Set Based probability Questions in English

(C) Given that the probability that at least one of $A$ and $B$ occurs is $P(A \cup B) = 0.6$.
Also,the probability that $A$ and $B$ occur simultaneously is $P(A \cap B) = 0.3$.
We know that $P(A \cup B) = 1 - P(A' \cap B')$,so $P(A' \cap B') = 1 - 0.6 = 0.4$.
Using De Morgan's Law,$P(A' \cap B') = P((A \cup B)')$.
We need to find $P(A') + P(B')$.
Using the formula $P(A' \cup B') = P(A') + P(B') - P(A' \cap B')$,we have:
$P(A' \cup B') = P((A \cap B)') = 1 - P(A \cap B) = 1 - 0.3 = 0.7$.
Substituting the values into the formula:
$0.7 = P(A') + P(B') - 0.4$.
Therefore,$P(A') + P(B') = 0.7 + 0.4 = 1.1$.

(C) Using the distributive law of sets,we have $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.
Applying the addition theorem of probability,$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$,where $X = (A \cap B)$ and $Y = (A \cap C)$:
$P(A \cap (B \cup C)) = P((A \cap B) \cup (A \cap C))$
$= P(A \cap B) + P(A \cap C) - P((A \cap B) \cap (A \cap C))$
$= P(A \cap B) + P(A \cap C) - P(A \cap B \cap C)$.

(C) For three events $E_1, E_2, E_3$:
$1.$ $P(\text{only one of them occurs}) = P(E_1\bar{E}_2\bar{E}_3 + \bar{E}_1E_2\bar{E}_3 + \bar{E}_1\bar{E}_2E_3)$. Option $(A)$ is incorrect as it lists the probability of exactly two events occurring.
$2.$ $P(\text{none of them occurs}) = P(\bar{E}_1 \cap \bar{E}_2 \cap \bar{E}_3) \neq P(\bar{E}_1 + \bar{E}_2 + \bar{E}_3)$. Option $(B)$ is incorrect.
$3.$ $P(\text{at least one of them occurs}) = P(E_1 \cup E_2 \cup E_3)$. This is the definition of the union of events. Option $(C)$ is correct.
$4.$ $P(\text{all three occur}) = P(E_1 \cap E_2 \cap E_3) \neq P(E_1 + E_2 + E_3)$. Option $(D)$ is incorrect.
Therefore,the correct statement is $(C)$.

(C) We know that the event $\bar{A} \cap B$ represents the occurrence of event $B$ but not event $A$.
This can be written as $P(\bar{A} \cap B) = P(B) - P(A \cap B)$.
Given $P(B) = y$ and $P(A \cap B) = z$,we substitute these values into the formula.
Therefore,$P(\bar{A} \cap B) = y - z$.

(A) Let $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{4}$,and $P(C) = \frac{1}{6}$ be the probabilities of the three students solving the question.
The question is solved if at least one of them solves it. The probability that the question is solved is $P(A \cup B \cup C) = 1 - P(\text{none of them solve it})$.
The probability that student $A$ does not solve it is $P(\bar{A}) = 1 - \frac{1}{2} = \frac{1}{2}$.
The probability that student $B$ does not solve it is $P(\bar{B}) = 1 - \frac{1}{4} = \frac{3}{4}$.
The probability that student $C$ does not solve it is $P(\bar{C}) = 1 - \frac{1}{6} = \frac{5}{6}$.
Since the events are independent,the probability that none of them solve it is $P(\bar{A}) \times P(\bar{B}) \times P(\bar{C}) = \frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} = \frac{15}{48}$.
Therefore,the probability that the question is solved is $1 - \frac{15}{48} = \frac{48 - 15}{48} = \frac{33}{48}$.

(D) We know that the event $B$ can be expressed as the union of two disjoint events: $B = (B \cap A) \cup (B \cap \bar{A})$.
Since these events are mutually exclusive,we have $P(B) = P(B \cap A) + P(B \cap \bar{A})$.
Rearranging the terms to solve for $P(\bar{A} \cap B)$,we get $P(\bar{A} \cap B) = P(B) - P(A \cap B)$.

