Probability Questions in English

(C) When $4$ coins are tossed,the total number of possible outcomes is $2^4 = 16$.
The number of ways to get exactly $2$ tails out of $4$ coins is given by the combination formula $^nC_r$,where $n=4$ and $r=2$.
$^4C_2 = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6$.
The favourable outcomes are: $HHTT, HTHT, HTTH, THHT, THTH, TTHH$.
Therefore,the probability of getting exactly $2$ tails is $P = \frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \frac{6}{16} = \frac{3}{8}$.

(B) When $4$ coins are tossed, the total number of possible outcomes is $2^4 = 16$.
The event 'at least $1$ tail' is the complement of the event 'no tails' (i.e., all heads).
The only outcome with no tails is $(H, H, H, H)$, so there is only $1$ such outcome.
$P(\text{all heads}) = \frac{1}{16}$.
$P(\text{at least } 1 \text{ tail}) = 1 - P(\text{all heads})$.
$P(\text{at least } 1 \text{ tail}) = 1 - \frac{1}{16} = \frac{15}{16}$.

(C) When $2$ dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
Let $E$ be the event of getting a total of $3$ or $5$.
The outcomes resulting in a total of $3$ are: $(1, 2)$ and $(2, 1)$.
The outcomes resulting in a total of $5$ are: $(1, 4), (2, 3), (3, 2),$ and $(4, 1)$.
Total favorable outcomes $= 2 + 4 = 6$.
Probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}$.

(A) When $2$ dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
The favorable outcome to get a total of $12$ is only $(6, 6)$.
Thus,the number of favorable outcomes is $1$.
Therefore,the required probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{36}$.

(B) When $2$ dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
The outcomes that result in a sum of $11$ are $(5, 6)$ and $(6, 5)$.
Thus,the number of favorable outcomes is $2$.
The probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
$\text{Required probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{36} = \frac{1}{18}$.

(A) When $2$ dice are thrown,the total number of possible outcomes is $6 \times 6 = 36.$
$A$ total of $8$ can be obtained in the following $5$ ways: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2).$
Therefore,the number of favorable outcomes is $5.$
The required probability is given by the formula: $\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}.$
$\text{Probability} = \frac{5}{36}.$

(C) When $2$ dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
The favorable outcomes for getting a total of $7$ are: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$.
There are $6$ such favorable outcomes.
Therefore,the required probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}$.

(A) 'doublet' means that both the dice show the same number on the uppermost faces.
When $2$ dice are thrown, the total number of possible outcomes is $6 \times 6 = 36$.
The outcomes favourable to the event of getting a doublet are:
$(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$
Thus, the number of favourable cases $= 6$.
Therefore, the probability of getting a doublet is:
$P(\text{doublet}) = \frac{\text{Number of favourable cases}}{\text{Total number of possible outcomes}} = \frac{6}{36} = \frac{1}{6}$.

(C) When $2$ dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
Let $A$ be the event of getting a multiple of $2$ on the first die: $A = \{2, 4, 6\}$.
Let $B$ be the event of getting a multiple of $3$ on the second die: $B = \{3, 6\}$.
We need to find the probability of getting a multiple of $2$ on one die and a multiple of $3$ on the other.
Let $E_1$ be the set of outcomes where the first die is a multiple of $2$ and the second is a multiple of $3$: $E_1 = \{(2,3), (2,6), (4,3), (4,6), (6,3), (6,6)\}$.
Let $E_2$ be the set of outcomes where the first die is a multiple of $3$ and the second is a multiple of $2$: $E_2 = \{(3,2), (3,4), (3,6), (6,2), (6,4), (6,6)\}$.
The union $E_1 \cup E_2$ represents the favourable outcomes. Note that $(6,6)$ is common to both sets.
$E_1 \cup E_2 = \{(2,3), (2,6), (4,3), (4,6), (6,3), (6,6), (3,2), (3,4), (6,2), (6,4)\}$.
Counting the unique elements,the number of favourable outcomes is $11$.
Therefore,the required probability is $\frac{11}{36}$.

