Odds in favor and odds against, Addition theorem on probability Questions in English

(C) We use the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $\frac{5}{6} = \frac{2}{3} + \frac{1}{2} - P(A \cap B)$.
Calculating the sum: $\frac{2}{3} + \frac{1}{2} = \frac{4+3}{6} = \frac{7}{6}$.
Thus,$\frac{5}{6} = \frac{7}{6} - P(A \cap B)$,which implies $P(A \cap B) = \frac{7}{6} - \frac{5}{6} = \frac{2}{6} = \frac{1}{3}$.
Since $P(A \cap B) \neq 0$,the events are not mutually exclusive.
Now,check for independence: $P(A) \times P(B) = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3}$.
Since $P(A \cap B) = P(A) \times P(B) = \frac{1}{3}$,the events $A$ and $B$ are independent.

(B) Two events $A$ and $B$ are said to be mutually exclusive if they cannot occur simultaneously,which means $A \cap B = \phi$.
By the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since $A$ and $B$ are mutually exclusive,$P(A \cap B) = 0$.
Therefore,$P(A \cup B) = P(A) + P(B)$.

(C) Total number of cards $n(S) = 52$.
Let $A$ be the event that the card is a king and $B$ be the event that the card is a diamond.
Number of kings $n(A) = 4$.
Number of diamonds $n(B) = 13$.
The king of diamonds is common to both,so $n(A \cap B) = 1$.
Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52}$.
Simplifying the fraction: $\frac{16}{52} = \frac{4}{13}$.

(B) leap year has $366$ days,which consists of $52$ weeks and $2$ extra days.
The possible combinations for these $2$ days are: (Sunday,Monday),(Monday,Tuesday),(Tuesday,Wednesday),(Wednesday,Thursday),(Thursday,Friday),(Friday,Saturday),and (Saturday,Sunday).
There are $7$ total possible outcomes.
Let $A$ be the event of having $53$ Fridays and $B$ be the event of having $53$ Saturdays.
$P(A) = \frac{2}{7}$ (since Friday occurs in (Thursday,Friday) and (Friday,Saturday)).
$P(B) = \frac{2}{7}$ (since Saturday occurs in (Friday,Saturday) and (Saturday,Sunday)).
$P(A \cap B) = \frac{1}{7}$ (since (Friday,Saturday) is the only common outcome).
Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{2}{7} + \frac{2}{7} - \frac{1}{7} = \frac{3}{7}$.

(C) Let $P(A)$ be the probability that $A$ speaks the truth and $P(B)$ be the probability that $B$ speaks the truth.
Given: $P(A) = \frac{4}{5}$ and $P(B) = \frac{3}{4}$.
Then,the probability that $A$ lies is $P(\bar{A}) = 1 - \frac{4}{5} = \frac{1}{5}$.
The probability that $B$ lies is $P(\bar{B}) = 1 - \frac{3}{4} = \frac{1}{4}$.
They contradict each other if ($A$ speaks the truth and $B$ lies) $OR$ ($A$ lies and $B$ speaks the truth).
Required probability $= P(A) \times P(\bar{B}) + P(\bar{A}) \times P(B)$.
$= (\frac{4}{5} \times \frac{1}{4}) + (\frac{1}{5} \times \frac{3}{4})$.
$= \frac{4}{20} + \frac{3}{20} = \frac{7}{20}$.

(C) Given that the odds against an event are $2 : 3$.
This means the number of unfavorable outcomes is $2$ and the number of favorable outcomes is $3$.
Total outcomes = $2 + 3 = 5$.
The probability of the occurrence of the event is given by the ratio of favorable outcomes to the total number of outcomes.
Probability $P(E) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{3}{5}$.

(D) Given the odds in favour of an event are $3 : 5$.
Let the number of favourable outcomes be $3x$ and the number of unfavourable outcomes be $5x$.
The total number of outcomes is $3x + 5x = 8x$.
The probability of occurrence of the event $P(E) = \frac{3x}{8x} = \frac{3}{8}$.
The probability of non-occurrence of the event $P(E') = 1 - P(E) = 1 - \frac{3}{8} = \frac{5}{8}$.