(C) Given $P(A \cup B) = 3/5$ and $P(A \cap B) = 1/5$.
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Thus,$P(A) + P(B) = P(A \cup B) + P(A \cap B) = 3/5 + 1/5 = 4/5$.
We need to find $P(A') + P(B')$.
Since $P(A') = 1 - P(A)$ and $P(B') = 1 - P(B)$,
$P(A') + P(B') = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))$.
Substituting the value of $P(A) + P(B)$,
$P(A') + P(B') = 2 - 4/5 = 6/5$.

(A) Given: $P(A \cup B) = 3/4,$ $P(A \cap B) = 1/4,$ and $P(\bar{A}) = 2/3.$
First,find $P(A):$
$P(A) = 1 - P(\bar{A}) = 1 - 2/3 = 1/3.$
Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B).$
$3/4 = 1/3 + P(B) - 1/4.$
$P(B) = 3/4 + 1/4 - 1/3 = 1 - 1/3 = 2/3.$
Now,calculate $P(\bar{A} \cap B):$
$P(\bar{A} \cap B) = P(B) - P(A \cap B).$
$P(\bar{A} \cap B) = 2/3 - 1/4 = (8 - 3)/12 = 5/12.$

(A) Given $P(A) = P(B) = x$ and $P(A \cap B) = \frac{1}{3}$.
Also,$P(A' \cap B') = \frac{1}{3}$.
By De Morgan's Law,$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) = \frac{1}{3}$.
Therefore,$P(A \cup B) = 1 - \frac{1}{3} = \frac{2}{3}$.
Using the addition theorem,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values,$\frac{2}{3} = x + x - \frac{1}{3}$.
$\frac{2}{3} + \frac{1}{3} = 2x$.
$1 = 2x \Rightarrow x = \frac{1}{2}$.

(A) Given $P(\overline{A \cup B}) = \frac{1}{6}$ and $P(A \cap B) = \frac{1}{4}$.
Since $P(\bar{A}) = \frac{1}{4}$,we have $P(A) = 1 - \frac{1}{4} = \frac{3}{4}$.
We know that $P(\overline{A \cup B}) = 1 - P(A \cup B) = 1 - [P(A) + P(B) - P(A \cap B)]$.
Substituting the values: $\frac{1}{6} = 1 - [\frac{3}{4} + P(B) - \frac{1}{4}] = 1 - [\frac{1}{2} + P(B)] = \frac{1}{2} - P(B)$.
Thus,$P(B) = \frac{1}{2} - \frac{1}{6} = \frac{3-1}{6} = \frac{2}{6} = \frac{1}{3}$.
Now,check for independence: $P(A) \times P(B) = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4}$.
Since $P(A \cap B) = P(A) \times P(B) = \frac{1}{4}$,the events $A$ and $B$ 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.
Therefore,the events $A$ and $B$ are independent but not equally likely.

(D) For each element $x \in S$,there are $4$ possibilities for its membership in $A$ and $B$:
$1$. $x \in A$ and $x \notin B$
$2$. $x \notin A$ and $x \in B$
$3$. $x \in A$ and $x \in B$
$4$. $x \notin A$ and $x \notin B$
Since there are $n$ elements,the total number of ways to choose two subsets $A$ and $B$ is $4^n = (2^2)^n = (2^n)^2$.
For the condition $A \cup B = S$ and $A \cap B = \phi$ to hold,for every element $x \in S$,it must be that either ($x \in A$ and $x \notin B$) or ($x \notin A$ and $x \in B$).
This means for each of the $n$ elements,there are exactly $2$ choices.
Thus,the number of favorable cases is $2^n$.
The required probability is $\frac{2^n}{(2^n)^2} = \frac{2^n}{2^{2n}} = \frac{1}{2^n}$.

(B) Given $P(A') = 0.3$,we have $P(A) = 1 - P(A') = 1 - 0.3 = 0.7$.
Given $P(B) = 0.4$,we have $P(B') = 1 - P(B) = 1 - 0.4 = 0.6$.
We are given $P(A \cap B') = 0.5$.
Using the addition theorem for probability,$P(A \cup B') = P(A) + P(B') - P(A \cap B')$.
Substituting the values,$P(A \cup B') = 0.7 + 0.6 - 0.5 = 0.8$.

(A) The total number of outcomes when a coin is tossed $3$ times is $2^3 = 8$.
The sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
The outcomes with exactly two heads are $\{HHT, HTH, THH\}$.
The number of favorable outcomes is $3$.
Therefore,the probability of getting exactly two heads is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{8}$.

