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

(D) Let $p$ be the probability of the other event. Then the probability of the first event is $\frac{2}{3}p$.
Since one of the two events must occur and they are mutually exclusive,the sum of their probabilities is $1$:
$p + \frac{2}{3}p = 1$
$\frac{5}{3}p = 1 \Rightarrow p = \frac{3}{5}$.
The probability of the first event is $1 - \frac{3}{5} = \frac{2}{5}$.
The odds in favour of the other event are given by the ratio of its probability to the probability of its complement (the first event):
$\text{Odds} = \frac{p}{1-p} = \frac{3/5}{2/5} = \frac{3}{2}$.
Thus,the odds in favour of the other event are $3:2$.

(C) leap year consists of $366$ days,which is equal to $52$ weeks and $2$ extra days.
The $7$ possible pairs 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 contains $53$ Sundays.
Let $B$ be the event that the leap year contains $53$ Mondays.
From the sample space,the favorable outcomes for $A$ are $(i)$ and $(vii)$,so $P(A) = \frac{2}{7}$.
The favorable outcomes for $B$ are $(i)$ and $(ii)$,so $P(B) = \frac{2}{7}$.
The intersection $A \cap B$ (containing both $53$ Sundays and $53$ Mondays) occurs only in case $(i)$,so $P(A \cap B) = \frac{1}{7}$.
Using the addition theorem of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) = \frac{2}{7} + \frac{2}{7} - \frac{1}{7} = \frac{3}{7}$.

(A) Let $p_1$ and $p_2$ be the probabilities of the first and second events respectively.
According to the given conditions:
$p_1 = p_2^2$
The odds against an event with probability $p$ are given by $\frac{1-p}{p}$.
Therefore,$\frac{1-p_1}{p_1} = \left(\frac{1-p_2}{p_2}\right)^3$.
Substituting $p_1 = p_2^2$:
$\frac{1-p_2^2}{p_2^2} = \frac{(1-p_2)^3}{p_2^3}$
$\frac{(1-p_2)(1+p_2)}{p_2^2} = \frac{(1-p_2)^3}{p_2^3}$
Assuming $p_2 \neq 1$ and $p_2 \neq 0$,we divide by $\frac{1-p_2}{p_2^2}$:
$1+p_2 = \frac{1-p_2}{p_2}$
$p_2 + p_2^2 = 1 - p_2$
$p_2^2 + 2p_2 - 1 = 0$
Solving for $p_2$ using the quadratic formula:
$p_2 = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}$.
Since $0 \le p_2 \le 1$,we take $p_2 = \sqrt{2} - 1$.
This does not match the provided options. Re-evaluating the problem statement: if the odds against the first are the cube of the odds against the second,and $p_1 = 1/9, p_2 = 1/3$:
Odds against $p_1 = \frac{1-1/9}{1/9} = 8$.
Odds against $p_2 = \frac{1-1/3}{1/3} = 2$.
Since $8 = 2^3$,the condition holds. Thus,the correct option is $A$.

(B) Let $A$ and $B$ be the two independent events.
For event $A$,the odds against are $5 : 2$,so the probability of $A$ occurring is $P(A) = \frac{2}{5+2} = \frac{2}{7}$.
The probability of $A$ not occurring is $P(A') = 1 - \frac{2}{7} = \frac{5}{7}$.
For event $B$,the odds in favor are $6 : 5$,so the probability of $B$ occurring is $P(B) = \frac{6}{6+5} = \frac{6}{11}$.
The probability of $B$ not occurring is $P(B') = 1 - \frac{6}{11} = \frac{5}{11}$.
The probability that at least one event occurs is $1 - P(A' \cap B')$.
Since $A$ and $B$ are independent,$P(A' \cap B') = P(A') \times P(B') = \frac{5}{7} \times \frac{5}{11} = \frac{25}{77}$.
Therefore,the required probability is $1 - \frac{25}{77} = \frac{77-25}{77} = \frac{52}{77}$.

