Probability Questions in English

(C) When three fair coins are tossed,the total sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
Since the condition is that both heads and tails must appear,we exclude the cases $HHH$ and $TTT$.
Thus,the reduced sample space is $S' = \{HHT, HTH, THH, HTT, THT, TTH\}$,where $n(S') = 6$.
We want the probability that exactly one head appears. The favorable outcomes are $E = \{HTT, THT, TTH\}$,where $n(E) = 3$.
Therefore,the required probability is $P = \frac{n(E)}{n(S')} = \frac{3}{6} = \frac{1}{2}$.

(D) Total number of balls in the bag $= 4 + 5 + 6 = 15$.
Number of white balls $= 4$.
Number of red balls $= 6$.
Since the events of drawing a white ball and a red ball are mutually exclusive,the probability of drawing either a white or a red ball is the sum of their individual probabilities.
Probability of drawing a white ball $P(W) = \frac{4}{15}$.
Probability of drawing a red ball $P(R) = \frac{6}{15}$.
Probability (white or red) $= P(W) + P(R) = \frac{4}{15} + \frac{6}{15} = \frac{10}{15} = \frac{2}{3}$.

(B) Total number of cards in a pack = $52$.
Number of hearts = $13$.
Number of kings = $4$.
Since one king is also a heart,the number of cards that are either a heart or a king = $13 + 4 - 1 = 16$.
Number of cards that are neither a heart nor a king = $52 - 16 = 36$.
Required probability = $\frac{36}{52} = \frac{9}{13}$.

(A) When two dice are thrown,the possible outcomes for each die range from $1$ to $6$.
The maximum possible sum of the numbers on the two dice is $6 + 6 = 12$.
Since the sum $13$ is greater than the maximum possible sum of $12$,it is impossible to obtain a total of $13$.
Therefore,the number of favorable outcomes is $0$.
The probability is given by the formula: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$.
$P(13) = \frac{0}{36} = 0$.

(C) The total number of outcomes when three dice are thrown simultaneously is $6 \times 6 \times 6 = 216$.
$A$ total of $17$ can be obtained in the following ways: $(5, 6, 6), (6, 5, 6), (6, 6, 5)$. There are $3$ such outcomes.
$A$ total of $18$ can be obtained in the following way: $(6, 6, 6)$. There is $1$ such outcome.
The total number of favorable outcomes is $3 + 1 = 4$.
Therefore,the required probability is $\frac{4}{216} = \frac{1}{54}$.

(D) Total number of articles in the box $= 10 + 6 = 16$.
Let $E$ be the event that the chosen article is either good or has a defect.
Since every article in the box is either good or defective,the event $E$ is a sure event (sample space).
Therefore,the number of favorable outcomes $= 16$.
The probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{16}{16} = 1$.

(B) An impossible event is an event that cannot occur in any trial of a random experiment.
By definition,the probability of an impossible event is always $0$.
Therefore,$P(\phi) = 0$.

(C) The experiment stops when a head appears or after $5$ tosses.
Given that a head does not occur on the first two tosses,we have already observed $TT$ (Tail,Tail).
For the coin to be tossed $5$ times,we must not get a head on the $3^{rd}$ toss $AND$ we must not get a head on the $4^{th}$ toss.
If we get a head on the $3^{rd}$ or $4^{th}$ toss,the experiment would stop earlier.
Therefore,the sequence of outcomes for the $3^{rd}$ and $4^{th}$ tosses must be $T$ and $T$.
The probability of getting a tail on any single toss is $1/2$.
Thus,the probability that the coin is tossed $5$ times given the first two were tails is $P(T_3) \cdot P(T_4) = (1/2) \cdot (1/2) = 1/4$.

