Set Based probability Questions in English

(A) Let $P(A)$ be the probability that athlete $A$ wins and $P(B)$ be the probability that athlete $B$ wins.
Given $P(A) = \frac{1}{5}$ and $P(B) = \frac{1}{4}$.
The probability that $A$ does not win is $P(A') = 1 - P(A) = 1 - \frac{1}{5} = \frac{4}{5}$.
The probability that $B$ does not win is $P(B') = 1 - P(B) = 1 - \frac{1}{4} = \frac{3}{4}$.
Since the events are independent,the probability that neither of them wins is $P(A' \cap B') = P(A') \times P(B')$.
$P(A' \cap B') = \frac{4}{5} \times \frac{3}{4} = \frac{3}{5}$.

(B) The sample space $S$ for tossing three coins is: $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.
Total number of outcomes $n(S) = 8$.
The outcomes where heads and tails appear alternately are $HTH$ and $THT$.
Number of favourable outcomes $n(E) = 2$.
Therefore,the required probability $P(E) = \frac{n(E)}{n(S)} = \frac{2}{8} = \frac{1}{4}$.

(A) sure event is an event that is certain to occur,so $P(A) = 1$.
By the complement rule,$P(\text{not } A) = 1 - P(A)$.
Substituting the value,$P(\text{not } A) = 1 - 1 = 0$.
Therefore,the correct option is $A$.

(D) The first ten natural numbers are $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
Total number of outcomes $= 10$.
The numbers that are both odd and perfect squares in this set are $\{1, 9\}$.
Number of favorable outcomes $= 2$.
Therefore,the required probability $= \frac{2}{10} = \frac{1}{5}$.

(D) When two dice are thrown,the total number of outcomes where the first die shows $5$ is $6$. These outcomes are $(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)$.
We are looking for the outcomes where the sum of the numbers is $8$ or more.
The sums for the given outcomes are:
$(5, 1) \rightarrow 6$
$(5, 2) \rightarrow 7$
$(5, 3) \rightarrow 8$
$(5, 4) \rightarrow 9$
$(5, 5) \rightarrow 10$
$(5, 6) \rightarrow 11$
The favorable outcomes are $(5, 3), (5, 4), (5, 5), (5, 6)$,which are $4$ in total.
The required probability is $\frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{6} = \frac{2}{3}$.

(A) Total number of cards in a pack $= 52$.
Number of aces in a pack $= 4$.
Number of kings in a pack $= 4$.
Total number of cards that are either an ace or a king $= 4 + 4 = 8$.
Number of cards that are neither an ace nor a king $= 52 - 8 = 44$.
Required probability $= \frac{44}{52} = \frac{11}{13}$.

(C) An event $A$ is independent of itself if and only if $P(A \cap A) = P(A) \cdot P(A)$.
Since $A \cap A = A$,the condition becomes $P(A) = P(A)^2$.
Rearranging the equation,we get $P(A)^2 - P(A) = 0$.
Factoring gives $P(A)(P(A) - 1) = 0$.
Therefore,$P(A) = 0$ or $P(A) = 1$.

(A) Total number of cards in a pack = $52$.
There is only one $2$ of hearts and one $2$ of diamonds in a deck.
Number of favorable outcomes = $1 + 1 = 2$.
Required probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{52} = \frac{1}{26}$.

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

(B) Let $P(A) = \frac{1}{3}, P(B) = \frac{2}{7}, P(C) = \frac{3}{8}$.
Then the probabilities of not solving the problem are $P(A') = 1 - \frac{1}{3} = \frac{2}{3}, P(B') = 1 - \frac{2}{7} = \frac{5}{7},$ and $P(C') = 1 - \frac{3}{8} = \frac{5}{8}$.
Exactly one of them solves the problem if ($A$ solves and $B, C$ do not) $OR$ ($B$ solves and $A, C$ do not) $OR$ ($C$ solves and $A, B$ do not).
Required probability $= P(A)P(B')P(C') + P(A')P(B)P(C') + P(A')P(B')P(C)$
$= (\frac{1}{3} \times \frac{5}{7} \times \frac{5}{8}) + (\frac{2}{3} \times \frac{2}{7} \times \frac{5}{8}) + (\frac{2}{3} \times \frac{5}{7} \times \frac{3}{8})$
$= \frac{25}{168} + \frac{20}{168} + \frac{30}{168} = \frac{75}{168} = \frac{25}{56}$.