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

(C) According to the addition theorem of probability,for any two events $A$ and $B$,the probability of the union of the events is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

(D) The total number of cards is $52$.
There are $4$ kings in a deck of cards.
The number of favorable outcomes (drawing a king) is $m = 4$.
The number of unfavorable outcomes (not drawing a king) is $n = 52 - 4 = 48$.
The favorable odds in favor of an event are given by the ratio of favorable outcomes to unfavorable outcomes,which is $m : n$.
Therefore,the favorable odds are $4 : 48$,which simplifies to $1 : 12$.

(C) The month of November has $30$ days. The total number of outcomes is $n(S) = 30$.
Let $A$ be the event that the date is a multiple of $5$. The multiples of $5$ in ${1, 2, ..., 30}$ are ${5, 10, 15, 20, 25, 30}$. So,$n(A) = 6$.
Let $B$ be the event that the date is a multiple of $6$. The multiples of $6$ in ${1, 2, ..., 30}$ are ${6, 12, 18, 24, 30}$. So,$n(B) = 5$.
The intersection $A \cap B$ contains dates that are multiples of both $5$ and $6$ (i.e.,multiples of $30$). Thus,$A \cap B = {30}$,and $n(A \cap B) = 1$.
Using the addition rule for probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{6}{30} + \frac{5}{30} - \frac{1}{30} = \frac{10}{30} = \frac{1}{3}$.

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

(D) Given: $P(A) = 0.25$,$P(B) = 0.50$,and $P(A \cap B) = 0.12$.
First,we find the probability of the union of the two events using the addition theorem:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) = 0.25 + 0.50 - 0.12 = 0.63$.
We need to find the probability that neither event occurs,which is $P(A' \cap B')$.
By De Morgan's Law,$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$.
Therefore,$P(A' \cap B') = 1 - 0.63 = 0.37$.

(D) The probability of an event occurring is given as $P(E) = \frac{3}{8}$.
We know that the sum of the probability of an event occurring and not occurring is $1$,i.e.,$P(E) + P(\text{not } E) = 1$.
Therefore,the probability of the event not occurring is $P(\text{not } E) = 1 - P(E)$.
$P(\text{not } E) = 1 - \frac{3}{8} = \frac{8-3}{8} = \frac{5}{8}$.

(D) Let $P(A) = 1/3$ and $P(B) = 1/4$ be the probabilities of $A$ and $B$ solving the problem.
Since the events are independent,the probability that the problem is solved is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since $A$ and $B$ are independent,$P(A \cap B) = P(A) \times P(B) = (1/3) \times (1/4) = 1/12$.
Therefore,$P(A \cup B) = 1/3 + 1/4 - 1/12 = (4 + 3 - 1)/12 = 6/12 = 1/2$.
Statement $- I$ claims the probability is $7/12$,which is incorrect.
Statement $- II$ states that the events are independent,which is a standard assumption for such problems,making Statement $- II$ true.
Thus,Statement $- I$ is false and Statement $- II$ is true.

(C) Let $P(A) = 1/5$ be the probability of $A$ failing and $P(B) = 3/10$ be the probability of $B$ failing.
Assuming $A$ and $B$ are independent events,the probability that either $A$ or $B$ fails (but not both) is given by $P(A \cup B) - P(A \cap B)$.
Alternatively,the probability that exactly one of them fails is $P(A \cap \overline{B}) + P(\overline{A} \cap B)$.
$P(A \cap \overline{B}) = P(A) \times (1 - P(B)) = \frac{1}{5} \times (1 - \frac{3}{10}) = \frac{1}{5} \times \frac{7}{10} = \frac{7}{50}$.
$P(\overline{A} \cap B) = (1 - P(A)) \times P(B) = (1 - \frac{1}{5}) \times \frac{3}{10} = \frac{4}{5} \times \frac{3}{10} = \frac{12}{50}$.
Total probability $= \frac{7}{50} + \frac{12}{50} = \frac{19}{50}$.

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

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

(D) Let $A$ be the event that the card is red and $B$ be the event that the card is a queen.
Total number of cards $= 52$.
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: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{26}{52} + \frac{4}{52} - \frac{2}{52} = \frac{28}{52}$.
Simplifying the fraction,we get $\frac{28}{52} = \frac{7}{13}$.