(B) When two dice are tossed,the total number of possible outcomes is $6 \times 6 = 36$.
The possible sums range from $2$ to $12$. The prime numbers in this range are $\{2, 3, 5, 7, 11\}$.
Now,we count the number of outcomes for each prime sum:
- Sum $= 2$: $(1, 1)$ $\rightarrow 1$ outcome
- Sum $= 3$: $(1, 2), (2, 1)$ $\rightarrow 2$ outcomes
- Sum $= 5$: $(1, 4), (2, 3), (3, 2), (4, 1)$ $\rightarrow 4$ outcomes
- Sum $= 7$: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$ $\rightarrow 6$ outcomes
- Sum $= 11$: $(5, 6), (6, 5)$ $\rightarrow 2$ outcomes
Total favorable outcomes $= 1 + 2 + 4 + 6 + 2 = 15$.
Therefore,the probability $= \frac{15}{36} = \frac{5}{12}$.

(A) Let $A$,$B$,and $C$ be the events that the three persons solve the problem independently.
Given probabilities are $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{4}$,and $P(C) = \frac{1}{5}$.
The probability that they do not solve the problem is given by $P(A') = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3}$,$P(B') = 1 - P(B) = 1 - \frac{1}{4} = \frac{3}{4}$,and $P(C') = 1 - P(C) = 1 - \frac{1}{5} = \frac{4}{5}$.
Since the events are independent,the probability that none of them solve the problem is $P(A' \cap B' \cap C') = P(A') \times P(B') \times P(C')$.
$P(A' \cap B' \cap C') = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{2}{5}$.

(B) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
The outcomes where the sum of the points is $7$ are: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$.
The number of favourable outcomes is $6$.
The probability $P$ is given by the ratio of favourable outcomes to total outcomes:
$P = \frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \frac{6}{36}$.

(D) Let $P(\bar{A})$ be the probability that the event fails to happen and $P(A)$ be the probability that the event takes place.
Given $P(\bar{A}) = 0.05$.
Since $P(A) + P(\bar{A}) = 1$,we have $P(A) = 1 - 0.05 = 0.95$.
The probability that the event takes place on $4$ consecutive occasions is given by $(P(A))^4$.
Therefore,$(0.95)^4 = 0.95 \times 0.95 \times 0.95 \times 0.95 = 0.81450625$.

(B) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
We need to find the probability of getting a sum of at least $9$,which means the sum can be $9, 10, 11,$ or $12$.
The favorable outcomes are:
- Sum $= 9$: $(3,6), (4,5), (5,4), (6,3)$ ($4$ outcomes)
- Sum $= 10$: $(4,6), (5,5), (6,4)$ ($3$ outcomes)
- Sum $= 11$: $(5,6), (6,5)$ ($2$ outcomes)
- Sum $= 12$: $(6,6)$ ($1$ outcome)
Total favorable outcomes $= 4 + 3 + 2 + 1 = 10$.
The required probability $= \frac{\text{Total favorable outcomes}}{\text{Total possible outcomes}} = \frac{10}{36} = \frac{5}{18}$.

(B) The word $POSSESSIVE$ contains the following letters: $P, O, S, S, E, S, S, I, V, E$.
Total number of letters in the word = $10$.
The letters are: $P, O, S, S, E, S, S, I, V, E$.
Number of times the letter $S$ appears in the word = $4$.
The probability of choosing the letter $S$ is given by the ratio of the number of favorable outcomes to the total number of outcomes.
Probability $P(S) = \frac{\text{Number of } S}{\text{Total number of letters}} = \frac{4}{10}$.

(B) When three dice are rolled,the total number of possible outcomes is $6 \times 6 \times 6 = 216$.
The favorable outcomes where the same number appears on each die 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 $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{216} = \frac{1}{36}$.

(B) The probability of an event not occurring is given by the complement rule: $P(\text{not } A) = P(\bar{A}) = 1 - P(A)$.
Given that $P(A) = 0.38$,we substitute this value into the formula:
$P(\bar{A}) = 1 - 0.38 = 0.62$.
Therefore,the correct option is $B$.

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

(B) When three coins are tossed,the total number of possible outcomes is $2^3 = 8$.
The sample space is $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.
We are looking for outcomes where heads and tails appear alternately.
The favorable outcomes are $HTH$ and $THT$.
Thus,the number of favorable outcomes is $2$.
The probability is calculated as $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{8} = \frac{1}{4}$.

