Probability Questions in English

(B) When two dice are thrown,the total number of possible outcomes is $6^2 = 36$.
The sum of the numbers on the two dice can range from $2$ to $12$.
The sum is an even number if it is $2, 4, 6, 8, 10,$ or $12$.
Let us list the favorable outcomes for each even sum:
- Sum $= 2: (1, 1) \rightarrow 1$ outcome
- Sum $= 4: (1, 3), (2, 2), (3, 1) \rightarrow 3$ outcomes
- Sum $= 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) \rightarrow 5$ outcomes
- Sum $= 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) \rightarrow 5$ outcomes
- Sum $= 10: (4, 6), (5, 5), (6, 4) \rightarrow 3$ outcomes
- Sum $= 12: (6, 6) \rightarrow 1$ outcome
Total number of favorable outcomes $= 1 + 3 + 5 + 5 + 3 + 1 = 18$.
Probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{18}{36} = \frac{1}{2}$.

(A) Let the boxes be $A$,$B$,and $C$.
Box $A$ contains $3$ white and $1$ black ball. Probability of drawing white $P(W_A) = \frac{3}{4}$,black $P(B_A) = \frac{1}{4}$.
Box $B$ contains $2$ white and $2$ black balls. Probability of drawing white $P(W_B) = \frac{2}{4}$,black $P(B_B) = \frac{2}{4}$.
Box $C$ contains $1$ white and $3$ black balls. Probability of drawing white $P(W_C) = \frac{1}{4}$,black $P(B_C) = \frac{3}{4}$.
To get $2$ white and $1$ black ball,the possible cases are $(W, W, B)$,$(W, B, W)$,and $(B, W, W)$.
Probability $= P(W_A) \times P(W_B) \times P(B_C) + P(W_A) \times P(B_B) \times P(W_C) + P(B_A) \times P(W_B) \times P(W_C)$
$= (\frac{3}{4} \times \frac{2}{4} \times \frac{3}{4}) + (\frac{3}{4} \times \frac{2}{4} \times \frac{1}{4}) + (\frac{1}{4} \times \frac{2}{4} \times \frac{1}{4})$
$= \frac{18}{64} + \frac{6}{64} + \frac{2}{64} = \frac{26}{64} = \frac{13}{32}$.

(B) Total number of balls $= 10 + 12 + 11 = 33$.
The number of ways to select $2$ balls out of $33$ is given by the combination formula $nC_r = \frac{n!}{r!(n-r)!}$.
Total ways $= 33C_2 = \frac{33 \times 32}{2 \times 1} = 33 \times 16 = 528$.
The number of ways to select $2$ green balls out of $10$ green balls is $10C_2$.
Favorable ways $= 10C_2 = \frac{10 \times 9}{2 \times 1} = 45$.
The required probability is the ratio of favorable outcomes to total outcomes.
Probability $= \frac{45}{528}$.
Dividing both numerator and denominator by $3$,we get $\frac{15}{176}$.

(B) Total number of balls $= 10 + 12 + 11 = 33$.
Number of non-white balls (green + yellow) $= 10 + 12 = 22$.
When selecting $2$ balls without replacement,the probability that none is white is the probability that both selected balls are non-white.
Probability $= \frac{22}{33} \times \frac{21}{32}$.
Simplifying the fractions: $\frac{22}{33} = \frac{2}{3}$ and $\frac{21}{32}$ remains as is.
Probability $= \frac{2}{3} \times \frac{21}{32} = \frac{1}{1} \times \frac{7}{16} = \frac{7}{16}$.

(D) Total number of balls = $10 + 12 + 11 = 33$.
Number of ways to select $2$ balls from $33$ is $^33C_2 = \frac{33 \times 32}{2 \times 1} = 528$.
Number of balls that are not white = $10 + 12 = 22$.
Number of ways to select $2$ balls such that none are white = $^22C_2 = \frac{22 \times 21}{2 \times 1} = 231$.
Probability of selecting no white balls = $\frac{231}{528} = \frac{7}{16}$.
Probability of selecting at least one white ball = $1 - P(\text{no white ball}) = 1 - \frac{7}{16} = \frac{9}{16}$.

