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

(A) The probability that ship $A$ reaches safely is $P(A) = \frac{2}{2+5} = \frac{2}{7}$.
The probability that ship $B$ reaches safely is $P(B) = \frac{3}{3+7} = \frac{3}{10}$.
The probability that ship $C$ reaches safely is $P(C) = \frac{6}{6+11} = \frac{6}{17}$.
Since the events are independent,the probability that all of them arrive safely is $P(A \cap B \cap C) = P(A) \times P(B) \times P(C)$.
$P(A \cap B \cap C) = \frac{2}{7} \times \frac{3}{10} \times \frac{6}{17} = \frac{36}{1190} = \frac{18}{595}$.

(C) Let $P(C_1), P(C_2), P(C_3)$ be the probabilities that the three critics are in favour of the book.
Given the odds in favour are $(5: 2), (4: 3), (3: 4)$,we have:
$P(C_1) = \frac{5}{5+2} = \frac{5}{7}, P(\bar{C}_1) = \frac{2}{7}$
$P(C_2) = \frac{4}{4+3} = \frac{4}{7}, P(\bar{C}_2) = \frac{3}{7}$
$P(C_3) = \frac{3}{3+4} = \frac{3}{7}, P(\bar{C}_3) = \frac{4}{7}$
For the majority to be in favour,at least two critics must be in favour.
Required probability $= P(C_1)P(C_2)P(\bar{C}_3) + P(C_1)P(\bar{C}_2)P(C_3) + P(\bar{C}_1)P(C_2)P(C_3) + P(C_1)P(C_2)P(C_3)$
$= (\frac{5}{7} \times \frac{4}{7} \times \frac{4}{7}) + (\frac{5}{7} \times \frac{3}{7} \times \frac{3}{7}) + (\frac{2}{7} \times \frac{4}{7} \times \frac{3}{7}) + (\frac{5}{7} \times \frac{4}{7} \times \frac{3}{7})$
$= \frac{80}{343} + \frac{45}{343} + \frac{24}{343} + \frac{60}{343} = \frac{209}{343}$

(A) Given that $A, B, C$ are mutually exclusive and exhaustive events,we have $P(A) + P(B) + P(C) = 1$.
Odds in favour of $A$ are $4:6$,so $P(A) = \frac{4}{4+6} = \frac{4}{10} = \frac{2}{5}$.
Odds against $B$ are $7:3$,which means odds in favour of $B$ are $3:7$,so $P(B) = \frac{3}{3+7} = \frac{3}{10}$.
Substituting these into the equation: $\frac{2}{5} + \frac{3}{10} + P(C) = 1$.
$\frac{4}{10} + \frac{3}{10} + P(C) = 1$ $\Rightarrow \frac{7}{10} + P(C) = 1$ $\Rightarrow P(C) = 1 - \frac{7}{10} = \frac{3}{10}$.
Odds against $C$ are given by $\frac{1 - P(C)}{P(C)} = \frac{1 - \frac{3}{10}}{\frac{3}{10}} = \frac{\frac{7}{10}}{\frac{3}{10}} = \frac{7}{3}$.
Thus,the odds against $C$ are $7:3$.

(A) Total number of cards $= 52$.
Number of kings in a pack $= 4$.
Number of non-king cards $= 52 - 4 = 48$.
The odds in favour of an event $E$ are defined as the ratio of the number of favourable outcomes to the number of unfavourable outcomes.
Odds in favour of drawing a king $= \frac{\text{Number of kings}}{\text{Number of non-king cards}} = \frac{4}{48} = \frac{1}{12}$.
Thus,the odds in favour are $1:12$.