(A) Two events $A$ and $B$ are said to be mutually exclusive if they cannot occur simultaneously,meaning $A \cap B = \emptyset$.
For any two events $A$ and $B$,the addition rule of probability is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since $A$ and $B$ are mutually exclusive,$P(A \cap B) = 0$.
Therefore,$P(A \cup B) = P(A) + P(B) - 0 = P(A) + P(B)$.

(A) sure event is an event that is certain to occur. The probability of a sure event is $P(A) = 1$.
The complement of event $A$,denoted as $A^c$ or $A$ not,represents the event that $A$ does not occur.
By the property of complementary events,$P(A^c) = 1 - P(A)$.
Substituting the value of $P(A)$,we get $P(A^c) = 1 - 1 = 0$.
Therefore,the probability of the complement of a sure event is $0$.

(D) The set of the first ten natural numbers is $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
The total number of outcomes is $n(S) = 10$.
We need to find numbers that are both odd and perfect squares within this set.
The perfect squares in the set are $1, 4, 9$.
Among these,the odd numbers are $1$ and $9$.
Thus,the number of favorable outcomes is $n(E) = 2$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{2}{10} = \frac{1}{5}$.

(D) Let $S$ be the set of the first $100$ natural numbers,so $n(S) = 100$.
Let $A$ be the event that the number is even. The even numbers are $2, 4, 6, \dots, 100$. Thus,$n(A) = 50$.
Let $B$ be the event that the number is divisible by $5$. The numbers divisible by $5$ are $5, 10, 15, \dots, 100$. Thus,$n(B) = 20$.
The intersection $A \cap B$ represents numbers that are both even and divisible by $5$,which means they are divisible by $10$. These are $10, 20, 30, \dots, 100$. Thus,$n(A \cap B) = 10$.
Using the inclusion-exclusion principle,the number of favorable outcomes is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 50 + 20 - 10 = 60$.
The probability is $P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{60}{100} = \frac{3}{5}$.

(D) When two dice are thrown,the total number of outcomes is $36$.
Given that the first die shows $5$,the possible outcomes are $(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)$.
There are $6$ such outcomes.
We need the sum of the numbers to be $8$ or more than $8$.
The outcomes satisfying this condition are $(5, 3), (5, 4), (5, 5), (5, 6)$.
There are $4$ such outcomes.
Therefore,the required probability is $\frac{4}{6} = \frac{2}{3}$.

(A) standard pack of playing cards contains $52$ cards.
There are $4$ aces and $4$ kings in the deck.
The total number of cards that are either an ace or a king is $4 + 4 = 8$.
The number of cards that are neither an ace nor a king is $52 - 8 = 44$.
The probability of drawing a card that is neither an ace nor a king is given by the ratio of the number of favorable outcomes to the total number of outcomes.
Probability $P = \frac{44}{52}$.
Simplifying the fraction by dividing both numerator and denominator by $4$,we get $P = \frac{11}{13}$.

(C) The total number of ways to arrange $4$ letters in $4$ envelopes is $4! = 24$.
The number of ways in which all letters go into their correct envelopes is $1$.
The probability that all letters go into their correct envelopes is $P(\text{correct}) = \frac{1}{4!} = \frac{1}{24}$.
The probability that no letter goes into its correct envelope (a derangement) is $1 - P(\text{correct})$.
Therefore,the required probability is $1 - \frac{1}{24} = \frac{23}{24}$.

(A) The total number of ways to place $n$ letters into $n$ envelopes is $n!$.
Only one specific arrangement corresponds to the case where every letter is placed in its correct envelope.
Therefore,the number of favorable outcomes is $1$.
The probability $P$ is given by the ratio of favorable outcomes to total outcomes:
$P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{n!}$.

(B) The probability of a boy being born is $P(B) = \frac{1}{2}$ and the probability of a girl being born is $P(G) = \frac{1}{2}$.
In a family of $4$ children,the total number of outcomes is $2^4 = 16$.
The event of having 'at least one girl' is the complement of the event of having 'no girls' (i.e.,all boys).
The probability of having no girls is $P(\text{all boys}) = (\frac{1}{2})^4 = \frac{1}{16}$.
Therefore,the probability of having at least one girl is $1 - P(\text{all boys}) = 1 - \frac{1}{16} = \frac{15}{16}$.