(A) Total number of balls $= 15 + 20 = 35$.
Total number of ways to select $2$ balls from $35$ is given by ${}^{35}C_{2} = \frac{35 \times 34}{2 \times 1} = 35 \times 17 = 595$.
Number of ways to select $2$ white balls from $20$ white balls is given by ${}^{20}C_{2} = \frac{20 \times 19}{2 \times 1} = 10 \times 19 = 190$.
The probability of selecting both white balls is $P = \frac{{}^{20}C_{2}}{{}^{35}C_{2}} = \frac{190}{595}$.
Dividing both numerator and denominator by $5$,we get $P = \frac{38}{119}$.

(C) Total number of balls = $15 + 20 = 35$.
Number of ways to select $2$ balls out of $35$ is given by $^{35}C_{2} = \frac{35 \times 34}{2 \times 1} = 35 \times 17 = 595$.
We need to select $1$ white ball and $1$ black ball.
Number of ways to select $1$ white ball from $20$ is $^{20}C_{1} = 20$.
Number of ways to select $1$ black ball from $15$ is $^{15}C_{1} = 15$.
Number of favorable outcomes = $^{20}C_{1} \times ^{15}C_{1} = 20 \times 15 = 300$.
Probability = $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{300}{595}$.
Dividing both numerator and denominator by $5$,we get $\frac{60}{119}$.

(A) Total number of balls = $15 + 20 = 35$.
Number of ways to select $2$ balls from $35$ is $^{35}C_2 = \frac{35 \times 34}{2} = 595$.
'At most one white ball' means either $0$ white balls (both black) or $1$ white ball (one black and one white).
Case $1$: Both balls are black. Number of ways = $^{15}C_2 = \frac{15 \times 14}{2} = 105$.
Case $2$: One white and one black ball. Number of ways = $^{20}C_1 \times ^{15}C_1 = 20 \times 15 = 300$.
Total favorable outcomes = $105 + 300 = 405$.
Probability = $\frac{405}{595} = \frac{81}{119}$.

(D) Total number of marbles $= 6 + 4 + 3 + 2 = 15$.
The number of ways to choose $2$ marbles out of $15$ is given by $^{15}C_2 = \frac{15 \times 14}{2 \times 1} = 105$.
The number of ways to choose $2$ blue marbles out of $6$ blue marbles is given by $^{6}C_2 = \frac{6 \times 5}{2 \times 1} = 15$.
The probability of picking both blue marbles is $\frac{^{6}C_2}{^{15}C_2} = \frac{15}{105} = \frac{1}{7}$.

(C) Total number of marbles = $6 + 4 + 3 + 2 = 15$.
We need to choose $3$ marbles out of $15$. The total number of ways to select $3$ marbles is given by $^{15}C_{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455$.
The number of ways to choose $2$ red marbles from $4$ red marbles is $^{4}C_{2} = \frac{4 \times 3}{2 \times 1} = 6$.
The number of ways to choose $1$ green marble from $3$ green marbles is $^{3}C_{1} = 3$.
The number of favorable outcomes is $^{4}C_{2} \times ^{3}C_{1} = 6 \times 3 = 18$.
Therefore,the probability is $\frac{18}{455}$.

(C) Total number of marbles $= 6 + 4 + 3 + 2 = 15$.
Number of ways to choose $4$ marbles out of $15$ is $^{15}C_4 = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365$.
Number of non-red marbles $= 15 - 4 = 11$.
Number of ways to choose $4$ marbles such that none is red is $^{11}C_4 = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$.
Probability of getting no red marble $= \frac{330}{1365} = \frac{22}{91}$.
Probability of getting at least one red marble $= 1 - P(\text{no red}) = 1 - \frac{22}{91} = \frac{69}{91}$.