(C) Total number of cards = $52$.
Number of spades = $13$.
Number of aces = $4$.
Since one card is an ace of spades,the number of favorable outcomes (spade or ace) = $13 + 4 - 1 = 16$.
Probability of winning $P(E) = \frac{16}{52} = \frac{4}{13}$.
Probability of losing $P(E') = 1 - \frac{4}{13} = \frac{9}{13}$.
Odds in favor = $P(E) : P(E') = \frac{4}{13} : \frac{9}{13} = 4 : 9$.
Odds against winning = $P(E') : P(E) = 9 : 4$.

(C) The odds in favour of an event are given as $a : b = 4 : 5$.
The probability $P(E)$ of the event occurring is given by the formula $P(E) = \frac{a}{a + b}$.
Substituting the values,we get $P(E) = \frac{4}{4 + 5} = \frac{4}{9}$.
Therefore,the correct option is $C$.

(B) Given that the odds against an event are $a : b = 6 : 5$.
If the odds against an event are $a : b$,then the probability that the event does not occur is given by $P(E') = \frac{a}{a + b}$.
Here,$a = 6$ and $b = 5$.
Therefore,the probability that the event does not occur is $P(E') = \frac{6}{6 + 5} = \frac{6}{11}$.

(B) The odds in favour of the three horses are $1:2$,$1:3$,and $1:4$.
The probabilities of winning for the three horses are $P(A) = \frac{1}{1+2} = \frac{1}{3}$,$P(B) = \frac{1}{1+3} = \frac{1}{4}$,and $P(C) = \frac{1}{1+4} = \frac{1}{5}$.
Since these are mutually exclusive events (only one horse can win),the probability that one of the horses will win is the sum of their individual probabilities:
$P(A \cup B \cup C) = P(A) + P(B) + P(C) = \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$.
Taking the least common multiple of $3, 4, 5$,which is $60$:
$P = \frac{20 + 15 + 12}{60} = \frac{47}{60}$.

(B) Let $A$ and $B$ be two independent events.
The odds against $A$ are $5 : 2$,so the probability of $A$ occurring is $P(A) = \frac{2}{5 + 2} = \frac{2}{7}$.
Thus,$P(\bar{A}) = 1 - \frac{2}{7} = \frac{5}{7}$.
The odds in favour of $B$ are $6 : 5$,so the probability of $B$ occurring is $P(B) = \frac{6}{6 + 5} = \frac{6}{11}$.
Thus,$P(\bar{B}) = 1 - \frac{6}{11} = \frac{5}{11}$.
The probability that at least one of the events will happen is $P(A \cup B) = 1 - P(\bar{A}) \times P(\bar{B})$.
$P(A \cup B) = 1 - (\frac{5}{7} \times \frac{5}{11}) = 1 - \frac{25}{77} = \frac{77 - 25}{77} = \frac{52}{77}$.

(C) Let $A, B, C$ be the events that the three students solve the question respectively.
Given odds against are $2:1, 5:2, 5:3$.
Therefore,the probabilities of solving the question are:
$P(A) = \frac{1}{1+2} = \frac{1}{3}$,$P(\bar{A}) = \frac{2}{3}$
$P(B) = \frac{2}{2+5} = \frac{2}{7}$,$P(\bar{B}) = \frac{5}{7}$
$P(C) = \frac{3}{3+5} = \frac{3}{8}$,$P(\bar{C}) = \frac{5}{8}$
The probability that the question is solved by only one student is given by:
$P = P(A \cap \bar{B} \cap \bar{C}) + P(\bar{A} \cap B \cap \bar{C}) + P(\bar{A} \cap \bar{B} \cap C)$
$P = (\frac{1}{3} \times \frac{5}{7} \times \frac{5}{8}) + (\frac{2}{3} \times \frac{2}{7} \times \frac{5}{8}) + (\frac{2}{3} \times \frac{5}{7} \times \frac{3}{8})$
$P = \frac{25}{168} + \frac{20}{168} + \frac{30}{168}$
$P = \frac{75}{168} = \frac{25}{56}$.