(B) determinant of order $2$ is of the form $\Delta = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$.
Each element $a, b, c, d$ can be chosen in $2$ ways ($0$ or $1$).
Total number of possible determinants = $2^4 = 16$.
The determinant $\Delta$ is non-zero if $ad - bc \neq 0$,which means $ad \neq bc$.
Since $a, b, c, d \in \{0, 1\}$,the possible values for $ad$ and $bc$ are $0$ or $1$.
$ad - bc = 0$ occurs when $ad = bc$. This happens in the following cases:
$1$. $ad = 0$ and $bc = 0$: $ad=0$ occurs in $3$ ways $((0,0), (0,1), (1,0))$,and $bc=0$ occurs in $3$ ways. Total ways = $3 \times 3 = 9$.
$2$. $ad = 1$ and $bc = 1$: $ad=1$ occurs in $1$ way $((1,1))$,and $bc=1$ occurs in $1$ way. Total ways = $1 \times 1 = 1$.
Total cases where $\Delta = 0$ is $9 + 1 = 10$.
Total cases where $\Delta \neq 0$ is $16 - 10 = 6$.
Therefore,the required probability = $\frac{6}{16} = \frac{3}{8}$.

(C) By definition,two events $A$ and $B$ are independent if $P(A \cap B) = P(A) \cdot P(B)$.
For event $A$ to be independent of itself,we must have $P(A \cap A) = P(A) \cdot P(A)$.
Since $A \cap A = A$,the equation becomes $P(A) = P(A)^2$.
Rearranging the terms,we get $P(A)^2 - P(A) = 0$.
Factoring the expression,we get $P(A)(P(A) - 1) = 0$.
Therefore,$P(A) = 0$ or $P(A) = 1$.

(B) The total number of possible codes from $000$ to $999$ is $1000$.
Let $E_i$ be the event that the stranger fails at the $i^{th}$ trial.
The probability of failing at the first trial is $P(E_1) = \frac{999}{1000}$.
If the stranger fails at the first trial,there are $999$ codes remaining,and $998$ of them are incorrect. Thus,the probability of failing at the second trial given failure at the first is $P(E_2|E_1) = \frac{998}{999}$.
For the stranger to succeed at the $k^{th}$ trial,they must fail the first $(k-1)$ trials and then succeed on the $k^{th}$ trial.
The probability of failing the first $(k-1)$ trials is $P(E_1 \cap E_2 \cap ... \cap E_{k-1}) = \frac{999}{1000} \times \frac{998}{999} \times ... \times \frac{1000-(k-1)}{1000-(k-2)} = \frac{1000-k+1}{1000}$.
Given that the stranger has failed $(k-1)$ times,there are $1000-(k-1) = 1001-k$ codes remaining,one of which is the correct code.
The probability of succeeding on the $k^{th}$ trial given failure in the previous $(k-1)$ trials is $\frac{1}{1001-k}$.
Therefore,the probability of success at the $k^{th}$ trial is $\frac{1000-k+1}{1000} \times \frac{1}{1001-k} = \frac{1}{1000}$.

(B) The total number of outcomes when throwing three dice is $6^3 = 216$.
Let $E$ be the event that at least one die shows up $1$.
It is easier to calculate the probability of the complement event $E'$,which is the event that no die shows up $1$.
For a single die,the probability of not getting a $1$ is $\frac{5}{6}$.
Since the three dice are independent,the probability that none of the three dice show $1$ is $\left( \frac{5}{6} \right)^3 = \frac{125}{216}$.
Therefore,the probability of the event $E$ is $P(E) = 1 - P(E') = 1 - \frac{125}{216} = \frac{216 - 125}{216} = \frac{91}{216}$.