(C) Total number of marbles $= 6 + 4 + 3 + 2 = 15$.
The total number of ways to pick $2$ marbles from $15$ is given by ${}^{15}C_{2} = \frac{15 \times 14}{2 \times 1} = 105$.
Probability of picking $2$ yellow marbles: There are $2$ yellow marbles,so the number of ways is ${}^{2}C_{2} = 1$. Thus,$P(\text{both yellow}) = \frac{1}{105}$.
Probability of picking $2$ green marbles: There are $3$ green marbles,so the number of ways is ${}^{3}C_{2} = 3$. Thus,$P(\text{both green}) = \frac{3}{105}$.
Since these events are mutually exclusive,the required probability is $P(\text{both yellow or both green}) = \frac{1}{105} + \frac{3}{105} = \frac{4}{105}$.

(D) Total number of marbles $= 6 + 4 + 3 + 2 = 15$.
We need to select $4$ marbles out of $15$. The total number of ways to select $4$ marbles is given by ${}^{15}C_{4} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365$.
The number of ways to select $1$ yellow marble from $2$ is ${}^{2}C_{1} = 2$.
The number of ways to select $2$ red marbles from $4$ is ${}^{4}C_{2} = \frac{4 \times 3}{2 \times 1} = 6$.
The number of ways to select $1$ blue marble from $6$ is ${}^{6}C_{1} = 6$.
The number of favorable outcomes $= {}^{2}C_{1} \times {}^{4}C_{2} \times {}^{6}C_{1} = 2 \times 6 \times 6 = 72$.
Required Probability $= \frac{72}{1365} = \frac{24}{455}$.

(D) Total number of marbles $= 4 + 5 + 3 = 12$.
Total ways to pick $3$ marbles from $12$ is given by $^{12}C_{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
Probability that all $3$ marbles are green $= \frac{^{3}C_{3}}{^{12}C_{3}} = \frac{1}{220}$.
Probability that all $3$ marbles are red $= \frac{^{4}C_{3}}{^{12}C_{3}} = \frac{4}{220} = \frac{1}{55}$.
Since these are mutually exclusive events,the required probability is the sum of the individual probabilities:
Required probability $= \frac{1}{220} + \frac{4}{220} = \frac{5}{220} = \frac{1}{44}$.

(B) Total number of marbles $= 4 + 5 + 3 = 12$.
Total ways to pick $3$ marbles from $12$ is given by $^{12}C_{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
Number of non-blue marbles $= 4 + 3 = 7$.
Ways to pick $3$ marbles such that none is blue $= ^{7}C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.
Probability that none is blue $= \frac{35}{220} = \frac{7}{44}$.
Probability that at least one is blue $= 1 - P(\text{none is blue}) = 1 - \frac{7}{44} = \frac{37}{44}$.

(C) Total number of marbles $= 4 + 5 + 3 = 12$.
We need to select $3$ marbles out of $12$. The total number of ways to select $3$ marbles is given by ${}^{12}C_{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
The number of ways to select $2$ red marbles out of $4$ is ${}^{4}C_{2} = \frac{4 \times 3}{2 \times 1} = 6$.
The number of ways to select $1$ green marble out of $3$ is ${}^{3}C_{1} = 3$.
The number of favorable outcomes is ${}^{4}C_{2} \times {}^{3}C_{1} = 6 \times 3 = 18$.
Required Probability $= \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{18}{220} = \frac{9}{110}$.

(B) Let $E$ be the event of drawing a coloured ball (red or blue).
Total number of balls = $30$.
Number of red balls = $10$.
Number of blue balls = $5$.
Number of coloured balls (red or blue) = $10 + 5 = 15$.
The probability $P(E)$ of drawing a coloured ball is given by the ratio of the number of favourable outcomes to the total number of outcomes.
$P(E) = \frac{\text{Number of coloured balls}}{\text{Total number of balls}} = \frac{15}{30} = \frac{1}{2}$.