(A) The probabilities of ships $A, B$ and $C$ arriving safely are calculated from the given odds in favor.
For ship $A$,the probability $P(A) = \frac{2}{2 + 5} = \frac{2}{7}$.
For ship $B$,the probability $P(B) = \frac{3}{3 + 7} = \frac{3}{10}$.
For ship $C$,the probability $P(C) = \frac{6}{6 + 11} = \frac{6}{17}$.
Since the events are independent,the probability of all ships arriving safely is $P(A \cap B \cap C) = P(A) \times P(B) \times P(C)$.
$P(A \cap B \cap C) = \frac{2}{7} \times \frac{3}{10} \times \frac{6}{17} = \frac{36}{1190} = \frac{18}{595}$.

(A) The total number of ways to arrange $23$ persons at a round table is $(23-1)! = 22!$.
To find the number of ways two specific persons sit together,we treat them as a single unit. Now we have $22$ units to arrange in a circle,which can be done in $(22-1)! = 21!$ ways. The two persons can be arranged among themselves in $2!$ ways.
So,the number of favorable ways for them to sit together is $21! \times 2!$.
The probability $P$ that they sit together is $\frac{21! \times 2!}{22!} = \frac{2}{22} = \frac{1}{11}$.
The probability that they do not sit together is $1 - \frac{1}{11} = \frac{10}{11}$.
The odds against them sitting together are given by the ratio of the probability of not sitting together to the probability of sitting together,which is $\frac{10}{11} : \frac{1}{11} = 10 : 1$.

(B) For mutually exclusive events $A, B$,and $C$,the sum of their probabilities must be less than or equal to $1$.
Calculate the sum: $P(A \cup B \cup C) = P(A) + P(B) + P(C) = \frac{2}{3} + \frac{1}{4} + \frac{1}{6}$.
Find a common denominator: $\frac{8}{12} + \frac{3}{12} + \frac{2}{12} = \frac{13}{12}$.
Since $\frac{13}{12} > 1$,the given probabilities cannot represent mutually exclusive events.
Therefore,the statement is wrong.

(C) We use the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Given values are $P(A) = 0.4$,$P(A \cup B) = 0.7$,and $P(A \cap B) = 0.2$.
Substituting these values into the formula:
$0.7 = 0.4 + P(B) - 0.2$
$0.7 = 0.2 + P(B)$
$P(B) = 0.7 - 0.2 = 0.5$.
Thus,the correct option is $C$.

(D) The probability of the union of two events $A$ and $B$ is given by the addition theorem of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Given that $P(A) = \frac{1}{4}$,$P(B) = \frac{1}{4}$,and $P(AB) = 0$:
$P(A \cup B) = \frac{1}{4} + \frac{1}{4} - 0$
$P(A \cup B) = \frac{2}{4} = \frac{1}{2} = 0.5$
Thus,the correct option is $D$.

(D) Total number of cards = $52$.
Let $A$ be the event of drawing a red card and $B$ be the event of drawing a queen.
Number of red cards $n(A) = 26$.
Number of queens $n(B) = 4$.
Number of red queens $n(A \cap B) = 2$.
Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{26}{52} + \frac{4}{52} - \frac{2}{52}$.
$P(A \cup B) = \frac{28}{52} = \frac{7}{13}$.

(B) Given,$P(X) = 0.3$ and $P(Y) = 0.2$.
Since the events are independent,the probability that both $X$ and $Y$ fail is $P(X \cap Y) = P(X) \times P(Y) = 0.3 \times 0.2 = 0.06$.
The probability that either $X$ or $Y$ fails is given by the addition theorem of probability:
$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$
Substituting the values:
$P(X \cup Y) = 0.3 + 0.2 - 0.06 = 0.44$.