(A) Total number of cards in a well-shuffled pack = $52$.
There is only one 'two of hearts' and one 'two of diamonds' in the deck.
Therefore,the number of favorable outcomes = $1 + 1 = 2$.
The probability of an event is given by the ratio of favorable outcomes to the total number of possible outcomes.
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 $P(W) = \frac{1}{5}$.
Probability that the husband is not selected: $P(H') = 1 - \frac{1}{7} = \frac{6}{7}$.
Probability that the wife is not selected: $P(W') = 1 - \frac{1}{5} = \frac{4}{5}$.
Only one of them is selected if (husband is selected $AND$ wife is not) $OR$ (wife is selected $AND$ husband is not).
Probability = $P(H) \times P(W') + P(W) \times P(H')$.
Probability = $(\frac{1}{7} \times \frac{4}{5}) + (\frac{1}{5} \times \frac{6}{7})$.
Probability = $\frac{4}{35} + \frac{6}{35} = \frac{10}{35} = \frac{2}{7}$.

(A) Total number of balls $= 5 + 7 + 8 = 20$.
We need to draw $4$ white balls one by one without replacement.
The probability of drawing the first white ball is $\frac{5}{20}$.
Since the ball is not replaced,the number of white balls becomes $4$ and the total number of balls becomes $19$. The probability of drawing the second white ball is $\frac{4}{19}$.
Similarly,the probability of drawing the third white ball is $\frac{3}{18}$,and the probability of drawing the fourth white ball is $\frac{2}{17}$.
Therefore,the required probability is $\frac{5}{20} \times \frac{4}{19} \times \frac{3}{18} \times \frac{2}{17} = \frac{1}{4} \times \frac{4}{19} \times \frac{1}{6} \times \frac{2}{17} = \frac{1}{19} \times \frac{1}{3} \times \frac{1}{17} = \frac{1}{969}$.

(B) Let $P(A) = p_1 = \frac{1}{3}$,$P(B) = p_2 = \frac{2}{7}$,and $P(C) = p_3 = \frac{3}{8}$.
Then the probabilities of not solving the problem are $q_1 = 1 - p_1 = \frac{2}{3}$,$q_2 = 1 - p_2 = \frac{5}{7}$,and $q_3 = 1 - p_3 = \frac{5}{8}$.
Exactly one person 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 $= p_1 q_2 q_3 + q_1 p_2 q_3 + q_1 q_2 p_3$.
$= (\frac{1}{3} \times \frac{5}{7} \times \frac{5}{8}) + (\frac{2}{3} \times \frac{2}{7} \times \frac{5}{8}) + (\frac{2}{3} \times \frac{5}{7} \times \frac{3}{8})$.
$= \frac{25}{168} + \frac{20}{168} + \frac{30}{168} = \frac{75}{168}$.
Simplifying by dividing by $3$,we get $\frac{25}{56}$.

(A) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
Let $E_1$ be the event of getting a sum of $7$. The favorable outcomes are: $(6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6)$. The number of outcomes is $6$.
Let $E_2$ be the event of getting a sum of $9$. The favorable outcomes are: $(6, 3), (5, 4), (4, 5), (3, 6)$. The number of outcomes is $4$.
Since these events are mutually exclusive,the total number of favorable outcomes is $6 + 4 = 10$.
The required probability is $P = \frac{\text{Total favorable outcomes}}{\text{Total possible outcomes}} = \frac{10}{36} = \frac{5}{18}$.

(D) The total number of tickets is $19$ (numbered $1$ to $19$).
The even numbers between $1$ and $19$ are $2, 4, 6, 8, 10, 12, 14, 16, 18$. There are $9$ even numbers.
The probability of drawing an even number in the first draw is $P(E_1) = \frac{9}{19}$.
Since the ticket is drawn without replacement,there are now $18$ tickets left in the bag,and $8$ of them are even.
The probability of drawing an even number in the second draw,given that the first was even,is $P(E_2|E_1) = \frac{8}{18}$.
The probability that both tickets show even numbers is $P(E_1 \cap E_2) = P(E_1) \times P(E_2|E_1) = \frac{9}{19} \times \frac{8}{18} = \frac{9}{19} \times \frac{4}{9} = \frac{4}{19}$.