(A) 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: $\{(6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6)\}$. There are $6$ such outcomes.
The outcomes that result in a sum of $9$ are: $\{(6, 3), (5, 4), (4, 5), (3, 6)\}$. There are $4$ such outcomes.
Since these events are mutually exclusive,the total number of favorable outcomes is $6 + 4 = 10$.
The required probability is $\frac{10}{36} = \frac{5}{18}$.

(A) Let $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$,and $P(C) = \frac{1}{4}$ be the probabilities of hitting the target.
Then,the probabilities of missing the target 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 one and only one of them hits the target is given by:
$P = P(A)P(\bar{B})P(\bar{C}) + P(\bar{A})P(B)P(\bar{C}) + P(\bar{A})P(\bar{B})P(C)$
$P = (\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}) + (\frac{1}{2} \times \frac{1}{3} \times \frac{3}{4}) + (\frac{1}{2} \times \frac{2}{3} \times \frac{1}{4})$
$P = \frac{6}{24} + \frac{3}{24} + \frac{2}{24} = \frac{11}{24}$.

(D) Total number of balls = $3 + 3 + 2 = 8$.
Let $R_3$ be the event that the third ball drawn is red.
By the principle of symmetry in probability,the probability of drawing a red ball at any specific position (first,second,or third) without replacement is equal to the initial proportion of red balls in the bag.
$P(R_3) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{2}{8} = \frac{1}{4}$.
Alternatively,summing all possible mutually exclusive cases for the first two draws:
$P(R_3) = P(W_1 W_2 R_3) + P(W_1 B_2 R_3) + P(B_1 W_2 R_3) + P(B_1 B_2 R_3) + P(R_1 W_2 R_3) + P(R_1 B_2 R_3) + P(W_1 R_2 R_3) + P(B_1 R_2 R_3)$
$= \frac{3}{8} \times \frac{2}{7} \times \frac{2}{6} + \frac{3}{8} \times \frac{3}{7} \times \frac{2}{6} + \frac{3}{8} \times \frac{3}{7} \times \frac{2}{6} + \frac{3}{8} \times \frac{2}{7} \times \frac{2}{6} + \frac{2}{8} \times \frac{3}{7} \times \frac{1}{6} + \frac{2}{8} \times \frac{3}{7} \times \frac{1}{6} + \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6} + \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6}$
$= \frac{12 + 18 + 18 + 12 + 6 + 6 + 6 + 6}{336} = \frac{84}{336} = \frac{1}{4}$.

(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 $8$ are: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$.
The number of favourable outcomes is $5$.
Therefore,the required probability is $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{5}{36}$.

(B) For any event $A$,the event $\bar{A}$ represents the complement of $A$.
Since the sum of the probability of an event and its complement is always equal to $1$,we have $P(A) + P(\bar{A}) = 1$.
Therefore,the correct option is $B$.

(C) Total number of cards = $100$.
The perfect squares between $1$ and $100$ are: $1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64, 9^2=81, 10^2=100$.
Number of favourable outcomes = $10$.
Probability = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{10}{100} = \frac{1}{10}$.

(C) The probability of non-occurrence of event $A_i$ is given by $P(A_i^c) = 1 - P(A_i) = 1 - \frac{1}{i + 1} = \frac{i}{i + 1}$.
Since the events are independent,the probability that none of the events occur is the product of their individual probabilities of non-occurrence:
$P(\text{none occur}) = P(A_1^c) \times P(A_2^c) \times \dots \times P(A_n^c)$
$= \left( \frac{1}{2} \right) \times \left( \frac{2}{3} \right) \times \left( \frac{3}{4} \right) \times \dots \times \left( \frac{n}{n + 1} \right)$
$= \frac{1 \times 2 \times 3 \times \dots \times n}{2 \times 3 \times 4 \times \dots \times (n + 1)}$
$= \frac{1}{n + 1}$.