(C) When a coin is tossed $3$ times,the total number of outcomes is $2^3 = 8$.
The sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
'At least two heads' means getting $2$ heads or $3$ heads.
The favorable outcomes are $\{HHH, HHT, HTH, THH\}$.
The number of favorable outcomes is $4$.
Therefore,the required probability is $\frac{4}{8} = \frac{1}{2}$.

(A) Since $\frac{1 + 3p}{3}, \frac{1 - p}{4}$ and $\frac{1 - 2p}{2}$ are the probabilities of three events,each probability must lie in the interval $[0, 1]$.
$1$) $0 \le \frac{1 + 3p}{3} \le 1 \Rightarrow 0 \le 1 + 3p \le 3 \Rightarrow -1 \le 3p \le 2 \Rightarrow -\frac{1}{3} \le p \le \frac{2}{3}$
$2$) $0 \le \frac{1 - p}{4} \le 1 \Rightarrow 0 \le 1 - p \le 4 \Rightarrow -1 \le -p \le 3 \Rightarrow -3 \le p \le 1$
$3$) $0 \le \frac{1 - 2p}{2} \le 1 \Rightarrow 0 \le 1 - 2p \le 2 \Rightarrow -1 \le -2p \le 1 \Rightarrow -\frac{1}{2} \le p \le \frac{1}{2}$
For mutually exclusive events,the sum of probabilities must be $\le 1$:
$\frac{1 + 3p}{3} + \frac{1 - p}{4} + \frac{1 - 2p}{2} \le 1$
Multiply by $12$: $4(1 + 3p) + 3(1 - p) + 6(1 - 2p) \le 12$
$4 + 12p + 3 - 3p + 6 - 12p \le 12$
$13 - 3p \le 12 \Rightarrow -3p \le -1 \Rightarrow p \ge \frac{1}{3}$
Taking the intersection of all conditions:
$p \in [-\frac{1}{3}, \frac{2}{3}] \cap [-3, 1] \cap [-\frac{1}{2}, \frac{1}{2}] \cap [\frac{1}{3}, \infty)$
Intersection is $\frac{1}{3} \le p \le \frac{1}{2}$.

(A) The total number of subsets of a set $X$ containing $n$ elements is $2^n$.
Since two subsets $A$ and $B$ are chosen,the total number of ways to choose the pair $(A, B)$ is $2^n \times 2^n = 2^{2n}$.
For $A$ and $B$ to have the same number of elements,let this number be $r$,where $0 \le r \le n$.
The number of ways to choose a subset with $r$ elements is $^nC_r$.
Thus,the number of ways to choose $A$ and $B$ such that $|A| = |B| = r$ is $(^nC_r)^2$.
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 number of favorable ways is $^{2n}C_n$.
Therefore,the required probability is $\frac{^{2n}C_n}{2^{2n}}$.

(B) Given that $P(A) = 0.5$ and $P(B) = 0.3$.
Since $A$ and $B$ are mutually exclusive events,$P(A \cap B) = 0$.
The probability of happening neither $A$ nor $B$ is $P(\overline{A} \cap \overline{B})$.
By De Morgan's Law,$P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.
For mutually exclusive events,$P(A \cup B) = P(A) + P(B) = 0.5 + 0.3 = 0.8$.
Therefore,the required probability is $1 - 0.8 = 0.2$.

(A) Let $X_1, X_2, X_3, X_4$ be the outcomes of the four rolls.
We want the probability that $2 \le X_i \le 5$ for all $i = 1, 2, 3, 4$.
For each roll,the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$,so there are $6$ total outcomes.
The favorable outcomes for each roll are $\{2, 3, 4, 5\}$,which gives $4$ favorable outcomes.
The probability of a single roll resulting in a value between $2$ and $5$ inclusive is $p = \frac{4}{6} = \frac{2}{3}$.
Since the four rolls are independent,the probability that all four rolls result in values between $2$ and $5$ is $p^4 = (\frac{2}{3})^4 = \frac{16}{81}$.

(A) The probability of getting no tails in $4$ tosses is calculated as:
$P(\text{no tail}) = (1/2) \times (1/2) \times (1/2) \times (1/2) = 1/16$.
The probability of getting at least one tail is:
$P(\text{at least one tail}) = 1 - P(\text{no tail}) = 1 - 1/16 = 15/16$.