(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 by the three marksmen.
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 event that one and only one of them hits the target can happen in three mutually exclusive ways: ($A$ hits,$B$ misses,$C$ misses) or ($A$ misses,$B$ hits,$C$ misses) or ($A$ misses,$B$ misses,$C$ hits).
Required probability $= P(A)P(\bar{B})P(\bar{C}) + P(\bar{A})P(B)P(\bar{C}) + P(\bar{A})P(\bar{B})P(C)$
$= (\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}) + (\frac{1}{2} \times \frac{1}{3} \times \frac{3}{4}) + (\frac{1}{2} \times \frac{2}{3} \times \frac{1}{4})$
$= \frac{6}{24} + \frac{3}{24} + \frac{2}{24} = \frac{11}{24}$.

(A) determinant of order $2$ is given by $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$.
Each element can be either $0$ or $1$. Since there are $4$ positions,the total number of possible determinants is $2^4 = 16$.
The value of the determinant is $ad - bc$. For this to be positive,$ad - bc > 0$,which means $ad > bc$.
Since $a, b, c, d \in \{0, 1\}$,the possible values for $ad$ and $bc$ are $0$ or $1$.
$ad > bc$ is possible only if $ad = 1$ and $bc = 0$.
This implies $a=1, d=1$ and $bc=0$.
The condition $bc=0$ is satisfied if $(b, c)$ is $(0, 0), (0, 1),$ or $(1, 0)$.
Thus,the favorable cases are:
$1. \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1(1) - 0(0) = 1$
$2. \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} = 1(1) - 0(1) = 1$
$3. \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} = 1(1) - 1(0) = 1$
There are $3$ favorable outcomes.
Therefore,the probability is $\frac{3}{16}$.

(C) Total number of cards in a pack = $52$.
Let $A$ be the event of drawing a king and $B$ be the event of drawing a diamond.
Number of kings,$n(A) = 4$.
Number of diamonds,$n(B) = 13$.
Number of cards that are both a king and a diamond (king of diamonds),$n(A \cap B) = 1$.
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,we get $\frac{16}{52} = \frac{4}{13}$.

(D) The total number of balls is $3 + 3 + 2 = 8$.
Since we are drawing three balls one by one without replacement,the probability that the $k$-th ball is of a certain color is equal to the initial proportion of that color in the bag.
Let $R_3$ be the event that the third ball drawn is red.
By the principle of symmetry in sampling without replacement,the probability of any specific draw being a certain color is the same as the probability of the first draw being that color.
Therefore,$P(R_3) = P(R_1) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{2}{8} = \frac{1}{4}$.

(B) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
The favourable outcomes for obtaining a sum of $8$ are:
$(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$.
There are $5$ such favourable outcomes.
Therefore,the required probability is $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{5}{36}$.

(B) The probability of an event $A$ occurring is denoted by $P(A)$,and the probability of the event $A$ not occurring (the complement of $A$) is denoted by $P(\bar{A})$.
Since an event either occurs or does not occur,these two events are complementary and exhaustive.
Therefore,the sum of their probabilities is always equal to $1$,i.e.,$P(A) + P(\bar{A}) = 1$.

(B) The total number of balls is $3 + 2 = 5$.
Let $W_1$ be the event that the first ball is white and $R_1$ be the event that the first ball is red.
Let $R_2$ be the event that the second ball is red.
The second ball can be red in two mutually exclusive ways:
$(i)$ First ball is white and second ball is red: $P(W_1 \cap R_2) = P(W_1) \times P(R_2|W_1) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20}$.
$(ii)$ First ball is red and second ball is red: $P(R_1 \cap R_2) = P(R_1) \times P(R_2|R_1) = \frac{2}{5} \times \frac{1}{4} = \frac{2}{20}$.
The total probability of the second ball being red is $P(R_2) = P(W_1 \cap R_2) + P(R_1 \cap R_2) = \frac{6}{20} + \frac{2}{20} = \frac{8}{20} = \frac{2}{5}$.