(A) When a pair of dice is thrown,the total number of outcomes is $6 \times 6 = 36$.
The sums that are multiples of $3$ are $3, 6, 9,$ and $12$.
The favourable outcomes are:
Sum $3$: $(1, 2), (2, 1)$ ($2$ outcomes)
Sum $6$: $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$ ($5$ outcomes)
Sum $9$: $(3, 6), (4, 5), (5, 4), (6, 3)$ ($4$ outcomes)
Sum $12$: $(6, 6)$ ($1$ outcome)
Total number of favourable outcomes $= 2 + 5 + 4 + 1 = 12$.
Number of unfavourable outcomes $= 36 - 12 = 24$.
The odds in favour are defined as $\frac{\text{Number of favourable outcomes}}{\text{Number of unfavourable outcomes}} = \frac{12}{24} = \frac{1}{2}$ or $1: 2$.

(A) The odds against $A$ solving the problem are $3:2$,so the probability that $A$ solves it is $P(A) = \frac{2}{3+2} = \frac{2}{5}$. The probability that $A$ fails is $P(\bar{A}) = 1 - \frac{2}{5} = \frac{3}{5}$.
The odds against $B$ solving the problem are $2:1$,so the probability that $B$ solves it is $P(B) = \frac{1}{2+1} = \frac{1}{3}$. The probability that $B$ fails is $P(\bar{B}) = 1 - \frac{1}{3} = \frac{2}{3}$.
The problem is solved if at least one of them solves it. This is the complement of the event that both fail to solve the problem.
$P(\text{solved}) = 1 - P(\bar{A} \cap \bar{B}) = 1 - P(\bar{A}) \cdot P(\bar{B})$
$P(\text{solved}) = 1 - (\frac{3}{5} \cdot \frac{2}{3}) = 1 - \frac{6}{15} = 1 - \frac{2}{5} = \frac{3}{5}$.

(B) Given that events $A, B, C$ are mutually exclusive and exhaustive,we have $P(A) + P(B) + P(C) = 1$.
Odds against $A$ are $8:3$,so $P(A) = \frac{3}{8+3} = \frac{3}{11}$.
Odds against $B$ are $5:2$,so $P(B) = \frac{2}{5+2} = \frac{2}{7}$.
Since $P(A) + P(B) + P(C) = 1$,we have $P(C) = 1 - (\frac{3}{11} + \frac{2}{7}) = 1 - (\frac{21+22}{77}) = 1 - \frac{43}{77} = \frac{34}{77}$.
The odds against $C$ are given by $\frac{P(C^c)}{P(C)} = \frac{1 - P(C)}{P(C)} = \frac{1 - 34/77}{34/77} = \frac{43/77}{34/77} = \frac{43}{34}$.
Given the odds against $C$ are $43:17k$,we equate: $\frac{43}{17k} = \frac{43}{34}$.
Thus,$17k = 34$,which implies $k = 2$.

(C) Given that the odds in favour of $A$ are $2:3$,so $P(A) = \frac{2}{2+3} = \frac{2}{5}$.
Given that the odds against $B$ are $4:5$,so the odds in favour of $B$ are $5:4$,which means $P(B) = \frac{5}{5+4} = \frac{5}{9}$.
Since $A$ and $B$ are independent events,the probability of their intersection is given by $P(A \cap B) = P(A) \times P(B)$.
Substituting the values,we get $P(A \cap B) = \frac{2}{5} \times \frac{5}{9} = \frac{2}{9}$.

(A) Given that $A$,$B$,and $C$ are mutually exclusive and exhaustive events,we have $P(A) + P(B) + P(C) = 1$.
Odds in favor of $A$ are $4 : 6$,so $P(A) = \frac{4}{4+6} = \frac{4}{10}$.
Odds against $B$ are $7 : 3$,which means odds in favor of $B$ are $3 : 7$,so $P(B) = \frac{3}{3+7} = \frac{3}{10}$.
Substituting these into the sum: $\frac{4}{10} + \frac{3}{10} + P(C) = 1$.
$\frac{7}{10} + P(C) = 1 \implies P(C) = 1 - \frac{7}{10} = \frac{3}{10}$.
The probability of the complement event $C'$ is $P(C') = 1 - P(C) = 1 - \frac{3}{10} = \frac{7}{10}$.
The odds against $C$ are defined as $P(C') : P(C) = \frac{7}{10} : \frac{3}{10} = 7 : 3$.