(B) Given that events $A$ and $B$ are independent,we have $P(A \cap B) = P(A) \times P(B) = 0.4x$.
Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $0.7 = 0.4 + x - 0.4x$.
$0.7 - 0.4 = x(1 - 0.4)$.
$0.3 = 0.6x$.
$x = \frac{0.3}{0.6} = \frac{1}{2}$.

(A) Since events $A$ and $B$ are mutually exclusive,the probability of their intersection is $P(A \cap B) = 0$.
By the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values,we get $0.7 = 0.4 + x - 0$.
Therefore,$x = 0.7 - 0.4 = 0.3$.
Thus,$x = 3/10$.

(D) The sample space for tossing a coin twice is $S = \{(HH, HT, TH, TT)\}$.
Event $A$ (head on first toss) $= \{(HH, HT)\}$,so $P(A) = \frac{2}{4} = \frac{1}{2}$.
Event $B$ (head on second toss) $= \{(HH, TH)\}$,so $P(B) = \frac{2}{4} = \frac{1}{2}$.
Event $A \cap B$ (head on both tosses) $= \{(HH)\}$,so $P(A \cap B) = \frac{1}{4}$.
Using the addition theorem of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) = \frac{1}{2} + \frac{1}{2} - \frac{1}{4} = 1 - \frac{1}{4} = \frac{3}{4}$.

(C) For any two events $A$ and $B$,the addition theorem of probability states that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since $A$ and $B$ are mutually exclusive events,they cannot occur simultaneously,which means $A \cap B = \emptyset$.
Therefore,$P(A \cap B) = 0$.
Substituting this into the formula,we get $P(A \cup B) = P(A) + P(B)$.

(A) We use the addition theorem of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Substituting the given values:
$\frac{3}{4} = \frac{1}{4} + \frac{5}{8} - P(A \cap B)$
$\frac{6}{8} = \frac{2}{8} + \frac{5}{8} - P(A \cap B)$
$\frac{6}{8} = \frac{7}{8} - P(A \cap B)$
$P(A \cap B) = \frac{7}{8} - \frac{6}{8} = \frac{1}{8}$

(A) For any two events $A$ and $B$,the addition theorem of probability states that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since $A$ and $B$ are independent events,$P(A \cap B) = P(A)P(B)$.
Substituting this into the addition theorem,we get $P(A \cup B) = P(A) + P(B) - P(A)P(B)$.

(B) Let $A$ be the event that the chosen number is a multiple of $4$,and $B$ be the event that the chosen number is a multiple of $6$.
The number of multiples of $4$ in the first $100$ integers is $\frac{100}{4} = 25$. So,$P(A) = \frac{25}{100}$.
The number of multiples of $6$ in the first $100$ integers is $\lfloor \frac{100}{6} \rfloor = 16$. So,$P(B) = \frac{16}{100}$.
The multiples of both $4$ and $6$ are multiples of $\text{lcm}(4, 6) = 12$. The number of multiples of $12$ in the first $100$ integers is $\lfloor \frac{100}{12} \rfloor = 8$. So,$P(A \cap B) = \frac{8}{100}$.
Using the addition theorem of probability,the required probability is:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) = \frac{25}{100} + \frac{16}{100} - \frac{8}{100} = \frac{33}{100}$.

(B) Let $P(A)$ be the probability that horse $A$ wins and $P(B)$ be the probability that horse $B$ wins.
Given $P(A) = 1/4$ and $P(B) = 1/5$.
Since only one horse can win the race,the events $A$ and $B$ are mutually exclusive.
Therefore,the probability that either of them wins is given by $P(A \cup B) = P(A) + P(B)$.
$P(A \cup B) = 1/4 + 1/5 = (5 + 4) / 20 = 9/20$.
Thus,the correct option is $B$.