(C) Let $A$ be the event that the number on the ticket is a multiple of $3$.
Let $B$ be the event that the number on the ticket is a multiple of $5$.
The multiples of $3$ up to $100$ are $A = \{3, 6, 9, \dots, 99\}$,so $n(A) = 33$.
The multiples of $5$ up to $100$ are $B = \{5, 10, 15, \dots, 100\}$,so $n(B) = 20$.
The multiples of both $3$ and $5$ (i.e.,multiples of $15$) are $A \cap B = \{15, 30, 45, 60, 75, 90\}$,so $n(A \cap B) = 6$.
Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{33}{100} + \frac{20}{100} - \frac{6}{100} = \frac{47}{100}$.

(C) Let $P(A)$,$P(B)$,and $P(C)$ be the probabilities of passing in the first,second,and third grade respectively.
Given $P(A) = 1/10$,$P(B) = 3/5$,and $P(C) = 1/4$.
Since these events are mutually exclusive,the probability of passing in any of these grades is $P(A \cup B \cup C) = P(A) + P(B) + P(C)$.
$P(A \cup B \cup C) = 1/10 + 3/5 + 1/4 = (2 + 12 + 5) / 20 = 19/20$.
The probability of failing (getting a fourth grade) is $P(D) = 1 - P(A \cup B \cup C)$.
$P(D) = 1 - 19/20 = 1/20$.
Converting to decimal,$1/20 = 0.05 = 5/100 = 1/20$.

(A) The total number of outcomes when two dice are thrown is $n(S) = 6 \times 6 = 36$.
The favorable outcomes for getting a sum of $7$ are $E = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\}$.
The number of favorable outcomes is $n(E) = 6$.
The number of unfavorable outcomes is $n(S) - n(E) = 36 - 6 = 30$.
The odds against the event are given by the ratio of unfavorable outcomes to favorable outcomes: $30 : 6 = 5 : 1$.

(A) Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $\frac{2}{3} = \frac{1}{3} + P(B) - \frac{1}{6}$.
$\frac{2}{3} = \frac{1}{6} + P(B) \implies P(B) = \frac{2}{3} - \frac{1}{6} = \frac{4-1}{6} = \frac{3}{6} = \frac{1}{2}$.
Now,check for independence: $P(A) \times P(B) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$.
Since $P(A \cap B) = P(A) \times P(B) = \frac{1}{6}$,the events $A$ and $B$ are independent.

(A) Given that the probability of at least one event occurring is $P(A \cup B) = 0.6$ and the probability that both events occur is $P(A \cap B) = 0.2$.
Using the addition theorem of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Substituting the given values:
$0.6 = P(A) + P(B) - 0.2$
Therefore,$P(A) + P(B) = 0.6 + 0.2 = 0.8$.

(A) Let $P(A)$ be the probability that $A$ speaks the truth,so $P(A) = 75/100 = 3/4$.
Then $P(A') = 1 - 3/4 = 1/4$ is the probability that $A$ lies.
Let $P(B)$ be the probability that $B$ speaks the truth,so $P(B) = 80/100 = 4/5$.
Then $P(B') = 1 - 4/5 = 1/5$ is the probability that $B$ lies.
They contradict each other if one speaks the truth and the other lies.
This can happen in two ways:
$1$. $A$ speaks the truth and $B$ lies: $P(A) \times P(B') = (3/4) \times (1/5) = 3/20$.
$2$. $A$ lies and $B$ speaks the truth: $P(A') \times P(B) = (1/4) \times (4/5) = 4/20$.
The total probability of contradiction is the sum of these probabilities:
$P(\text{contradiction}) = 3/20 + 4/20 = 7/20$.

(C) Let $P(A)$ be the probability that $A$ solves the problem and $P(B)$ be the probability that $B$ solves the problem.
Given odds against $A$ are $4:3$,so $P(A') = 4/7$ and $P(A) = 1 - 4/7 = 3/7$.
Given odds in favor of $B$ are $7:5$,so $P(B) = 7/(7+5) = 7/12$ and $P(B') = 1 - 7/12 = 5/12$.
The probability that only one of them solves the problem is $P(A \cap B') + P(A' \cap B)$.
$P(A \cap B') = P(A) \times P(B') = (3/7) \times (5/12) = 15/84$.
$P(A' \cap B) = P(A') \times P(B) = (4/7) \times (7/12) = 28/84$.
Total probability = $15/84 + 28/84 = 43/84$.