(D) The probability that the first critic favours the book is $P(A) = \frac{2}{2+5} = \frac{2}{7}$.
$\therefore P(A') = 1 - \frac{2}{7} = \frac{5}{7}$.
The probability that the second critic favours the book is $P(B) = \frac{3}{3+4} = \frac{3}{7}$.
$\therefore P(B') = 1 - \frac{3}{7} = \frac{4}{7}$.
The probability that the third critic favours the book is $P(C) = \frac{4}{4+3} = \frac{4}{7}$.
$\therefore P(C') = 1 - \frac{4}{7} = \frac{3}{7}$.
Majority will be in favour of the book if at least two critics favour the book.
Hence,the probability is $P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A' \cap B \cap C) + P(A \cap B \cap C)$.
$= P(A) \cdot P(B) \cdot P(C') + P(A) \cdot P(B') \cdot P(C) + P(A') \cdot P(B) \cdot P(C) + P(A) \cdot P(B) \cdot P(C)$.
$= \left(\frac{2}{7} \times \frac{3}{7} \times \frac{3}{7}\right) + \left(\frac{2}{7} \times \frac{4}{7} \times \frac{4}{7}\right) + \left(\frac{5}{7} \times \frac{3}{7} \times \frac{4}{7}\right) + \left(\frac{2}{7} \times \frac{3}{7} \times \frac{4}{7}\right)$.
$= \frac{18}{343} + \frac{32}{343} + \frac{60}{343} + \frac{24}{343} = \frac{134}{343}$.

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

(C) Since $A$ and $B$ are mutually exclusive events,$P(A \cap B) = 0$.
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $\frac{3}{5} = \frac{1}{2} + P(B) - 0$.
$P(B) = \frac{3}{5} - \frac{1}{2} = \frac{6 - 5}{10} = \frac{1}{10}$.
We are given $P(B') = p$.
Since $P(B') = 1 - P(B)$,we have $p = 1 - \frac{1}{10} = \frac{9}{10}$.
Thus,the correct option is $C$.

(B) Given: $P(A)=0.5$,$P(A \cup B)=0.6$,and $P(B)=K$.
Since $A$ and $B$ are independent events,the probability of their intersection is given by $P(A \cap B) = P(A) \times P(B) = 0.5K$.
Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $0.6 = 0.5 + K - 0.5K$.
Simplifying the equation: $0.6 = 0.5 + 0.5K$.
Subtracting $0.5$ from both sides: $0.1 = 0.5K$.
Solving for $K$: $K = \frac{0.1}{0.5} = 0.2$.

(A) Given that $A$ and $B$ are mutually exclusive events,the probability of $A$ or $B$ is given by the addition rule for mutually exclusive events:
$P(A \cup B) = P(A) + P(B)$
Substituting the given values:
$P(A \cup B) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$
Converting the fraction to decimal:
$\frac{4}{5} = 0.8$

(C) We know that for any two events $A$ and $B$,the addition theorem of probability states:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Substituting the given values:
$0.65 = P(A) + P(B) - 0.15$
$P(A) + P(B) = 0.65 + 0.15 = 0.8$
We are asked to find $P(\overline{A}) + P(\overline{B})$.
Using the complement rule $P(\overline{E}) = 1 - P(E)$:
$P(\overline{A}) + P(\overline{B}) = (1 - P(A)) + (1 - P(B))$
$= 2 - (P(A) + P(B))$
$= 2 - 0.8 = 1.2$

(A) Total number of cards in a pack $= 52$.
Number of sample space $n(S) = 52$.
Let $A$ be the event of drawing an ace and $B$ be the event of drawing a spade.
Number of aces $n(A) = 4$.
Number of spades $n(B) = 13$.
The number of cards that are both an ace and a spade is $n(A \cap B) = 1$.
Using the addition rule of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52}$.
Simplifying the fraction,we get $\frac{16}{52} = \frac{4}{13}$.