(C) Given that $P(A \cup B) = \frac{5}{6}$,$P(A \cap B) = \frac{1}{3}$,and $P(\bar{B}) = \frac{1}{3}$.
Since $P(B) = 1 - P(\bar{B})$,we have $P(B) = 1 - \frac{1}{3} = \frac{2}{3}$.
Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values: $\frac{5}{6} = P(A) + \frac{2}{3} - \frac{1}{3}$.
$\frac{5}{6} = P(A) + \frac{1}{3}$.
$P(A) = \frac{5}{6} - \frac{1}{3} = \frac{5-2}{6} = \frac{3}{6} = \frac{1}{2}$.

(A) We know that for any two events $A$ and $B$,the addition theorem of probability states that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Rearranging this,we get $P(A \cup B) + P(A \cap B) = P(A) + P(B)$.
Given that $P(A \cup B) + P(A \cap B) = \frac{7}{8}$ and $P(A) = 2P(B)$,which implies $P(B) = \frac{P(A)}{2}$.
Substituting these into the equation: $\frac{7}{8} = P(A) + \frac{P(A)}{2}$.
$\frac{7}{8} = \frac{3P(A)}{2}$.
$P(A) = \frac{7}{8} \times \frac{2}{3} = \frac{14}{24} = \frac{7}{12}$.

(B) Given that $A$ and $B$ are independent events,we have $P(A \cap B) = P(A) \times P(B)$.
Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Let $P(B) = x$. Substituting the given values:
$0.8 = 0.3 + x - (0.3 \times x)$
$0.8 - 0.3 = x - 0.3x$
$0.5 = 0.7x$
$x = \frac{0.5}{0.7} = \frac{5}{7}$.
Therefore,$P(B) = \frac{5}{7}$.

(D) The addition theorem of probability states that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Rearranging this gives $P(A \cap B) = P(A) + P(B) - P(A \cup B)$,which matches option $C$.
Since $0 \leq P(A \cup B) \leq 1$,we have $P(A \cup B) \leq 1 \implies P(A) + P(B) - P(A \cap B) \leq 1 \implies P(A \cap B) \geq P(A) + P(B) - 1$,which matches option $A$.
Also,since $P(A \cup B) \geq 0$,we have $P(A) + P(B) - P(A \cap B) \geq 0 \implies P(A \cap B) \leq P(A) + P(B)$,which matches option $B$.
Since options $A$,$B$,and $C$ are all mathematically correct,the correct choice is $D$.

(C) We know that the addition theorem of probability states that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Given the condition $P(A \cup B) = P(A \cap B)$,we substitute this into the formula:
$P(A \cap B) = P(A) + P(B) - P(A \cap B)$.
Rearranging the terms,we get $2P(A \cap B) = P(A) + P(B)$.
However,for the specific case where $P(A \cup B) = P(A \cap B)$,it implies that $P(A \setminus B) = 0$ and $P(B \setminus A) = 0$,which means $P(A) = P(A \cap B) = P(B)$.
Therefore,the correct relation is $P(A) = P(B)$.

(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 of Probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = 0.25 + 0.50 - 0.14 = 0.61$.
Therefore,$P(A^c \cap B^c) = 1 - 0.61 = 0.39$.

(A) The total number of outcomes is $n(S) = 12$.
Let $A$ be the event that the number is divisible by $2$. The numbers are ${2, 4, 6, 8, 10, 12}$,so $n(A) = 6$.
Let $B$ be the event that the number is divisible by $3$. The numbers are ${3, 6, 9, 12}$,so $n(B) = 4$.
The event $A \cap B$ is the set of numbers divisible by both $2$ and $3$ (i.e.,divisible by $6$). The numbers are ${6, 12}$,so $n(A \cap B) = 2$.
Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{6}{12} + \frac{4}{12} - \frac{2}{12} = \frac{8}{12} = \frac{2}{3}$.