(B) The total number of ways to select $3$ cards from $52$ cards is given by ${}^{52}C_{3} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22100$.
Let $E$ be the event of drawing one king,one queen,and one jack.
The number of ways to select $1$ king from $4$ kings,$1$ queen from $4$ queens,and $1$ jack from $4$ jacks is:
$n(E) = {}^{4}C_{1} \times {}^{4}C_{1} \times {}^{4}C_{1} = 4 \times 4 \times 4 = 64$.
The probability $P(E)$ is given by:
$P(E) = \frac{n(E)}{n(S)} = \frac{64}{22100}$.
Simplifying the fraction by dividing both numerator and denominator by $4$:
$P(E) = \frac{16}{5525}$.

(A) The total number of ways to obtain a sum of $15$ with $3$ dice is $10$. The possible outcomes are: $(6,6,3), (6,3,6), (3,6,6), (6,5,4), (6,4,5), (5,6,4), (5,4,6), (4,6,5), (4,5,6), (5,5,5)$.
The number of ways in which $4$ appears on the first die is $2$,which are $(4,6,5)$ and $(4,5,6)$.
Therefore,the required probability $= \frac{2}{10} = \frac{1}{5}$.

(D) The total number of outcomes when three dice are thrown is $N = 6^3 = 216$.
We need to find the number of outcomes where the sum of the numbers on the three dice is $16$.
The possible combinations of numbers that sum to $16$ are:
$1) (6, 6, 4)$ which can be arranged in $\frac{3!}{2!} = 3$ ways.
$2) (6, 5, 5)$ which can be arranged in $\frac{3!}{2!} = 3$ ways.
Total favorable outcomes $= 3 + 3 = 6$.
Therefore,the required probability $= \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{6}{216} = \frac{1}{36}$.

(B) Let $W_1$ be the event of drawing a white ball from the first bag and $B_1$ be the event of drawing a black ball from the first bag.
Total balls in the first bag $= 5 + 3 = 8$.
$P(W_1) = \frac{5}{8}$ and $P(B_1) = \frac{3}{8}$.
After transferring one ball,the second bag contains $2 + 4 + 1 = 7$ balls.
Case $1$: If a white ball is transferred,the second bag now contains $2 + 1 = 3$ white and $4$ black balls.
Probability of drawing a black ball from the second bag given $W_1$ is $P(B_2|W_1) = \frac{4}{7}$.
Case $2$: If a black ball is transferred,the second bag now contains $2$ white and $4 + 1 = 5$ black balls.
Probability of drawing a black ball from the second bag given $B_1$ is $P(B_2|B_1) = \frac{5}{7}$.
Using the Law of Total Probability:
$P(B_2) = P(W_1) \times P(B_2|W_1) + P(B_1) \times P(B_2|B_1)$
$P(B_2) = (\frac{5}{8} \times \frac{4}{7}) + (\frac{3}{8} \times \frac{5}{7})$
$P(B_2) = \frac{20}{56} + \frac{15}{56} = \frac{35}{56} = \frac{5}{8}$.

(B) The probability of winning a prize with a single ticket is $p = \frac{4}{16} = \frac{1}{4}$.
The probability of not winning a prize with a single ticket is $q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}$.
For $n = 4$ tickets,the probability of winning at least one prize is $1 - P(\text{winning zero prizes})$.
$P(\text{winning zero prizes}) = q^4 = \left(\frac{3}{4}\right)^4 = \frac{81}{256}$.
Therefore,the probability of winning at least one prize is $1 - \frac{81}{256} = \frac{256 - 81}{256} = \frac{175}{256}$.

(C) The total number of cards is $52$.
Let $S$ be the event that the card drawn is a spade,and $A$ be the event that the card drawn is an ace.
Number of spades $= 13$,so $P(S) = \frac{13}{52}$.
Number of aces $= 4$,so $P(A) = \frac{4}{52}$.
The card that is both a spade and an ace is the ace of spades,so $P(S \cap A) = \frac{1}{52}$.
The probability of winning the bet (drawing a spade or an ace) is $P(S \cup A) = P(S) + P(A) - P(S \cap A) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$.
The probability of losing the bet is $1 - P(S \cup A) = 1 - \frac{4}{13} = \frac{9}{13}$.
Odds against winning are defined as the ratio of the probability of losing to the probability of winning.
Odds against $= \frac{P(\text{losing})}{P(\text{winning})} = \frac{9/13}{4/13} = \frac{9}{4}$,which is $9$ to $4$.