(C) Let the two integers be $x$ and $y$. The possible parity combinations for $(x, y)$ are (Even,Even),(Even,Odd),(Odd,Even),and (Odd,Odd).
Each combination has a probability of $1/4$.
$1$. (Even,Even): Product is Even.
$2$. (Even,Odd): Product is Even.
$3$. (Odd,Even): Product is Even.
$4$. (Odd,Odd): Product is Odd.
The product is even in $3$ out of $4$ cases.
Therefore,the probability is $3/4$.

(A) Let $P(A) = 4/5$,$P(B) = 3/4$,and $P(C) = 2/3$. The probabilities of missing the target are $P(\overline{A}) = 1/5$,$P(\overline{B}) = 1/4$,and $P(\overline{C}) = 1/3$.
The event that exactly two hit the target occurs in three mutually exclusive cases: $(A \cap B \cap \overline{C})$,$(A \cap \overline{B} \cap C)$,and $(\overline{A} \cap B \cap C)$.
Since $A, B,$ and $C$ are independent events:
$P(A \cap B \cap \overline{C}) = P(A) \times P(B) \times P(\overline{C}) = (4/5) \times (3/4) \times (1/3) = 12/60 = 1/5$.
$P(A \cap \overline{B} \cap C) = P(A) \times P(\overline{B}) \times P(C) = (4/5) \times (1/4) \times (2/3) = 8/60 = 2/15$.
$P(\overline{A} \cap B \cap C) = P(\overline{A}) \times P(B) \times P(C) = (1/5) \times (3/4) \times (2/3) = 6/60 = 1/10$.
The total probability is the sum of these probabilities:
$P = 12/60 + 8/60 + 6/60 = 26/60 = 13/30$.

(C) The total number of outcomes when three tetrahedra are thrown is $n = 4 \times 4 \times 4 = 64$.
Let $A$ be the event that the sum of the numbers is $5$.
The possible outcomes are $(2, 2, 1), (2, 1, 2), (1, 2, 2), (1, 1, 3), (1, 3, 1), (3, 1, 1)$.
The number of favorable outcomes is $r = 6$.
Therefore,the probability $P(A) = \frac{r}{n} = \frac{6}{64} = \frac{3}{32}$.

(B) When three dice are thrown,the total number of possible outcomes is $6 \times 6 \times 6 = 216$.
The favorable outcomes where all three dice show the same number are: $(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6, 6, 6)$.
There are $6$ such favorable outcomes.
Therefore,the required probability is $P = \frac{6}{216} = \frac{1}{36}$.

(A) Given $P(A') = \frac{2}{3}$,we find $P(A) = 1 - P(A') = 1 - \frac{2}{3} = \frac{1}{3}$.
Using the addition theorem for probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values: $\frac{3}{4} = \frac{1}{3} + P(B) - \frac{1}{4}$.
Solving for $P(B)$: $P(B) = \frac{3}{4} + \frac{1}{4} - \frac{1}{3} = 1 - \frac{1}{3} = \frac{2}{3}$.
We know that $P(A' \cap B) = P(B) - P(A \cap B)$.
Therefore,$P(A' \cap B) = \frac{2}{3} - \frac{1}{4} = \frac{8 - 3}{12} = \frac{5}{12}$.

(B) non-leap year has $365$ days.
$365$ days $= 52$ weeks and $1$ day.
The $52$ weeks contain $52$ Sundays.
The remaining $1$ day can be any of the $7$ days of the week (Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday).
For the year to have $53$ Sundays,the remaining $1$ day must be a Sunday.
Therefore,the probability $= 1/7$.

(B) The probability that exactly one of the events $A$ or $B$ occurs is given by the probability that $A$ occurs and $B$ does not,or $B$ occurs and $A$ does not.
This is expressed as: $P(A \cap \bar{B}) + P(\bar{A} \cap B)$
Using the property $P(A \cap \bar{B}) = P(A) - P(A \cap B)$ and $P(\bar{A} \cap B) = P(B) - P(A \cap B)$,
We get: $P(A) - P(A \cap B) + P(B) - P(A \cap B)$
$= P(A) + P(B) - 2P(A \cap B)$

(B) The sample space $S$ for tossing a coin and a die is given by:
$S = \{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6\}$
Total number of outcomes $n(S) = 12$.
Let $E$ be the event of getting a head $(H)$ on the coin and $6$ on the die.
$E = \{H6\}$
Number of favorable outcomes $n(E) = 1$.
Therefore,the probability $P(E) = \frac{n(E)}{n(S)} = \frac{1}{12}$.