(D) Let $P(T)$ be the probability of a test occurring in a class,where $P(T) = 1/5$. The probability of no test is $P(T') = 1 - 1/5 = 4/5$.
The student is absent for $2$ classes. Let $T_1$ and $T_2$ be the events that a test occurs in the first and second class,respectively.
The student misses at least one test if a test occurs in the first class,the second class,or both.
This is equivalent to the complement of the event that no test occurs in either class.
Probability of no test in both classes $P(\text{No test}) = P(T_1') \times P(T_2') = (4/5) \times (4/5) = 16/25$.
Probability of missing at least one test $= 1 - P(\text{No test}) = 1 - 16/25 = 9/25$.

(C) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
Favorable outcomes for a sum of $3$: $(1, 2), (2, 1)$ (Total $2$ cases).
Favorable outcomes for a sum of $5$: $(1, 4), (2, 3), (3, 2), (4, 1)$ (Total $4$ cases).
Favorable outcomes for a sum of $11$: $(5, 6), (6, 5)$ (Total $2$ cases).
Total favorable outcomes $= 2 + 4 + 2 = 8$.
Required probability $= \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{8}{36} = \frac{2}{9}$.

(B) Total number of cards in a pack $= 52$.
Number of kings in a pack $= 4$.
Number of queens in a pack $= 4$.
Since the events are mutually exclusive,the probability of drawing a king or a queen is the sum of their individual probabilities.
$P(\text{king}) = \frac{4}{52} = \frac{1}{13}$.
$P(\text{queen}) = \frac{4}{52} = \frac{1}{13}$.
$P(\text{king or queen}) = P(\text{king}) + P(\text{queen}) = \frac{1}{13} + \frac{1}{13} = \frac{2}{13}$.

(D) Let $A, B, C$ be the events of getting $I, II$ and $III$ division respectively,and $D$ be the event of failing the examination.
Since these events are mutually exclusive and exhaustive,the sum of their probabilities must be $1$.
$P(A) = \frac{1}{10}, P(B) = \frac{3}{5}, P(C) = \frac{1}{4}$.
$P(A) + P(B) + P(C) + P(D) = 1$
$\frac{1}{10} + \frac{3}{5} + \frac{1}{4} + P(D) = 1$
To add the fractions,find the least common multiple of $10, 5, 4$,which is $20$.
$\frac{2}{20} + \frac{12}{20} + \frac{5}{20} + P(D) = 1$
$\frac{19}{20} + P(D) = 1$
$P(D) = 1 - \frac{19}{20} = \frac{1}{20}$.
Since $\frac{1}{20} = 0.05$,which is not among the given options,the correct answer is $D$ (None of these).

(A) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
Let $E$ be the event that at least one die shows $6$.
The outcomes where at least one die shows $6$ are: $(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6)$.
The number of favorable outcomes is $11$.
Therefore,the probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{11}{36}$.

(A) The outcome of each coin toss is an independent event.
The appearance of a head on the $5^{th}$ toss does not depend on the outcomes of the first four tosses.
Therefore,the probability of getting a head on the $5^{th}$ toss is $P(\text{Head}) = \frac{1}{2}$.

(B) The last digit of a square depends only on the last digit of the integer.
Let the last digit of the integer be $x \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
The squares of these digits end in:
$0^2 \to 0$
$1^2 \to 1$
$2^2 \to 4$
$3^2 \to 9$
$4^2 \to 6$
$5^2 \to 5$
$6^2 \to 6$
$7^2 \to 9$
$8^2 \to 4$
$9^2 \to 1$
The last digit of the square is $1$ if the integer ends in $1$ or $9$.
The last digit of the square is $5$ if the integer ends in $5$.
Thus,there are $3$ favorable outcomes out of $10$ possible digits.
Required probability $= \frac{3}{10}$.

(B) Let the two integers be $x$ and $y$. Each integer can be either even $(E)$ or odd $(O)$.
There are $4$ possible outcomes for the pair $(x, y)$: $(E, E), (E, O), (O, E), (O, O)$.
Each outcome has a probability of $\frac{1}{4}$.
$1$. If both are even $(E, E)$,the product is even.
$2$. If one is even and one is odd $(E, O)$ or $(O, E)$,the product is even.
$3$. If both are odd $(O, O)$,the product is odd.
The outcomes resulting in an even product are $(E, E), (E, O), (O, E)$.
Therefore,the required probability is $\frac{3}{4}$.