(C) Let $P(A)$,$P(B)$,and $P(C)$ be the probabilities of hitting the target for $A$,$B$,and $C$ respectively.
$P(A) = \frac{4}{5}$,$P(A') = \frac{1}{5}$
$P(B) = \frac{3}{4}$,$P(B') = \frac{1}{4}$
$P(C) = \frac{2}{3}$,$P(C') = \frac{1}{3}$
'At least two shots hit' means either exactly two hit or all three hit.
Probability = $P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A' \cap B \cap C) + P(A \cap B \cap C)$
$= (\frac{4}{5} \times \frac{3}{4} \times \frac{1}{3}) + (\frac{4}{5} \times \frac{1}{4} \times \frac{2}{3}) + (\frac{1}{5} \times \frac{3}{4} \times \frac{2}{3}) + (\frac{4}{5} \times \frac{3}{4} \times \frac{2}{3})$
$= \frac{12}{60} + \frac{8}{60} + \frac{6}{60} + \frac{24}{60} = \frac{50}{60} = \frac{5}{6}$

(D) leap year consists of $366$ days.
$366$ days are equivalent to $52$ weeks and $2$ extra days.
These $2$ extra days can be any of the following pairs: (Sunday,Monday),(Monday,Tuesday),(Tuesday,Wednesday),(Wednesday,Thursday),(Thursday,Friday),(Friday,Saturday),or (Saturday,Sunday).
There are $7$ possible outcomes for these $2$ extra days.
For the year to have $53$ Sundays,one of these $2$ extra days must be a Sunday.
From the list of pairs,the favorable outcomes are (Sunday,Monday) and (Saturday,Sunday).
Thus,there are $2$ favorable outcomes out of $7$ total possibilities.
Therefore,the required probability is $\frac{2}{7}$.

(B) The total number of possible outcomes when three $6$-faced dice are thrown is $6^3 = 216$.
The number of favorable outcomes where all three dice show the same number is $6$,which are: ${(1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6)}$.
Therefore,the required probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{216} = \frac{1}{36}$.

(B) When three $6$-faced dice are thrown,the total number of possible outcomes is $6 \times 6 \times 6 = 6^3 = 216$.
To find the number of favorable outcomes where no two dice show the same number:
The first die can show any of the $6$ numbers ($6$ choices).
The second die must show a number different from the first die ($5$ choices).
The third die must show a number different from both the first and second dice ($4$ choices).
Therefore,the number of favorable outcomes is $6 \times 5 \times 4 = 120$.
The required probability is $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{120}{216} = \frac{5}{9}$.

(B) The total number of outcomes when three $6$-faced dice are thrown is $6 \times 6 \times 6 = 216$.
To find the number of outcomes where exactly two dice show the same number,we follow these steps:
$1$. Choose which two dice show the same number: There are $^3C_2 = 3$ ways to choose the two dice.
$2$. Choose the number that appears on these two dice: There are $6$ choices $(1, 2, 3, 4, 5, 6)$.
$3$. Choose the number for the third die: It must be different from the number chosen for the first two dice,so there are $5$ choices.
Total favourable outcomes $= 3 \times 6 \times 5 = 90$.
Required probability $= \frac{90}{216} = \frac{5}{12}$.

(C) Total number of components = $10$.
Number of defective components = $4$.
Number of non-defective components = $10 - 4 = 6$.
We need to select $3$ components such that exactly $2$ are defective and $1$ is non-defective.
The number of ways to choose $2$ defective components from $4$ is ${}^{4}C_{2} = \frac{4 \times 3}{2 \times 1} = 6$.
The number of ways to choose $1$ non-defective component from $6$ is ${}^{6}C_{1} = 6$.
The total number of ways to choose $3$ components from $10$ is ${}^{10}C_{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
The required probability is $\frac{{}^{4}C_{2} \times {}^{6}C_{1}}{{}^{10}C_{3}} = \frac{6 \times 6}{120} = \frac{36}{120} = \frac{3}{10}$.