(B) By De Morgan's Law,$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$.
Using the Addition Theorem of Probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $P(A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{6+4-3}{12} = \frac{7}{12}$.
Therefore,$P(A' \cap B') = 1 - \frac{7}{12} = \frac{5}{12}$.

(C) The sample space of throwing a die is $S = \{1, 2, 3, 4, 5, 6\}$,so $n(S) = 6$.
The event $A$ is that the number obtained is greater than $3$,so $A = \{4, 5, 6\}$. Thus,$P(A) = \frac{3}{6}$.
The event $B$ is that the number obtained is less than $5$,so $B = \{1, 2, 3, 4\}$. Thus,$P(B) = \frac{4}{6}$.
The intersection $A \cap B$ is the event that the number is both greater than $3$ and less than $5$,so $A \cap B = \{4\}$. Thus,$P(A \cap B) = \frac{1}{6}$.
Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values: $P(A \cup B) = \frac{3}{6} + \frac{4}{6} - \frac{1}{6} = \frac{6}{6} = 1$.

(C) Let $A$ be the event that the sum is less than $7$ and $B$ be the event that the sum is odd.
Total outcomes when two dice are thrown $= 6 \times 6 = 36$.
Outcomes for $A$ (sum $< 7$): $(1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2), (5,1)$. Total $= 15$.
$P(A) = \frac{15}{36}$.
Outcomes for $B$ (sum is odd): Sums can be $3, 5, 7, 9, 11$. Total $= 18$.
$P(B) = \frac{18}{36}$.
Outcomes for $A \cap B$ (sum is odd $AND$ less than $7$): Sums can be $3, 5$. Outcomes are $(1,2), (2,1), (1,4), (2,3), (3,2), (4,1)$. Total $= 6$.
$P(A \cap B) = \frac{6}{36}$.
Using the addition theorem: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{15}{36} + \frac{18}{36} - \frac{6}{36} = \frac{27}{36} = \frac{3}{4}$.

(C) Let $H$ be the event that the husband lives for $20$ years and $W$ be the event that the wife lives for $20$ years.
Given the odds against the husband living for $20$ years are $5:2$,the probability that he does not live is $P(\overline{H}) = \frac{5}{5+2} = \frac{5}{7}$.
Thus,the probability that he lives is $P(H) = 1 - \frac{5}{7} = \frac{2}{7}$.
Given the odds against the wife living for $20$ years are $4:3$,the probability that she does not live is $P(\overline{W}) = \frac{4}{4+3} = \frac{4}{7}$.
Thus,the probability that she lives is $P(W) = 1 - \frac{4}{7} = \frac{3}{7}$.
The probability that at least one of them will be alive is $k = 1 - P(\overline{H} \cap \overline{W})$.
Since the events are independent,$k = 1 - P(\overline{H}) \times P(\overline{W})$.
$k = 1 - (\frac{5}{7} \times \frac{4}{7}) = 1 - \frac{20}{49} = \frac{29}{49}$.
Therefore,$49k = 49 \times \frac{29}{49} = 29$.

(A) Given $P(A) = 0.3$ and $P(A \cup B) = 0.8$.
Since $A$ and $B$ are independent,$P(A \cap B) = P(A)P(B)$.
Using the addition theorem: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$0.8 = 0.3 + P(B) - 0.3P(B)$.
$0.5 = 0.7P(B) \Rightarrow P(B) = \frac{0.5}{0.7} = \frac{5}{7}$.
The conditional statement $A \to B$ is equivalent to $\neg A \lor B$,so $P(A \to B) = P(\neg A \cup B) = 1 - P(A \cap \neg B)$.
Since $A$ and $B$ are independent,$A$ and $\neg B$ are also independent.
$P(A \cap \neg B) = P(A)P(\neg B) = 0.3 \times (1 - \frac{5}{7}) = 0.3 \times \frac{2}{7} = \frac{3}{10} \times \frac{2}{7} = \frac{6}{70} = \frac{3}{35}$.
Therefore,$P(A \to B) = 1 - \frac{3}{35} = \frac{32}{35}$.