(B) Let $P(K)$ be the probability of getting face $K$. Since $P(K) \propto K$,we have $P(K) = cK$ for some constant $c$.
Since the sum of all probabilities must be $1$,we have $\sum_{K=1}^{6} cK = 1$.
$c(1 + 2 + 3 + 4 + 5 + 6) = 1 \implies 21c = 1 \implies c = \frac{1}{21}$.
The outcomes that are even numbers are $2, 4, 6$.
The probability of getting an even number is $P(2) + P(4) + P(6) = c(2 + 4 + 6) = 12c$.
Substituting $c = \frac{1}{21}$,we get $\frac{12}{21} = \frac{4}{7}$.

(C) The sample space $S$ for throwing a die is $\{1, 2, 3, 4, 5, 6\}$.
Total number of outcomes $n(S) = 6$.
The even numbers on a die are $\{2, 4, 6\}$.
Number of favorable outcomes $n(E) = 3$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{3}{6} = \frac{1}{2}$.

(D) When three coins are tossed,the total number of possible outcomes is $2^3 = 8$.
The sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
Let $E$ be the event of getting at least one head.
The complement event $E'$ is the event of getting no heads,which means all tails.
$E' = \{TTT\}$,so $n(E') = 1$.
The probability of getting no heads is $P(E') = \frac{n(E')}{n(S)} = \frac{1}{8}$.
The probability of getting at least one head is $P(E) = 1 - P(E') = 1 - \frac{1}{8} = \frac{7}{8}$.

(C) Total number of outcomes when two dice are thrown is $6 \times 6 = 36$.
The favorable 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 required probability is $\frac{6}{36} = \frac{1}{6}$.

(C) The last digit of a product depends only on the last digits of the individual numbers.
For any single number,the possible last digits are $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$,so there are $10$ possibilities.
For the product of four numbers to have a last digit of $1, 3, 5$ or $7$,each of the four numbers must have a last digit from the set $\{1, 3, 5, 7\}$.
The probability that a single number has a last digit in $\{1, 3, 5, 7\}$ is $P = \frac{4}{10} = \frac{2}{5}$.
Since the four numbers are selected independently,the probability that all four have a last digit in $\{1, 3, 5, 7\}$ is $\left(\frac{2}{5}\right)^4 = \frac{16}{625}$.

(A) 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 $A$ is $P(A') = 1 - \frac{1}{2} = \frac{1}{2}$.
The probability that the problem is not solved by $B$ is $P(B') = 1 - \frac{1}{3} = \frac{2}{3}$.
The probability that the problem is not solved by $C$ is $P(C') = 1 - \frac{1}{4} = \frac{3}{4}$.
The probability that the problem is not solved by any of them is $P(\text{none}) = P(A') \times P(B') \times P(C') = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{6}{24} = \frac{1}{4}$.
The probability that the problem is solved is $P(\text{solved}) = 1 - P(\text{none}) = 1 - \frac{1}{4} = \frac{3}{4}$.

(B) Total number of outcomes when rolling $2$ dice is $6 \times 6 = 36$.
$A$ doublet occurs when both dice show the same number.
The favourable outcomes are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$.
Number of favourable outcomes $= 6$.
Probability $= \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}$.

(D) Total number of outcomes when throwing $2$ dice is $6 \times 6 = 36$.
For a sum of $7$,the favourable outcomes are $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$,which is $6$ outcomes.
For a sum of $12$,the only favourable outcome is $(6, 6)$,which is $1$ outcome.
Since these events are mutually exclusive,the total probability is $\frac{6}{36} + \frac{1}{36} = \frac{7}{36}$.

(B) leap year contains $366$ days,which is equal to $52$ weeks and $2$ extra days.
There are definitely $52$ Sundays in the $52$ full weeks.
For the remaining $2$ days,the possible combinations are:
$(i)$ Sunday and Monday,$(ii)$ Monday and Tuesday,$(iii)$ Tuesday and Wednesday,$(iv)$ Wednesday and Thursday,$(v)$ Thursday and Friday,$(vi)$ Friday and Saturday,$(vii)$ Saturday and Sunday.
There are $7$ total possible outcomes for the remaining $2$ days.
For the year to have $53$ Sundays,one of the two remaining days must be a Sunday.
This occurs in $2$ cases: $(i)$ Sunday and Monday,and $(vii)$ Saturday and Sunday.
Therefore,the required probability is $\frac{2}{7}$.