(C) The total number of possible outcomes when throwing $2$ dice is $6 \times 6 = 36$.
Let $A$ be the event of getting an odd number on the first die and a multiple of $3$ on the second die.
Odd numbers on a die are ${1, 3, 5}$. Multiples of $3$ on a die are ${3, 6}$.
Possible outcomes for $(A, B)$ where $A \in {1, 3, 5}$ and $B \in {3, 6}$ are: $(1,3), (1,6), (3,3), (3,6), (5,3), (5,6)$. (Total $6$ outcomes).
Let $C$ be the event of getting a multiple of $3$ on the first die and an odd number on the second die.
Possible outcomes for $(C, D)$ where $C \in {3, 6}$ and $D \in {1, 3, 5}$ are: $(3,1), (3,3), (3,5), (6,1), (6,3), (6,5)$. (Total $6$ outcomes).
The outcome $(3,3)$ is common to both sets.
Using the inclusion-exclusion principle,the total number of unique favourable outcomes is $6 + 6 - 1 = 11$.
The favourable outcomes are: $(1,3), (1,6), (3,3), (3,6), (5,3), (5,6), (3,1), (3,5), (6,1), (6,3), (6,5)$.
Therefore,the required probability is $\frac{11}{36}$.

(C) When $2$ dice are thrown,the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.
$A$ doublet of even numbers means both dice show the same even number.
The possible outcomes for this event $A$ are: $A = \{(2, 2), (4, 4), (6, 6)\}$.
The number of favorable outcomes is $n(A) = 3$.
The probability $P(A)$ is given by the formula: $P(A) = \frac{n(A)}{n(S)}$.
Substituting the values: $P(A) = \frac{3}{36} = \frac{1}{12}$.

(B) When $2$ dice are thrown,the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.
Let $A$ be the event of getting a sum less than $6$.
The possible outcomes for the sum being less than $6$ are:
Sum $= 2: (1, 1)$
Sum $= 3: (1, 2), (2, 1)$
Sum $= 4: (1, 3), (2, 2), (3, 1)$
Sum $= 5: (1, 4), (2, 3), (3, 2), (4, 1)$
Counting these,we get $n(A) = 1 + 2 + 3 + 4 = 10$.
The probability $P(A) = \frac{n(A)}{n(S)} = \frac{10}{36} = \frac{5}{18}$.

(D) When $2$ dice are thrown,the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.
Let $A$ be the event of getting a sum more than $7$. The possible outcomes are:
Sum $= 8: (2,6), (3,5), (4,4), (5,3), (6,2)$ ($5$ outcomes)
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 $n(A) = 5 + 4 + 3 + 2 + 1 = 15$.
Therefore,the required probability $P(A) = \frac{n(A)}{n(S)} = \frac{15}{36} = \frac{5}{12}$.

(A) When $2$ dice are thrown,the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.
Let $A$ be the event of getting a sum greater than $10$. The possible outcomes for this event are $(5, 6), (6, 5),$ and $(6, 6)$.
Thus,the number of favorable outcomes is $n(A) = 3$.
The probability of the event $A$ is given by $P(A) = \frac{n(A)}{n(S)}$.
$P(A) = \frac{3}{36} = \frac{1}{12}$.

(B) When $2$ dice are thrown,the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.
Let $A$ be the event of getting a sum of at least $10$ (i.e.,sum is $10, 11,$ or $12$).
The favorable outcomes are:
For sum $= 10$: $(4, 6), (5, 5), (6, 4)$
For sum $= 11$: $(5, 6), (6, 5)$
For sum $= 12$: $(6, 6)$
Thus,the set of favorable outcomes is $A = \{(4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)\}$.
The number of favorable outcomes is $n(A) = 6$.
The probability is $P(A) = \frac{n(A)}{n(S)} = \frac{6}{36} = \frac{1}{6}$.