(C) standard deck of cards contains $52$ cards in total.
Face cards are defined as Kings,Queens,and Jacks.
There are $3$ types of face cards in each of the $4$ suits (Hearts,Diamonds,Clubs,Spades).
Total number of face cards $= 3 \times 4 = 12$.
Probability of drawing a face card $= \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{12}{52}$.
Simplifying the fraction,we get $\frac{12}{52} = \frac{3}{13}$.

(B) Total number of cards in a deck $= 52$.
Number of black cards (spades and clubs) $= 26$.
Number of cards that are not black (red cards) $= 52 - 26 = 26$.
The probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
Required Probability $= \frac{\text{Number of non-black cards}}{\text{Total number of cards}} = \frac{26}{52} = \frac{1}{2}$.

(B) The total number of tickets is $30$,so the sample space $S = \{1, 2, 3, \ldots, 30\}$ and $n(S) = 30$.
Let $A$ be the event of drawing a ticket with a number that is a multiple of $5$.
$A = \{5, 10, 15, 20, 25, 30\}$,so $n(A) = 6$.
Let $B$ be the event of drawing a ticket with a number that is a multiple of $7$.
$B = \{7, 14, 21, 28\}$,so $n(B) = 4$.
Since $A$ and $B$ are mutually exclusive events (there are no common multiples of $5$ and $7$ up to $30$),the probability of $A$ or $B$ is given by $P(A \cup B) = P(A) + P(B)$.
$P(A) = \frac{n(A)}{n(S)} = \frac{6}{30}$.
$P(B) = \frac{n(B)}{n(S)} = \frac{4}{30}$.
Therefore,$P(A \cup B) = \frac{6}{30} + \frac{4}{30} = \frac{10}{30} = \frac{1}{3}$.

(A) The total number of possible outcomes when tossing $10$ coins is $2^{10} = 1024$.
The number of favorable outcomes for getting exactly $5$ heads is given by the combination formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$,where $n=10$ and $r=5$.
${}^{10}C_{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$.
The required probability is $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{252}{1024}$.
Simplifying the fraction by dividing both numerator and denominator by $4$,we get $\frac{252 \div 4}{1024 \div 4} = \frac{63}{256}$.

(A) The formula for the union of two events is given by: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Given values are $P(A) = \frac{2}{3}$,$P(B) = \frac{4}{9}$,and $P(A \cap B) = \frac{14}{45}$.
Substituting these values into the formula:
$P(A \cup B) = \frac{2}{3} + \frac{4}{9} - \frac{14}{45}$.
To add and subtract these fractions,find the least common multiple $(LCM)$ of the denominators $3, 9, 45$,which is $45$.
$P(A \cup B) = \frac{2 \times 15}{45} + \frac{4 \times 5}{45} - \frac{14}{45}$.
$P(A \cup B) = \frac{30}{45} + \frac{20}{45} - \frac{14}{45}$.
$P(A \cup B) = \frac{30 + 20 - 14}{45} = \frac{36}{45}$.
Simplifying the fraction by dividing both numerator and denominator by $9$,we get:
$P(A \cup B) = \frac{4}{5}$.

(B) The probability of getting a $6$ in a single throw is $p = \frac{1}{6}$.
The probability of not getting a $6$ in a single throw is $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.
For $n = 6$ independent throws, the probability of not getting a $6$ in any of the throws is $q^n = (\frac{5}{6})^{6}$.
The probability of getting a $6$ at least once is given by $1 - P(\text{no } 6) = 1 - (\frac{5}{6})^{6}$.

(B) The set of given digits is ${2, 3, 5, 7, 9}$. The total number of digits is $n = 5$.
To form a two-digit number without repetition,we need to select and arrange $2$ digits out of $5$.
The total number of possible outcomes (sample space $S$) is given by the permutation formula $^nP_r = \frac{n!}{(n-r)!}$.
Here,$n(S) = ^5P_2 = \frac{5!}{(5-2)!} = 5 \times 4 = 20$.
The event $E$ is that the number formed is $35$. Since $35$ is a specific outcome,$n(E) = 1$.
The probability $P(E)$ is given by $\frac{n(E)}{n(S)} = \frac{1}{20}$.