(A) Let $P(P)$ be the probability that person $P$ speaks the truth and $P(R)$ be the probability that person $R$ speaks the truth.
Given: $P(P) = \frac{75}{100} = \frac{3}{4}$ and $P(R) = \frac{80}{100} = \frac{4}{5}$.
Consequently,the probabilities of them lying are $P(P') = 1 - \frac{3}{4} = \frac{1}{4}$ and $P(R') = 1 - \frac{4}{5} = \frac{1}{5}$.
They contradict each other if one speaks the truth and the other lies.
Case $I$: $P$ speaks the truth and $R$ lies: $P(P) \times P(R') = \frac{3}{4} \times \frac{1}{5} = \frac{3}{20}$.
Case $II$: $R$ speaks the truth and $P$ lies: $P(R) \times P(P') = \frac{4}{5} \times \frac{1}{4} = \frac{4}{20}$.
Total probability of contradiction $= \frac{3}{20} + \frac{4}{20} = \frac{7}{20}$.

(B) Let $P(A)$ be the probability that candidate $A$ gets the job and $P(B)$ be the probability that candidate $B$ gets the job.
Given: $P(A) = 0.8$ and $P(B) = 0.7$.
Since the events are independent,the probability that both get the job is $P(A \cap B) = P(A) \times P(B) = 0.8 \times 0.7 = 0.56$.
The probability that at least one of them gets the job is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = 0.8 + 0.7 - 0.56 = 1.5 - 0.56 = 0.94$.

(B) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since $P(A) = 1 - P(A^C) = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting the values into the addition theorem of probability:
$\frac{5}{6} = \frac{1}{2} + P(B) - \frac{1}{3}$.
$\frac{5}{6} = \frac{3-2}{6} + P(B) = \frac{1}{6} + P(B)$.
$P(B) = \frac{5}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$.
Therefore,$P(B^C) = 1 - P(B) = 1 - \frac{2}{3} = \frac{1}{3}$.

(B) We know that for any two events $A$ and $B$,the addition theorem of probability is given by:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Comparing this with the given equation:
$P(A \cup B) = a P(A \cap B) + b P(A) + c P(B)$
We get the coefficients as:
$a = -1, b = 1, c = 1$
Now,substituting these values into the expression $3a + 2b + 5c$:
$3(-1) + 2(1) + 5(1) = -3 + 2 + 5 = 4$
Thus,the value is $4$.

(D) 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 $P(B) = \frac{3}{4}$.
The probability that $A$ does not speak the truth is $P(\overline{A}) = 1 - P(A) = 1 - \frac{4}{5} = \frac{1}{5}$.
The probability that $B$ does not speak the truth is $P(\overline{B}) = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4}$.
They contradict each other if one speaks the truth and the other lies.
Thus,the probability of contradiction is $P(A \cap \overline{B}) + P(\overline{A} \cap B) = P(A) \cdot P(\overline{B}) + P(\overline{A}) \cdot P(B)$.
Substituting the values: $(\frac{4}{5} \times \frac{1}{4}) + (\frac{1}{5} \times \frac{3}{4}) = \frac{4}{20} + \frac{3}{20} = \frac{7}{20}$.

(C) Let $P(A)$ be the probability that $A$ fails in an exam and $P(B)$ be the probability that $B$ fails in an exam.
We are given $P(A) = 0.2$ and $P(B) = 0.3$.
The probability that either $A$ or $B$ fails is given by the union of the two events,$P(A \cup B)$.
By the addition theorem of probability,we have:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since $P(A \cap B) \geq 0$,it follows that $P(A \cup B) \leq P(A) + P(B)$.
Substituting the given values:
$P(A \cup B) \leq 0.2 + 0.3$.
$P(A \cup B) \leq 0.5$.
Therefore,the probability that either $A$ or $B$ fails is $\leq 0.5$.
Hence,option $(C)$ is correct.