(A) Let $A$ be the event that the husband will be alive in $20$ years and $B$ be the event that the wife will be alive in $20$ years.
Given $P(A) = \frac{3}{5}$ and $P(B) = \frac{2}{3}$.
Since $A$ and $B$ are independent events,the probability that at least one of them will be alive is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since $A$ and $B$ are independent,$P(A \cap B) = P(A) \times P(B) = \frac{3}{5} \times \frac{2}{3} = \frac{6}{15} = \frac{2}{5}$.
Therefore,$P(A \cup B) = \frac{3}{5} + \frac{2}{3} - \frac{6}{15} = \frac{9 + 10 - 6}{15} = \frac{13}{15}$.
Alternatively,the probability that at least one is alive is $1 - P(\text{both die}) = 1 - (1 - P(A))(1 - P(B)) = 1 - (\frac{2}{5} \times \frac{1}{3}) = 1 - \frac{2}{15} = \frac{13}{15}$.

(D) For two mutually exclusive events $A$ and $B$,the probability of $A$ or $B$ occurring is given by the addition theorem:
$P(A \cup B) = P(A) + P(B)$
Substituting the given values:
$P(A \cup B) = 0.45 + 0.35 = 0.8$
Thus,the correct option is $D$.

(C) For any two events $A$ and $B$,we have the addition theorem of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Since $P(A \cap B) \ge 0$,it follows that $P(A \cup B) \le P(A) + P(B)$.
By the principle of mathematical induction,this inequality can be generalized for $n$ events as:
$P\left( \bigcup_{i=1}^{n} A_i \right) \le \sum_{i=1}^{n} P(A_i)$.

(C) Let $A$ be the event of drawing a queen and $B$ be the event of drawing a heart.
Total number of cards = $52$.
Number of queens $n(A) = 4$,so $P(A) = \frac{4}{52} = \frac{1}{13}$.
Number of hearts $n(B) = 13$,so $P(B) = \frac{13}{52} = \frac{1}{4}$.
The card that is both a queen and a heart is the queen of hearts,so $n(A \cap B) = 1$,which means $P(A \cap B) = \frac{1}{52}$.
Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$.

(B) Let $A$ and $B$ be the two events.
Given $P(A) = 0.21$,$P(B) = 0.49$,and $P(A \cap B) = 0.16$.
Using the addition theorem of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) = 0.21 + 0.49 - 0.16 = 0.54$.
The probability that none of the two events occurs is given by $P(A^c \cap B^c) = 1 - P(A \cup B)$.
$P(A^c \cap B^c) = 1 - 0.54 = 0.46$.

(B) Given $P(A \cap B) = \frac{1}{6}$ and $P(A^c \cap B^c) = \frac{1}{3}$.
Since $P(A^c \cap B^c) = P((A \cup B)^c) = 1 - P(A \cup B) = \frac{1}{3}$,we have $P(A \cup B) = \frac{2}{3}$.
Using the addition theorem,$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{2}{3}$.
Thus,$P(A) + P(B) = \frac{2}{3} + \frac{1}{6} = \frac{5}{6}$.
Since $A$ and $B$ are independent,$P(A \cap B) = P(A)P(B) = \frac{1}{6}$.
Let $x = P(A)$ and $y = P(B)$. Then $x + y = \frac{5}{6}$ and $xy = \frac{1}{6}$.
These are roots of the quadratic equation $t^2 - (x+y)t + xy = 0$,which is $t^2 - \frac{5}{6}t + \frac{1}{6} = 0$.
Multiplying by $6$,we get $6t^2 - 5t + 1 = 0$.
$(2t - 1)(3t - 1) = 0$,so $t = \frac{1}{2}$ or $t = \frac{1}{3}$.
Therefore,$P(A)$ is $\frac{1}{2}$ or $\frac{1}{3}$.

(C) Let $A$ be the event of drawing a king and $B$ be the event of drawing a spade.
Total number of cards $= 52$.
Number of kings $n(A) = 4$,so $P(A) = \frac{4}{52} = \frac{1}{13}$.
Number of spades $n(B) = 13$,so $P(B) = \frac{13}{52} = \frac{1}{4}$.
The card that is both a king and a spade is the king of spades,so $n(A \cap B) = 1$,and $P(A \cap B) = \frac{1}{52}$.
Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$.