(C) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
The favorable outcomes where the sum of the numbers is $5$ are: $\{(1, 4), (4, 1), (2, 3), (3, 2)\}$.
The number of favorable outcomes is $r = 4$.
The probability $P(A)$ is given by:
$P(A) = \frac{r}{n} = \frac{4}{36} = \frac{1}{9}$.

(B) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
The outcomes that result in a sum of $7$ are: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$.
There are $6$ such favorable outcomes.
The probability is given by the ratio of favorable outcomes to total outcomes: $P = \frac{6}{36}$.

(C) The total number of possible outcomes when rolling a fair die is $n = 6$,where the sample space is $S = \{1, 2, 3, 4, 5, 6\}$.
The favorable outcomes for getting $1$ or $6$ are $\{1, 6\}$,so the number of favorable outcomes is $r = 2$.
The probability $P$ is given by the formula $P = \frac{r}{n}$.
Substituting the values,$P = \frac{2}{6} = \frac{1}{3}$.

(C) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
For the first throw,the sum is $10$. The possible pairs are $(4, 6), (5, 5), (6, 4)$. Thus,there are $3$ favorable outcomes. The probability $P(10) = 3/36 = 1/12$.
For the second throw,the sum is $11$. The possible pairs are $(5, 6), (6, 5)$. Thus,there are $2$ favorable outcomes. The probability $P(11) = 2/36 = 1/18$.
For the third throw,the sum is $12$. The only possible pair is $(6, 6)$. Thus,there is $1$ favorable outcome. The probability $P(12) = 1/36$.
Since the throws are independent,the total probability is $P(10) \times P(11) \times P(12) = (1/12) \times (1/18) \times (1/36) = 1/7776$.

(A) When a die is thrown $2$ times,the total number of outcomes is $6^2 = 36$.
Let $E$ be the event of getting $4$ at least once.
It is easier to calculate the probability of the complement event $E'$,which is the event of not getting $4$ in either throw.
The probability of not getting $4$ in one throw is $\frac{5}{6}$.
Since the two throws are independent,the probability of not getting $4$ in both throws is $P(E') = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}$.
Therefore,the probability of getting $4$ at least once is $P(E) = 1 - P(E') = 1 - \frac{25}{36} = \frac{11}{36}$.

(B) Total number of balls in the bag $= 6 + 5 + 4 = 15$.
Let $A$ be the event of getting a white ball and $B$ be the event of getting a black ball.
The number of white balls is $6$,so $P(A) = \frac{6}{15}$.
The number of black balls is $5$,so $P(B) = \frac{5}{15}$.
Since the events $A$ and $B$ are mutually exclusive,the probability of getting a white or a black ball is $P(A \cup B) = P(A) + P(B)$.
$P(A \cup B) = \frac{6}{15} + \frac{5}{15} = \frac{11}{15}$.

(C) Initially,the container has $2$ black balls.
After adding $1$ white ball,the total number of balls in the container becomes $2 + 1 = 3$.
The composition of the container is now $2$ black balls and $1$ white ball.
The probability $P$ of drawing a white ball is given by the ratio of the number of white balls to the total number of balls.
$P(\text{white}) = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{1}{3}$.

(A) Given: $P(A \cup B) = \frac{3}{4}$,$P(A \cap B) = \frac{1}{4}$,and $P(\overline{A}) = \frac{2}{3}$.
Since $P(A) = 1 - P(\overline{A})$,we have $P(A) = 1 - \frac{2}{3} = \frac{1}{3}$.
Using the addition theorem: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values: $\frac{3}{4} = \frac{1}{3} + P(B) - \frac{1}{4}$.
$P(B) = \frac{3}{4} + \frac{1}{4} - \frac{1}{3} = 1 - \frac{1}{3} = \frac{2}{3}$.
We need to find $P(\overline{A} \cap B)$.
Note that $P(\overline{A} \cap B) = P(B) - P(A \cap B)$.
$P(\overline{A} \cap B) = \frac{2}{3} - \frac{1}{4} = \frac{8 - 3}{12} = \frac{5}{12}$.