(C) Let $W$ denote a win and $L$ denote a loss for India. The probability of winning $P(W) = \frac{1}{2}$ and the probability of losing $P(L) = 1 - \frac{1}{2} = \frac{1}{2}$.
For the second win to occur exactly at the third test match,India must have won exactly one match in the first two tests and then won the third test.
The possible sequences for the first three matches are $(L, W, W)$ and $(W, L, W)$.
The probability of the sequence $(L, W, W)$ is $P(L) \times P(W) \times P(W) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$.
The probability of the sequence $(W, L, W)$ is $P(W) \times P(L) \times P(W) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$.
The total probability is the sum of these two mutually exclusive events: $\frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$.

(C) The total number of cards is $100$. Thus,the total number of possible outcomes is $100$.
$A$ number is a perfect square if it is of the form $n^2$. The perfect squares between $1$ and $100$ are: $1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64, 9^2=81, 10^2=100$.
There are $10$ such numbers.
Therefore,the number of favourable outcomes is $10$.
The probability of drawing a perfect square is $\frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \frac{10}{100} = \frac{1}{10}$.

(D) Let the numbers on the three chits be $X_1, X_2, X_3$. Each $X_i \in \{1, 2, 3, 4, 5, 6, 7\}$.
We want the probability that $\min(X_1, X_2, X_3) = 5$.
This implies that all $X_i \ge 5$ $AND$ at least one $X_i = 5$.
$P(\min(X_1, X_2, X_3) = 5) = P(\min(X_1, X_2, X_3) \ge 5) - P(\min(X_1, X_2, X_3) \ge 6)$.
Since the draws are with replacement,the total number of outcomes is $7^3 = 343$.
For $\min(X_i) \ge 5$,each $X_i$ must be in $\{5, 6, 7\}$. There are $3$ choices for each draw,so $3^3 = 27$ outcomes.
For $\min(X_i) \ge 6$,each $X_i$ must be in $\{6, 7\}$. There are $2$ choices for each draw,so $2^3 = 8$ outcomes.
Therefore,the number of favorable outcomes is $27 - 8 = 19$.
The probability is $\frac{19}{343}$.

(B) We know that for any event $E$,the probability of its complement is given by $P(\bar{E}) = 1 - P(E)$.
Given $P(A) = 0.65$,therefore $P(\bar{A}) = 1 - 0.65 = 0.35$.
Given $P(B) = 0.15$,therefore $P(\bar{B}) = 1 - 0.15 = 0.85$.
Now,calculating the sum: $P(\bar{A}) + P(\bar{B}) = 0.35 + 0.85 = 1.2$.

(A) By De Morgan's Law,we know that $\bar{E}_1 \cap \bar{E}_2 = \overline{E_1 \cup E_2}$.
Substituting this into the expression,we get $(E_1 \cup E_2) \cap (\overline{E_1 \cup E_2})$.
Since the intersection of any set $A$ and its complement $\bar{A}$ is the empty set $\phi$,we have $(E_1 \cup E_2) \cap (\overline{E_1 \cup E_2}) = \phi$.
The probability of an empty set is $P(\phi) = 0$.
Since $0 < \frac{1}{4}$,the correct option is $A$.

(C) The probability of non-occurrence of an event $A_i$ is given by $P(A_i^c) = 1 - P(A_i)$.
Given $P(A_i) = \frac{1}{i + 1}$, we have $P(A_i^c) = 1 - \frac{1}{i + 1} = \frac{i}{i + 1}$.
Since the events $A_1, A_2, \dots, A_n$ are independent, their complements $A_1^c, A_2^c, \dots, A_n^c$ are also independent.
The probability that none of the events occur is the product of the probabilities of their non-occurrence:
$P(\text{none occur}) = P(A_1^c) \cdot P(A_2^c) \cdot \dots \cdot P(A_n^c)$
$= \left( \frac{1}{2} \right) \cdot \left( \frac{2}{3} \right) \cdot \left( \frac{3}{4} \right) \dots \left( \frac{n}{n + 1} \right)$
By canceling the terms in the numerator and denominator, we get:
$= \frac{1}{n + 1}$.