(C) The total number of outcomes when throwing two dice is $n(S) = 6 \times 6 = 36$.
Let $E$ be the event that the sum of the numbers on the two dice is $12$. The only outcome for this event is $(6, 6)$,so $n(E) = 1$.
The probability of getting a sum of $12$ is $P(E) = \frac{n(E)}{n(S)} = \frac{1}{36}$.
The event of getting a sum less than $12$ is the complement of the event of getting a sum of $12$,denoted as $\overline{E}$.
Therefore,the required probability is $P(\overline{E}) = 1 - P(E) = 1 - \frac{1}{36} = \frac{35}{36}$.

(B) The total number of tickets is $n(S) = 21$,where $S = \{1, 2, 3, \ldots, 21\}$.
The numbers divisible by $3$ between $1$ and $21$ are $E = \{3, 6, 9, 12, 15, 18, 21\}$.
The number of favorable outcomes is $n(E) = 7$.
The probability $P(E)$ is given by the ratio of favorable outcomes to total outcomes: $P(E) = \frac{n(E)}{n(S)} = \frac{7}{21} = \frac{1}{3}$.

(C) Given: In the first bag,the number of white balls $= 4$ and the number of black balls $= 2$. Total balls in the first bag $= 4 + 2 = 6$.
In the second bag,the number of white balls $= 3$ and the number of black balls $= 5$. Total balls in the second bag $= 3 + 5 = 8$.
We know that the probability of drawing one black ball from the first bag $= \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{2}{6} = \frac{1}{3}$.
Similarly,the probability of drawing one black ball from the second bag $= \frac{5}{8}$.
Since the events are independent,the probability that both balls are black $= P(\text{Black from bag } 1) \times P(\text{Black from bag } 2) = \frac{1}{3} \times \frac{5}{8} = \frac{5}{24}$.

(D) Total number of red balls $= 3$.
Total number of black balls $= 7$.
Total number of balls $= 3 + 7 = 10$.
We are given that the first ball drawn is red.
Since the balls are drawn without replacement,after drawing one red ball,the number of remaining red balls $= 3 - 1 = 2$.
The total number of remaining balls $= 10 - 1 = 9$.
The probability that the second ball is red,given that the first ball was red,is the ratio of the number of remaining red balls to the total number of remaining balls.
Required probability $= \frac{2}{9}$.

(D) Total number of balls $= 5 + 7 + 4 = 16$.
Number of ways to draw $3$ balls from $16$ balls is given by $^{16}C_{3} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 16 \times 5 \times 7 = 560$.
Number of ways to draw $3$ white balls from $5$ white balls is given by $^{5}C_{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$.
The probability that all three balls are white is $\frac{^{5}C_{3}}{^{16}C_{3}} = \frac{10}{560} = \frac{1}{56}$.

(C) Total number of balls $= 2 + 3 + 4 = 9$.
The total number of ways to draw $3$ balls from $9$ is given by ${}^{9}C_{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
Since there are only $2$ red balls,it is impossible to draw $3$ red balls.
The number of ways to draw $3$ black balls from $3$ is ${}^{3}C_{3} = 1$.
The number of ways to draw $3$ white balls from $4$ is ${}^{4}C_{3} = 4$.
Therefore,the number of favorable outcomes is ${}^{3}C_{3} + {}^{4}C_{3} = 1 + 4 = 5$.
The required probability is $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{5}{84}$.

(C) Total number of balls $= 9 + 7 + 4 = 20$.
The number of ways to choose $2$ balls out of $20$ is given by $^{20}C_{2} = \frac{20 \times 19}{2 \times 1} = 190$.
The number of ways to draw one white ball out of $7$ is $^{7}C_{1} = 7$.
The number of ways to draw one red ball out of $9$ is $^{9}C_{1} = 9$.
The number of favorable outcomes is $^{7}C_{1} \times ^{9}C_{1} = 7 \times 9 = 63$.
Therefore,the required probability $= \frac{63}{190}$.