(D) When $2$ dice are thrown,the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.
Let $A$ be the event of getting an odd number as the sum.
The sum of two dice is odd if one die shows an even number and the other shows an odd number.
The possible outcomes for an odd sum are:
$(1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5)$.
Counting these,we get $n(A) = 18$.
The probability of the event $A$ is given by $P(A) = \frac{n(A)}{n(S)}$.
$P(A) = \frac{18}{36} = \frac{1}{2}$.

(C) When two dice are thrown,the total number of outcomes is $n(S) = 6 \times 6 = 36$.
The sum of the numbers on the two dice is even if both numbers are even or both numbers are odd.
Let $A$ be the event of getting an even sum.
The outcomes resulting in an even sum are:
$(1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6)$.
Counting these,we get $n(A) = 18$.
The probability $P(A) = \frac{n(A)}{n(S)} = \frac{18}{36} = \frac{1}{2}$.

(A) When $2$ dice are thrown,the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.
Let $A$ be the event of getting $6$ as the product of the numbers on the two dice.
The pairs $(x, y)$ such that $x \times y = 6$ are $(1, 6), (6, 1), (2, 3),$ and $(3, 2)$.
Thus,the number of favorable outcomes is $n(A) = 4$.
The probability of the event $A$ is given by $P(A) = \frac{n(A)}{n(S)}$.
$P(A) = \frac{4}{36} = \frac{1}{9}$.

(B) When $2$ dice are thrown,the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.
Let $A$ be the event of getting a multiple of $3$ as the sum. The possible sums are $3, 6, 9, 12$.
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 favorable outcomes $n(A) = 2 + 5 + 4 + 1 = 12$.
Therefore,the required probability $P(A) = \frac{n(A)}{n(S)} = \frac{12}{36} = \frac{1}{3}$.

(B) When $2$ dice are thrown,the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.
Let $A$ be the event of getting a product that is a perfect square.
The pairs $(x, y)$ such that $x \times y$ is a perfect square are:
$(1, 1) \rightarrow 1 \times 1 = 1 = 1^2$
$(1, 4) \rightarrow 1 \times 4 = 4 = 2^2$
$(2, 2) \rightarrow 2 \times 2 = 4 = 2^2$
$(3, 3) \rightarrow 3 \times 3 = 9 = 3^2$
$(4, 1) \rightarrow 4 \times 1 = 4 = 2^2$
$(4, 4) \rightarrow 4 \times 4 = 16 = 4^2$
$(5, 5) \rightarrow 5 \times 5 = 25 = 5^2$
$(6, 6) \rightarrow 6 \times 6 = 36 = 6^2$
Thus,the set of favorable outcomes is $A = \{(1, 1), (1, 4), (2, 2), (3, 3), (4, 1), (4, 4), (5, 5), (6, 6)\}$.
The number of favorable outcomes is $n(A) = 8$.
The probability $P(A) = \frac{n(A)}{n(S)} = \frac{8}{36} = \frac{2}{9}$.

(C) When two dice are thrown,the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.
Let $A$ be the event of getting at least one of the two numbers as $4$.
The favorable outcomes are: $(1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (4,1), (4,2), (4,3), (4,5), (4,6)$.
Counting these,we get $n(A) = 11$.
The probability $P(A)$ is given by $\frac{n(A)}{n(S)} = \frac{11}{36}$.

(A) When $2$ dice are thrown,the total number of outcomes is $n(S) = 6 \times 6 = 36$.
The possible sums range from $2$ to $12$. The prime numbers in this range are $2, 3, 5, 7, 11$.
Let $A$ be the event of getting a sum as a prime number:
- Sum $= 2$: $(1, 1)$ $\rightarrow 1$ outcome
- Sum $= 3$: $(1, 2), (2, 1)$ $\rightarrow 2$ outcomes
- Sum $= 5$: $(1, 4), (4, 1), (2, 3), (3, 2)$ $\rightarrow 4$ outcomes
- Sum $= 7$: $(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)$ $\rightarrow 6$ outcomes
- Sum $= 11$: $(5, 6), (6, 5)$ $\rightarrow 2$ outcomes
Total favorable outcomes $n(A) = 1 + 2 + 4 + 6 + 2 = 15$.
Therefore,the required probability $P(A) = \frac{n(A)}{n(S)} = \frac{15}{36} = \frac{5}{12}$.