(A) We are given the probabilities $P(A)=\frac{6}{11}, P(B)=\frac{5}{11}$ and $P(A \cup B)=\frac{7}{11}.$
Using the addition theorem of probability,we know that:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Substituting the given values into the formula:
$\frac{7}{11} = \frac{6}{11} + \frac{5}{11} - P(A \cap B)$
$\frac{7}{11} = \frac{11}{11} - P(A \cap B)$
$P(A \cap B) = 1 - \frac{7}{11}$
$P(A \cap B) = \frac{11-7}{11} = \frac{4}{11}$

(A) We know that the probability of occurrence of at least one of $A$ and $B$ is $P(A \cup B)$.
By the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since $A$ and $B$ are independent events,$P(A \cap B) = P(A) P(B)$.
Thus,$P(A \cup B) = P(A) + P(B) - P(A) P(B)$.
We can write $P(A) = 1 - P(A')$,so:
$P(A \cup B) = 1 - P(A') + P(B) - (1 - P(A')) P(B)$
$= 1 - P(A') + P(B) - P(B) + P(A') P(B)$
$= 1 - P(A') + P(A') P(B)$
$= 1 - P(A') [1 - P(B)]$
$= 1 - P(A') P(B')$.
Therefore,the statement is true.

(A) It is given that $P(A) = \frac{1}{2}$,$P(A \cup B) = \frac{3}{5}$,and $P(B) = p$.
When events $A$ and $B$ are mutually exclusive,$A \cap B = \phi$.
Therefore,$P(A \cap B) = 0$.
Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $\frac{3}{5} = \frac{1}{2} + p - 0$.
Solving for $p$: $p = \frac{3}{5} - \frac{1}{2} = \frac{6-5}{10} = \frac{1}{10}$.

(A) Since events $A$ and $B$ are independent,we have $P(A \cap B) = P(A) \cdot P(B) = \frac{1}{2} p$.
Using the addition theorem of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Substituting the given values:
$\frac{3}{5} = \frac{1}{2} + p - \frac{1}{2} p$
$\frac{3}{5} = \frac{1}{2} + \frac{p}{2}$
Subtracting $\frac{1}{2}$ from both sides:
$\frac{p}{2} = \frac{3}{5} - \frac{1}{2} = \frac{6-5}{10} = \frac{1}{10}$
Multiplying by $2$:
$p = \frac{2}{10} = \frac{1}{5}$

(A) Given that $A$ and $B$ are independent events,we have $P(A \cap B) = P(A) \times P(B)$.
Substituting the given values: $P(A \cap B) = 0.3 \times 0.4 = 0.12$.
Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values: $P(A \cup B) = 0.3 + 0.4 - 0.12 = 0.58$.

(A) Given that $A$ and $B$ are independent events,$P(A) = 0.3$ and $P(B) = 0.6$.
For independent events,the probability of the intersection is $P(A \cap B) = P(A) \times P(B)$.
$P(A \cap B) = 0.3 \times 0.6 = 0.18$.
The probability of $A$ or $B$ is given by the addition theorem: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = 0.3 + 0.6 - 0.18$.
$P(A \cup B) = 0.9 - 0.18 = 0.72$.

(A) Since $A$ and $B$ are independent events,$P(A \cap B) = P(A) \times P(B) = 0.3 \times 0.6 = 0.18$.
We need to find $P(\text{neither } A \text{ nor } B)$,which is $P(A' \cap B')$.
By De Morgan's Law,$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$.
Using the addition theorem,$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.6 - 0.18 = 0.72$.
Therefore,$P(A' \cap B') = 1 - 0.72 = 0.28$.