(B) Given $P(A^c) = 5/6$. Since $P(A) = 1 - P(A^c)$,we have $P(A) = 1 - 5/6 = 1/6$.
Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $5/6 = 1/6 + 2/3 - P(A \cap B)$.
$5/6 = (1+4)/6 - P(A \cap B) = 5/6 - P(A \cap B)$.
This implies $P(A \cap B) = 0$.
Since the probability of the intersection of the two events is $0$,the events $A$ and $B$ are mutually exclusive.

(B) The addition theorem for probability states that for any two events $A$ and $B$,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since the probability of any event is non-negative,we have $P(A \cap B) \ge 0$.
Therefore,$P(A \cup B) = P(A) + P(B) - P(A \cap B) \le P(A) + P(B)$.

(D) We know that $P(\bar{A}) = 1 - P(A)$ and $P(\bar{B}) = 1 - P(B).$
Therefore,$P(\bar{A}) + P(\bar{B}) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B)).$
Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B).$
Substituting the given values: $0.8 = P(A) + P(B) - 0.3.$
Thus,$P(A) + P(B) = 0.8 + 0.3 = 1.1.$
Finally,$P(\bar{A}) + P(\bar{B}) = 2 - 1.1 = 0.9.$

(C) Let $R$ be the event that a person is rich and $F$ be the event that a person is famous.
Given: $P(R) = 10\% = 0.1$,$P(F) = 5\% = 0.05$,and $P(R \cap F) = 3\% = 0.03$.
We need to find the probability that a person is either rich or famous but not both,which is given by the symmetric difference $P(R \Delta F) = P(R \cup F) - P(R \cap F)$.
Alternatively,$P(R \Delta F) = P(R) + P(F) - 2P(R \cap F)$.
Substituting the values:
$P(R \Delta F) = 0.1 + 0.05 - 2(0.03)$
$P(R \Delta F) = 0.15 - 0.06 = 0.09$.

(C) Total number of cards = $52$.
Let $A$ be the event of drawing a queen and $B$ be the event of drawing a heart.
Number of queens $n(A) = 4$.
Number of hearts $n(B) = 13$.
Number of cards that are both a queen and a heart $n(A \cap B) = 1$.
Using the addition theorem of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$.

(B) Let $A$ be the event that the first person is alive after $30$ years and $B$ be the event that the second person is alive after $30$ years.
Given the odds against $A$ are $8:5$,the probability that the first person dies is $P(\bar{A}) = \frac{8}{8+5} = \frac{8}{13}$.
Thus,the probability that the first person is alive is $P(A) = 1 - \frac{8}{13} = \frac{5}{13}$.
Given the odds against $B$ are $4:3$,the probability that the second person dies is $P(\bar{B}) = \frac{4}{4+3} = \frac{4}{7}$.
Thus,the probability that the second person is alive is $P(B) = 1 - \frac{4}{7} = \frac{3}{7}$.
The event that exactly one of them is alive is $(\bar{A} \cap B) \cup (A \cap \bar{B})$.
Since these events are independent,the probability is $P(\bar{A})P(B) + P(A)P(\bar{B})$.
$= (\frac{8}{13} \times \frac{3}{7}) + (\frac{5}{13} \times \frac{4}{7})$
$= \frac{24}{91} + \frac{20}{91} = \frac{44}{91}$.

(D) Let $A$ be the event that the student is selected in the $IIT$ entrance test and $B$ be the event that the student is selected in the Roorkee entrance test.
Given: $P(A) = 0.2$,$P(B) = 0.5$,and $P(A \cap B) = 0.3$.
We need to find the probability that he does not succeed at either place,which is $P(\overline{A} \cap \overline{B})$.
By De Morgan's Law,$P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.
Using the Addition Theorem: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = 0.2 + 0.5 - 0.3 = 0.4$.
Therefore,$P(\overline{A} \cap \overline{B}) = 1 - 0.4 = 0.6$.