(C) Total number of pens in the first box = $4 + 2 = 6$.
Total number of pens in the second box = $3 + 5 = 8$.
The probability of selecting a white pen from the first box = $\frac{4}{6} = \frac{2}{3}$.
The probability of selecting a white pen from the second box = $\frac{3}{8}$.
Since the events are independent,the probability that both pens are white = $\frac{2}{3} \times \frac{3}{8} = \frac{6}{24} = \frac{1}{4}$.

(B) Let $A$ be the first bag and $B$ be the second bag.
For bag $A$: $P(\text{red}) = \frac{3}{8}$,$P(\text{black}) = \frac{5}{8}$.
For bag $B$: $P(\text{red}) = \frac{6}{10} = \frac{3}{5}$,$P(\text{black}) = \frac{4}{10} = \frac{2}{5}$.
The event that one ball is red and the other is black can happen in two ways:
$1$. Red from bag $A$ and black from bag $B$.
$2$. Black from bag $A$ and red from bag $B$.
Required probability $= (P(\text{red}_A) \times P(\text{black}_B)) + (P(\text{black}_A) \times P(\text{red}_B))$
$= (\frac{3}{8} \times \frac{4}{10}) + (\frac{5}{8} \times \frac{6}{10})$
$= \frac{12}{80} + \frac{30}{80} = \frac{42}{80} = \frac{21}{40}$.

(C) When four coins are tossed,the total number of outcomes is $2^4 = 16$.
The event of getting 'at least one head' is the complement of the event of getting 'no head'.
The probability of getting no head (i.e.,all tails) is $P(\text{no head}) = (1/2)^4 = 1/16$.
Therefore,the probability of getting at least one head is $1 - P(\text{no head}) = 1 - 1/16 = 15/16$.

(C) Let $P(X)$ be the probability that $X$ speaks the truth and $P(Y)$ be the probability that $Y$ speaks the truth.
Given $P(X) = 60\% = \frac{60}{100} = \frac{3}{5}$ and $P(Y) = 50\% = \frac{50}{100} = \frac{1}{2}$.
Then $P(\bar{X}) = 1 - \frac{3}{5} = \frac{2}{5}$ and $P(\bar{Y}) = 1 - \frac{1}{2} = \frac{1}{2}$.
They contradict each other if one speaks the truth and the other lies.
Required probability $= P(X) \cdot P(\bar{Y}) + P(\bar{X}) \cdot P(Y)$
$= (\frac{3}{5} \cdot \frac{1}{2}) + (\frac{2}{5} \cdot \frac{1}{2})$
$= \frac{3}{10} + \frac{2}{10} = \frac{5}{10} = \frac{1}{2}$.

(A) Let $A$ and $B$ be the events that the problem is solved by the first and second student,respectively.
Given $P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{3}$.
The probability that the problem is not solved by either student is $P(A^c \cap B^c) = P(A^c) \times P(B^c)$.
$P(A^c) = 1 - \frac{1}{2} = \frac{1}{2}$.
$P(B^c) = 1 - \frac{1}{3} = \frac{2}{3}$.
So,$P(A^c \cap B^c) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$.
The probability that the problem is solved is $P(A \cup B) = 1 - P(A^c \cap B^c) = 1 - \frac{1}{3} = \frac{2}{3}$.

(D) Total number of balls in the bag $= 3 + 7 = 10$.
Let $E$ be the event of drawing a ball that is either white or red.
Since all balls in the bag are either white or red,this is a sure event.
The number of favorable outcomes $= 3 + 7 = 10$.
The total number of possible outcomes $= 10$.
Therefore,the probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{10}{10} = 1$.

(B) Total number of white balls = $3$.
Total number of black balls = $5$.
Total number of balls = $3 + 5 = 8$.
The probability of drawing a black ball is given by the ratio of the number of black balls to the total number of balls.
Required probability = $\frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{5}{8}$.

(A) Total number of pencils $= 12 + 6 + 2 = 20$.
Number of good (non-defective) pencils $= 12$.
The probability that the chosen pencil is not defective is given by the ratio of the number of good pencils to the total number of pencils.
Required probability $= \frac{12}{20} = \frac{3}{5}$.