(A) Given:
$P(A) = \frac{1}{3}$,$P(B) = \frac{1}{5}$,$P(A \cap B) = \frac{1}{15}$
We know the addition theorem of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Substituting the values:
$P(A \cup B) = \frac{1}{3} + \frac{1}{5} - \frac{1}{15}$
Taking the least common multiple $(LCM)$ of $3, 5, 15$,which is $15$:
$P(A \cup B) = \frac{5 + 3 - 1}{15} = \frac{7}{15}$

(A) We use the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Given values are $P(A) = 0.35$,$P(A \cap B) = 0.25$,and $P(A \cup B) = 0.6$.
Substituting these values into the formula:
$0.6 = 0.35 + P(B) - 0.25$
Simplify the equation:
$0.6 = 0.1 + P(B)$
Solving for $P(B)$:
$P(B) = 0.6 - 0.1 = 0.5$.

(A) Given that $P(A) = 0.5$,$P(B) = 0.35$,and $P(A \cup B) = 0.7$.
We use the addition theorem of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Substituting the given values:
$0.7 = 0.5 + 0.35 - P(A \cap B)$
$0.7 = 0.85 - P(A \cap B)$
$P(A \cap B) = 0.85 - 0.7$
$P(A \cap B) = 0.15$

(A) Given that $P(A) = \frac{3}{5}$ and $P(B) = \frac{1}{5}$.
Since $A$ and $B$ are mutually exclusive events,the probability of $A$ or $B$ is given by the addition theorem:
$P(A \cup B) = P(A) + P(B)$.
Substituting the given values:
$P(A \cup B) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$.

(B) Given: $P(E) = \frac{1}{4}$,$P(F) = \frac{1}{2}$,and $P(E \cap F) = \frac{1}{8}$.
We use the Addition Theorem of Probability:
$P(E \cup F) = P(E) + P(F) - P(E \cap F)$
Substituting the values:
$P(E \cup F) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8}$
Finding a common denominator $(8)$:
$P(E \cup F) = \frac{2}{8} + \frac{4}{8} - \frac{1}{8} = \frac{2 + 4 - 1}{8} = \frac{5}{8}$.

(A) Given: $P(E) = \frac{1}{4}$,$P(F) = \frac{1}{2}$,and $P(E \cap F) = \frac{1}{8}$.
Using the addition theorem of probability,$P(E \cup F) = P(E) + P(F) - P(E \cap F)$.
$P(E \cup F) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2+4-1}{8} = \frac{5}{8}$.
We need to find $P(\text{not } E \text{ and not } F)$,which is $P(E' \cap F')$.
By De Morgan's Law,$E' \cap F' = (E \cup F)'$.
Therefore,$P(E' \cap F') = P((E \cup F)') = 1 - P(E \cup F)$.
$P(E' \cap F') = 1 - \frac{5}{8} = \frac{3}{8}$.

(A) Given: $P(A)=0.42$,$P(B)=0.48$,and $P(A \cap B)=0.16$.
We use the addition theorem of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values:
$P(A \cup B) = 0.42 + 0.48 - 0.16$.
$P(A \cup B) = 0.90 - 0.16 = 0.74$.

(A) Let $A$ be the event that the selected student studies Mathematics and $B$ be the event that the selected student studies Biology.
Given:
$P(A) = 40\% = \frac{40}{100} = 0.4$
$P(B) = 30\% = \frac{30}{100} = 0.3$
$P(A \cap B) = 10\% = \frac{10}{100} = 0.1$
We need to find the probability that the student studies Mathematics or Biology,which is $P(A \cup B)$.
Using the addition theorem of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) = 0.4 + 0.3 - 0.1$
$P(A \cup B) = 0.6$
Thus,the probability that the selected student studies Mathematics or Biology is $0.6$.

(A) Let $A$ and $B$ be the events of passing the first and second examinations,respectively.
Given: $P(A) = 0.8$,$P(B) = 0.7$,and $P(A \cup B) = 0.95$.
We know the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $0.95 = 0.8 + 0.7 - P(A \cap B)$.
$0.95 = 1.5 - P(A \cap B)$.
$P(A \cap B) = 1.5 - 0.95 = 0.55$.
Thus,the probability of passing both examinations is $0.55$.

