=2$,$4 \geq 4c \implies c \leq 1$ ($c=1$,$1$ case)
If $b=3$,$9 \geq 4c \implies c \leq 2.25$ ($c=1, 2$,$2$ cases)
If $b=4$,$16 \geq 4c \implies c \leq 4$ ($c=1, 2, 3, 4$,$4$ cases)
If $b=5$,$25 \geq 4c \implies c \leq 6.25$ ($c=1, 2, 3, 4, 5, 6$,$6$ cases)
If $b=6$,$36 \geq 4c \implies c \leq 9$ ($c=1, 2, 3, 4, 5, 6$,$6$ cases)
Total favorable outcomes $n(E) = 0 + 1 + 2 + 4 + 6 + 6 = 19$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{19}{36}$.

(A) Total number of balls = $3 + 5 + 8 = 16$.
Number of blue balls = $5$.
Probability of getting a blue ball,$P(B) = \frac{5}{16}$.
Probability of not getting a blue ball,$P(\bar{B}) = 1 - P(B) = 1 - \frac{5}{16} = \frac{11}{16}$.

(A) We know that for any two events $A$ and $B$,the probability of their union is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Given $P(A) = 0.9$ and $P(B) = 0.8$,we have $P(A \cup B) = 1.7 - P(A \cap B)$.
Since $P(A \cap B) \geq 0.7$,the maximum value of $P(A \cap B)$ is $0.9$ (because $P(A \cap B) \leq P(A)$ and $P(A \cap B) \leq P(B)$).
Also,$P(A \cup B) \leq 1$ must hold for any probability space.
Substituting the minimum value of $P(A \cap B) = 0.7$,we get $P(A \cup B) = 1.7 - 0.7 = 1.0$.
Since $P(A \cup B)$ can be at most $1$,and the condition $P(A \cap B) \geq 0.7$ allows for valid probability values,this case is always true.

(B) non-leap year has $365$ days.
$365 = 52 \times 7 + 1$.
This means a non-leap year consists of $52$ full weeks and $1$ extra day.
The sample space for this extra day is $S = \{\text{Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}\}$.
Total number of outcomes $n(S) = 7$.
For the year to have $53$ Mondays,the extra day must be a Monday.
Number of favorable outcomes $n(A) = 1$.
Therefore,the probability $P = \frac{n(A)}{n(S)} = \frac{1}{7}$.
Hence,option $B$ is correct.

(D) When two dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
Let $E$ be the event that at least one die shows a $6$.
It is easier to calculate the probability of the complement event $E'$,which is the event that no die shows a $6$.
If no die shows a $6$,each die can show any of the numbers from $1$ to $5$.
Thus,the number of outcomes for $E'$ is $5 \times 5 = 25$.
The probability of $E'$ is $P(E') = \frac{25}{36}$.
The probability of event $E$ is $P(E) = 1 - P(E') = 1 - \frac{25}{36} = \frac{11}{36}$.

(A) For any event $E$,the probability $P(E)$ must satisfy $0 \leq P(E) \leq 1$.
For mutually exclusive events $A$ and $B$,$P(A \cup B) = P(A) + P(B) \leq 1$.
Step $1$: Constraints on individual probabilities:
$0 \leq \frac{1+3P}{3} \leq 1$ $\Rightarrow 0 \leq 1+3P \leq 3$ $\Rightarrow -1 \leq 3P \leq 2$ $\Rightarrow -\frac{1}{3} \leq P \leq \frac{2}{3}$.
$0 \leq \frac{1-2P}{2} \leq 1$ $\Rightarrow 0 \leq 1-2P \leq 2$ $\Rightarrow -1 \leq -2P \leq 1$ $\Rightarrow -\frac{1}{2} \leq P \leq \frac{1}{2}$.
Step $2$: Constraint for mutually exclusive events:
$P(A) + P(B) \leq 1
$ $\Rightarrow \frac{1+3P}{3} + \frac{1-2P}{2} \leq 1
$ $\Rightarrow \frac{2(1+3P) + 3(1-2P)}{6} \leq 1
$ $\Rightarrow 2+6P+3-6P \leq 6
$ $\Rightarrow 5 \leq 6$.
This is always true.
Step $3$: Intersection of all intervals:
$P \in [-\frac{1}{3}, \frac{2}{3}] \cap [-\frac{1}{2}, \frac{1}{2}] = [-\frac{1}{3}, \frac{1}{2}]$.