(A) Given $P(A \cup B) = \frac{5}{6}$,$P(A \cap B) = \frac{1}{3}$,and $P(\bar{A}) = \frac{1}{2}$.
Since $P(A) = 1 - P(\bar{A}) = 1 - \frac{1}{2} = \frac{1}{2}$.
Using the addition theorem: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$\frac{5}{6} = \frac{1}{2} + P(B) - \frac{1}{3}$.
$\frac{5}{6} = \frac{1}{6} + P(B) \Rightarrow P(B) = \frac{4}{6} = \frac{2}{3}$.
Now,check for independence: $P(A) \times P(B) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$.
Since $P(A \cap B) = P(A) \times P(B) = \frac{1}{3}$,the events $A$ and $B$ are independent.

(A) We know that the event $A$ can be expressed as the union of two disjoint events: $A = (A \cap B) \cup (A \cap \bar{B})$.
By the addition theorem of probability,since these events are disjoint,we have $P(A) = P(A \cap B) + P(A \cap \bar{B})$.
Rearranging the formula to solve for $P(A \cap \bar{B})$,we get $P(A \cap \bar{B}) = P(A) - P(A \cap B)$.
Substituting the given values: $P(A \cap \bar{B}) = 0.25 - 0.15 = 0.1$.

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

(C) Given that $P(A \cup B) = 0.6$ and $P(A \cap B) = 0.2$.
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since $P(A) = 1 - P(\bar{A})$ and $P(B) = 1 - P(\bar{B})$,we can substitute these into the addition theorem:
$P(A \cup B) = (1 - P(\bar{A})) + (1 - P(\bar{B})) - P(A \cap B)$
$0.6 = 2 - (P(\bar{A}) + P(\bar{B})) - 0.2$
$0.6 = 1.8 - (P(\bar{A}) + P(\bar{B}))$
$P(\bar{A}) + P(\bar{B}) = 1.8 - 0.6 = 1.2$.

(A) Let $P(A)$ be the probability of failing in Physics and $P(B)$ be the probability of failing in Mathematics.
$P(A) = \frac{20}{100} = 0.2$
$P(B) = \frac{10}{100} = 0.1$
Since 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,$P(A \cap B) = P(A) \times P(B) = 0.2 \times 0.1 = 0.02$.
Therefore,$P(A \cup B) = 0.2 + 0.1 - 0.02 = 0.28$.
Converting to percentage,$0.28 \times 100 = 28\%$.

(B) Let $P(A)$ be the probability that $A$ dies $= p$,so $P(A') = 1 - p$ is the probability that $A$ is alive.
Let $P(B)$ be the probability that $B$ dies $= q$,so $P(B') = 1 - q$ is the probability that $B$ is alive.
We want the probability that only one of them is alive at the end of the year.
This happens if ($A$ is alive and $B$ dies) or ($B$ is alive and $A$ dies).
Required probability $= P(A' \cap B) + P(B' \cap A)$
Since the events are independent,this is equal to $P(A')P(B) + P(B')P(A)$.
$= (1 - p)q + (1 - q)p$
$= q - pq + p - pq$
$= p + q - 2pq$.

(C) Let $P(A), P(B),$ and $P(C)$ be the probabilities of winning for athletes $A, B,$ and $C$ respectively.
Given that $P(A) = 2P(C)$ and $P(B) = 2P(C)$.
Since the sum of probabilities of all mutually exclusive outcomes is $1$,we have:
$P(A) + P(B) + P(C) = 1$
Substituting the values,we get:
$2P(C) + 2P(C) + P(C) = 1$
$5P(C) = 1 \Rightarrow P(C) = \frac{1}{5}$
Thus,$P(A) = \frac{2}{5}$ and $P(B) = \frac{2}{5}$.
The probability that the race is won by $A$ or $B$ is $P(A \cup B) = P(A) + P(B)$.
$P(A \cup B) = \frac{2}{5} + \frac{2}{5} = \frac{4}{5}$.

(B) By De Morgan's Law,$P(A' \cap B') = P((A \cup B)')$.
$P(A' \cap B') = 1 - P(A \cup B)$.
Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $P(A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{4}$.
Finding a common denominator of $12$: $P(A \cup B) = \frac{6}{12} + \frac{4}{12} - \frac{3}{12} = \frac{7}{12}$.
Therefore,$P(A' \cap B') = 1 - \frac{7}{12} = \frac{5}{12}$.