(B) Total number of cards in a deck $= 52$.
Number of kings in a deck $= 4$.
Number of queens in a deck $= 4$.
Since the events are mutually exclusive,the number of favorable outcomes (king or queen) $= 4 + 4 = 8$.
Therefore,the required probability $= \frac{8}{52} = \frac{2}{13}$.

(C) Let $P(A) = 1/2, P(B) = 1/3$,and $P(C) = 1/4$ be the probabilities of students $A, B$,and $C$ solving the problem respectively.
The problem is solved if at least one of them solves it.
The probability that the problem is not solved by any of them is given by:
$P(\text{not solved}) = P(A') \times P(B') \times P(C')$
$P(\text{not solved}) = (1 - 1/2) \times (1 - 1/3) \times (1 - 1/4)$
$P(\text{not solved}) = (1/2) \times (2/3) \times (3/4) = 1/4$
The probability that the problem is solved is:
$P(\text{solved}) = 1 - P(\text{not solved})$
$P(\text{solved}) = 1 - 1/4 = 3/4$

(B) Let $A$ be the event of getting $4, 5,$ or $6$ on the first throw. The number of favorable outcomes is $3$. The total outcomes for one die is $6$. So,$P(A) = \frac{3}{6} = \frac{1}{2}$.
Let $B$ be the event of getting $1, 2, 3,$ or $4$ on the second throw. The number of favorable outcomes is $4$. The total outcomes for one die is $6$. So,$P(B) = \frac{4}{6} = \frac{2}{3}$.
Since the two throws are independent events,the probability of both occurring is $P(A \cap B) = P(A) \times P(B)$.
$P(A \cap B) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$.

(C) leap year consists of $366$ days,which is $52$ weeks and $2$ extra days. The $7$ possible outcomes for these $2$ extra days are:
$(i)$ (Sunday,Monday),$(ii)$ (Monday,Tuesday),$(iii)$ (Tuesday,Wednesday),$(iv)$ (Wednesday,Thursday),$(v)$ (Thursday,Friday),$(vi)$ (Friday,Saturday),$(vii)$ (Saturday,Sunday).
Let $A$ be the event that the leap year has $53$ Sundays,and $B$ be the event that it has $53$ Mondays.
Then $P(A) = 2/7$,$P(B) = 2/7$,and $P(A \cap B) = 1/7$ (since the only case with both is Sunday and Monday).
The required probability is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = 2/7 + 2/7 - 1/7 = 3/7$.

(C) Since $A, B$ and $C$ are mutually exclusive events,$P(A \cup B \cup C) = P(A) + P(B) + P(C)$.
Also,$P(A \cup B \cup C) \leq 1$.
Thus,$0 \leq \frac{3x + 1}{3} + \frac{1 - x}{4} + \frac{1 - 2x}{2} \leq 1$.
Multiplying by $12$,we get $0 \leq 4(3x + 1) + 3(1 - x) + 6(1 - 2x) \leq 12$.
$0 \leq 12x + 4 + 3 - 3x + 6 - 12x \leq 12$.
$0 \leq 13 - 3x \leq 12$.
Subtracting $13$,we get $-13 \leq -3x \leq -1$.
Dividing by $-3$ (reversing the inequality),we get $\frac{1}{3} \leq x \leq \frac{13}{3}$.
Additionally,for any event $E$,$0 \leq P(E) \leq 1$:
$1$) $0 \leq \frac{3x + 1}{3} \leq 1 \Rightarrow 0 \leq 3x + 1 \leq 3 \Rightarrow -1 \leq 3x \leq 2 \Rightarrow -\frac{1}{3} \leq x \leq \frac{2}{3}$.
$2$) $0 \leq \frac{1 - x}{4} \leq 1 \Rightarrow 0 \leq 1 - x \leq 4 \Rightarrow -1 \leq -x \leq 3 \Rightarrow -3 \leq x \leq 1$.
$3$) $0 \leq \frac{1 - 2x}{2} \leq 1 \Rightarrow 0 \leq 1 - 2x \leq 2 \Rightarrow -1 \leq -2x \leq 1 \Rightarrow -\frac{1}{2} \leq x \leq \frac{1}{2}$.
Taking the intersection of all conditions: $x \in [\frac{1}{3}, \frac{13}{3}] \cap [-\frac{1}{3}, \frac{2}{3}] \cap [-3, 1] \cap [-\frac{1}{2}, \frac{1}{2}] = [\frac{1}{3}, \frac{1}{2}]$.