(B) Total number of balls $= 6 + 4 + 8 = 18$.
Number of balls drawn $= 3$.
The total number of ways to draw $3$ balls from $18$ is given by $^{18}C_{3} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 3 \times 17 \times 16 = 816$.
The number of ways to draw $1$ red ball from $6$ is $^{6}C_{1} = 6$.
The number of ways to draw $2$ white balls from $4$ is $^{4}C_{2} = \frac{4 \times 3}{2 \times 1} = 6$.
The number of favorable outcomes $= ^{6}C_{1} \times ^{4}C_{2} = 6 \times 6 = 36$.
Required probability $= \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{36}{816}$.
Dividing both numerator and denominator by $12$,we get $\frac{36 \div 12}{816 \div 12} = \frac{3}{68}$.

(D) Total number of marbles $= 3 + 7 + 10 = 20$.
The number of favorable outcomes (choosing $1$ white marble) $= 10$.
The total number of possible outcomes (choosing $1$ marble from $20$) $= 20$.
Probability of drawing a white marble $= \frac{\text{Number of white marbles}}{\text{Total number of marbles}} = \frac{10}{20} = \frac{1}{2}$.

(D) When a balanced coin is tossed $3$ times,the total number of possible outcomes is $2^3 = 8$.
The sample space $S$ is given by: $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
The favorable outcome for getting heads in all three trials is only one: $\{HHH\}$.
Therefore,the required probability $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{8}$.

(A) When two dice are thrown, the total number of possible outcomes is $6 \times 6 = 36$.
The condition 'sum is at least $10$' means the sum can be $10, 11,$ or $12$.
Let us list the favorable outcomes for each sum:
For sum $= 10$: $(4, 6), (5, 5), (6, 4)$ (Total $3$ outcomes)
For sum $= 11$: $(5, 6), (6, 5)$ (Total $2$ outcomes)
For sum $= 12$: $(6, 6)$ (Total $1$ outcome)
Total number of favorable outcomes $= 3 + 2 + 1 = 6$.
The probability is given by the ratio of favorable outcomes to total outcomes:
$P(\text{sum} \ge 10) = \frac{6}{36} = \frac{1}{6}$.

(D) Total number of tickets $= 20$.
Multiples of $3$ between $1$ and $20$ are ${3, 6, 9, 12, 15, 18}$. There are $6$ such tickets.
Multiples of $5$ between $1$ and $20$ are ${5, 10, 15, 20}$. There are $4$ such tickets.
The number $15$ is common to both sets.
Using the inclusion-exclusion principle,the number of favorable outcomes $= 6 + 4 - 1 = 9$.
Therefore,the required probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{9}{20}$.

(A) Total number of balls $= 2 + 3 + 2 = 7$.
Let $S$ be the sample space.
Then,$N(S) =$ Number of ways of drawing $2$ balls out of $7 = {}^{7}C_{2} = \frac{7 \times 6}{2 \times 1} = 21$.
Let $E$ be the event of drawing two balls,none of which is blue.
This means the $2$ balls must be chosen from the $(2 + 3) = 5$ non-blue balls (red and green).
Therefore,$N(E) =$ Number of ways of drawing $2$ balls out of $5 = {}^{5}C_{2} = \frac{5 \times 4}{2 \times 1} = 10$.
$P(E) = \frac{N(E)}{N(S)} = \frac{10}{21}$.

(A) Total number of balls $= 8 + 7 + 6 = 21$.
Let $E$ be the event that the ball drawn is neither red nor green.
Since the ball is neither red nor green,it must be blue.
Number of blue balls $= 7$.
Therefore,the number of favorable outcomes $N(E) = 7$.
The total number of possible outcomes $N(S) = 21$.
The probability $P(E) = \frac{N(E)}{N(S)} = \frac{7}{21} = \frac{1}{3}$.