Consider $E = B \cup C$ so that
$P(A \cup B \cup C) = P(A \cup E)$
$= P(A) + P(E) - P(A \cap E)$ ...... $(1)$
Now
$P(E) = P(B \cup C)$
$= P(B) + P(C) - P(B \cap C)$ ......... $(2)$
Also $A \cap E = A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ [using the distributive property of intersection over union].
Thus,$P(A \cap E) = P(A \cap B) + P(A \cap C) - P[(A \cap B) \cap (A \cap C)]$
$= P(A \cap B) + P(A \cap C) - P(A \cap B \cap C)$ ......... $(3)$
Using $(2)$ and $(3)$ in $(1)$,we get
$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(B \cap C) - P(A \cap B) - P(A \cap C) + P(A \cap B \cap C)$.

(A) Given: $P(A) = 0.54$,$P(B) = 0.69$,and $P(A \cap B) = 0.35$.
We use the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $P(A \cup B) = 0.54 + 0.69 - 0.35$.
$P(A \cup B) = 1.23 - 0.35 = 0.88$.

(A) Given: $P(A)=0.54$,$P(B)=0.69$,and $P(A \cap B)=0.35$.
By the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = 0.54 + 0.69 - 0.35 = 0.88$.
By De Morgan's Law,$A' \cap B' = (A \cup B)'$.
Therefore,$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$.
$P(A' \cap B') = 1 - 0.88 = 0.12$.

(D) Let $E$ be the event that the spokesperson is a male and $F$ be the event that the spokesperson is over $35$ years of age.
The total number of persons is $5$.
The males are Harish,Rohan,and Salim. So,$P(E) = \frac{3}{5}$.
The persons over $35$ years are Sheetal $(46)$ and Salim $(41)$. So,$P(F) = \frac{2}{5}$.
The person who is both male and over $35$ years is Salim. So,$P(E \cap F) = \frac{1}{5}$.
Using the addition theorem of probability,$P(E \cup F) = P(E) + P(F) - P(E \cap F)$.
$P(E \cup F) = \frac{3}{5} + \frac{2}{5} - \frac{1}{5} = \frac{4}{5}$.
Thus,the probability that the spokesperson is either male or over $35$ years is $\frac{4}{5}$.

(A) The sample space $S$ for a single toss of a fair die is $\{1, 2, 3, 4, 5, 6\}$,so the total number of outcomes is $n(S) = 6$.
Let $E$ be the event that the number $4$ or $5$ turns up. Then $E = \{4, 5\}$,so $n(E) = 2$.
The probability of event $E$ is $P(E) = \frac{n(E)}{n(S)} = \frac{2}{6} = \frac{1}{3}$.
The probability of the complement event $E^c$ (the event that $4$ or $5$ does not turn up) is $P(E^c) = 1 - P(E) = 1 - \frac{1}{3} = \frac{2}{3}$.
The odds against an event $E$ are given by the ratio $P(E^c) : P(E)$.
Thus,the odds against $E$ are $\frac{2}{3} : \frac{1}{3} = 2 : 1$.

(D) Let $A, B, C$ be the events that the first,second,and third critics favor the book respectively.
The probabilities are:
$P(A) = \frac{2}{2+5} = \frac{2}{7}, P(A') = \frac{5}{7}$
$P(B) = \frac{3}{3+4} = \frac{3}{7}, P(B') = \frac{4}{7}$
$P(C) = \frac{4}{4+3} = \frac{4}{7}, P(C') = \frac{3}{7}$
The majority is in favor if at least two critics favor the book.
The probability is $P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A' \cap B \cap C) + P(A \cap B \cap C)$.
$= (\frac{2}{7} \times \frac{3}{7} \times \frac{3}{7}) + (\frac{2}{7} \times \frac{4}{7} \times \frac{4}{7}) + (\frac{5}{7} \times \frac{3}{7} \times \frac{4}{7}) + (\frac{2}{7} \times \frac{3}{7} \times \frac{4}{7})$
$= \frac{18}{343} + \frac{32}{343} + \frac{60}{343} + \frac{24}{343} = \frac{134}{343}$.