(B) Let $A, B$,and $C$ be the events of shooting down the plane in the first,second,and third shots respectively.
Given probabilities are $P(A) = 0.6, P(B) = 0.7, P(C) = 0.1$.
Since the events are independent,the probability of not hitting the plane in any of the three shots is $P(\overline{A}) \times P(\overline{B}) \times P(\overline{C})$.
$P(\overline{A}) = 1 - 0.6 = 0.4$
$P(\overline{B}) = 1 - 0.7 = 0.3$
$P(\overline{C}) = 1 - 0.1 = 0.9$
Probability of not hitting the plane = $0.4 \times 0.3 \times 0.9 = 0.108$.
Therefore,the probability of shooting down the plane is $1 - 0.108 = 0.892$.

(D) Total number of items = $10 + 6 = 16$.
Let $E$ be the event of selecting an item that is either good or defective.
Since every item in the box is either good or defective,the event $E$ is a sure event.
Therefore,the probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{16}{16} = 1$.

(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:
$P(\text{none}) = P(\text{not } A) \times P(\text{not } B) \times P(\text{not } C) = (1/2) \times (2/3) \times (3/4) = 1/4$.
The probability that the problem is solved is $1 - P(\text{none})$.
$P(\text{solved}) = 1 - 1/4 = 3/4$.

(A) Total number of pencils = $12 + 6 + 2 = 20$.
Number of non-defective (good) pencils = $12$.
The probability of selecting a non-defective pencil is given by:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{20}$.
Simplifying the fraction,we get:
$P(E) = \frac{3}{5}$.

(B) Let $H$ be the event that the husband is selected and $W$ be the event that the wife is selected.
Given $P(H) = 1/7$ and $P(W) = 1/5$.
The probability that the husband is not selected is $P(H') = 1 - 1/7 = 6/7$.
The probability that the wife is not selected is $P(W') = 1 - 1/5 = 4/5$.
The probability that only one of them is selected is $P(H \cap W') + P(H' \cap W)$.
Since the events are independent,this is $P(H) \times P(W') + P(H') \times P(W)$.
$= (1/7 \times 4/5) + (6/7 \times 1/5) = 4/35 + 6/35 = 10/35 = 2/7$.

(D) When three coins are tossed,the total number of possible outcomes is $2^3 = 8$. The sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
The event of getting 'at least one head' includes all outcomes except the one where no heads appear (i.e.,$TTT$).
The number of favorable outcomes is $8 - 1 = 7$.
Therefore,the probability $P$ of getting at least one head is:
$P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{7}{8}$.

(C) The total number of possible outcomes when two dice are thrown is $6 \times 6 = 36$.
The outcomes where the first die shows $1$ are: $(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$.
The number of favorable outcomes is $6$.
Therefore,the probability is $\frac{6}{36} = \frac{1}{6}$.

(B) The set of all odd-numbered overs in a $50$-over match is $U = \{1, 3, 5, 7, \dots, 49\}$.
Since there are $50$ overs,there are $25$ odd numbers in this set,so $n(U) = 25$.
We are looking for overs that are multiples of $9$ and are also odd numbers.
The multiples of $9$ are $9, 18, 27, 36, 45, \dots$.
From these,the odd multiples are $A = \{9, 27, 45\}$.
Thus,the number of favorable outcomes is $n(A) = 3$.
The probability $P(A)$ is given by $\frac{n(A)}{n(U)} = \frac{3}{25}$.

(A) The total number of outcomes when throwing three dice is $n = 6 \times 6 \times 6 = 216$.
The event $A$ of getting a sum of $16$ consists of the outcomes:
$A = \{(6, 6, 4), (6, 4, 6), (4, 6, 6), (5, 5, 6), (5, 6, 5), (6, 5, 5)\}$.
The number of favorable outcomes is $r = 6$.
Therefore,the probability $P(A) = \frac{r}{n} = \frac{6}{216} = \frac{1}{36}$.