(B) The prime numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$ are $2, 3, 5, 7$.
$P(E) = P(X=2) + P(X=3) + P(X=5) + P(X=7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62$.
The event $F = \{X < 4\}$ corresponds to $X \in \{1, 2, 3\}$.
$P(F) = P(X=1) + P(X=2) + P(X=3) = 0.15 + 0.23 + 0.12 = 0.50$.
The intersection $E \cap F$ consists of values that are both prime and less than $4$,which are $\{2, 3\}$.
$P(E \cap F) = P(X=2) + P(X=3) = 0.23 + 0.12 = 0.35$.
Using the addition theorem of probability,$P(E \cup F) = P(E) + P(F) - P(E \cap F)$.
$P(E \cup F) = 0.62 + 0.50 - 0.35 = 0.77$.

(B) We are given: $P(A \cup B) = \frac{5}{6}$,$P(\bar{A}) = \frac{1}{4}$,and $P(B) = \frac{1}{3}$.
First,calculate $P(A)$:
$P(A) = 1 - P(\bar{A}) = 1 - \frac{1}{4} = \frac{3}{4}$.
Using the addition theorem of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values:
$\frac{5}{6} = \frac{3}{4} + \frac{1}{3} - P(A \cap B)$.
$\frac{5}{6} = \frac{9 + 4}{12} - P(A \cap B) = \frac{13}{12} - P(A \cap B)$.
$P(A \cap B) = \frac{13}{12} - \frac{5}{6} = \frac{13 - 10}{12} = \frac{3}{12} = \frac{1}{4}$.
Now,check for independence by calculating $P(A) \cdot P(B)$:
$P(A) \cdot P(B) = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4}$.
Since $P(A \cap B) = P(A) \cdot P(B)$,the events $A$ and $B$ are independent.

(B) Given that $P(A) + P(B) = 2 P(A \cap B)$.
We know the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given condition into the formula:
$P(A \cup B) = 2 P(A \cap B) - P(A \cap B) = P(A \cap B)$.
Since $P(A \cap B) \leq P(A) \leq P(A \cup B)$ and $P(A \cap B) \leq P(B) \leq P(A \cup B)$,the equality $P(A \cup B) = P(A \cap B)$ implies that $P(A) = P(B) = P(A \cap B) = P(A \cup B)$.
Thus,$P(A) = P(B)$.

(D) The set of the first $30$ natural numbers is $S = \{1, 2, 3, \dots, 30\}$,so the total number of outcomes is $n(S) = 30$.
Let $A$ be the event that the number is divisible by $4$. The numbers are $\{4, 8, 12, 16, 20, 24, 28\}$,so $n(A) = 7$.
Let $B$ be the event that the number is divisible by $7$. The numbers are $\{7, 14, 21, 28\}$,so $n(B) = 4$.
The event $A \cap B$ represents numbers divisible by both $4$ and $7$ (i.e.,divisible by $28$). The only number is $\{28\}$,so $n(A \cap B) = 1$.
Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{7}{30} + \frac{4}{30} - \frac{1}{30} = \frac{10}{30} = \frac{1}{3}$.

(A) Given that $P(E_1 \cup E_2) = \frac{5}{8}$,$P(\bar{E}_1) = \frac{3}{4}$,and $P(E_2) = \frac{1}{2}$.
First,find $P(E_1)$ using the complement rule: $P(E_1) = 1 - P(\bar{E}_1) = 1 - \frac{3}{4} = \frac{1}{4}$.
Using the addition theorem of probability: $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$.
Substituting the values: $\frac{5}{8} = \frac{1}{4} + \frac{1}{2} - P(E_1 \cap E_2)$.
$\frac{5}{8} = \frac{3}{4} - P(E_1 \cap E_2)$.
$P(E_1 \cap E_2) = \frac{3}{4} - \frac{5}{8} = \frac{6-5}{8} = \frac{1}{8}$.
Now,check for independence: $P(E_1) \times P(E_2) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.
Since $P(E_1 \cap E_2) = P(E_1) \times P(E_2)$,the events $E_1$ and $E_2$ are independent events.