(A) The total number of possible outcomes when throwing $3$ dice is $6 \times 6 \times 6 = 216$.
To get a total of $18$,the only possible outcome is $(6, 6, 6)$,which is $1$ case.
To get a total of $17$,the possible outcomes are $(5, 6, 6), (6, 5, 6), (6, 6, 5)$,which are $3$ cases.
Therefore,the total number of favorable outcomes for a sum of $17$ or $18$ is $1 + 3 = 4$.
The probability is given by $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{216} = \frac{1}{54}$.

(C) The total number of possible outcomes when throwing $3$ dice is $6 \times 6 \times 6 = 216$.
We need to find the number of outcomes where the sum of the numbers on the $3$ dice is $5$.
The possible combinations are:
$(1, 1, 3)$ which can be arranged in $3$ ways: $(1, 1, 3), (1, 3, 1), (3, 1, 1)$.
$(1, 2, 2)$ which can be arranged in $3$ ways: $(1, 2, 2), (2, 1, 2), (2, 2, 1)$.
Total favorable outcomes $= 3 + 3 = 6$.
Therefore,the probability $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{216} = \frac{1}{36}$.

(A) The total number of outcomes when throwing $3$ dice is $6^3 = 216$.
$A$ total of at most $5$ means the sum of the numbers on the $3$ dice is $3, 4,$ or $5$.
Cases for a total of $3$: $(1, 1, 1)$ - $1$ case.
Cases for a total of $4$: $(1, 1, 2), (1, 2, 1), (2, 1, 1)$ - $3$ cases.
Cases for a total of $5$: $(1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1)$ - $6$ cases.
Total favourable cases = $1 + 3 + 6 = 10$.
Therefore,the probability of getting a total of at most $5$ is $\frac{10}{216} = \frac{5}{108}$.

(C) The total number of outcomes when throwing $3$ dice is $6^3 = 216$.
Let $E$ be the event of getting a total of at least $5$. It is easier to calculate the probability of the complement event $E'$,which is getting a total of less than $5$ (i.e.,a total of $3$ or $4$).
Possible outcomes for a total of $3$: $(1, 1, 1)$ - $1$ case.
Possible outcomes for a total of $4$: $(1, 1, 2), (1, 2, 1), (2, 1, 1)$ - $3$ cases.
Total outcomes for $E'$ (sum is $3$ or $4$) $= 1 + 3 = 4$.
$P(E') = \frac{4}{216} = \frac{1}{54}$.
Therefore,$P(E) = 1 - P(E') = 1 - \frac{1}{54} = \frac{53}{54}$.

(B) leap year consists of $366$ days,which is equivalent to $52$ weeks and $2$ extra days.
The possible combinations for these $2$ extra days are:
$(i)$ Sunday and Monday
$(ii)$ Monday and Tuesday
$(iii)$ Tuesday and Wednesday
$(iv)$ Wednesday and Thursday
$(v)$ Thursday and Friday
$(vi)$ Friday and Saturday
$(vii)$ Saturday and Sunday
Total number of exhaustive cases $n = 7$.
For the year to have $53$ Sundays,one of the $2$ extra days must be a Sunday. The favourable cases are $(i)$ (Sunday and Monday) and $(vii)$ (Saturday and Sunday).
Number of favourable cases $m = 2$.
Therefore,the probability of having $53$ Sundays is $P = m/n = 2/7$.

(A) Let $S$ be the sample space of drawing a card from $100$ cards. Thus,$n(S) = 100$.
Let $A$ be the event of drawing a number which is a perfect square.
The perfect squares between $1$ and $100$ are: $1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64, 9^2=81, 10^2=100$.
So,$A = \{1, 4, 9, 16, 25, 36, 49, 64, 81, 100\}$.
The number of favorable outcomes is $n(A) = 10$.
The probability $P(A)$ is given by $\frac{n(A)}{n(S)} = \frac{10}{100} = \frac{1}{10}$.

(C) The word $SOCIETY$ contains $7$ distinct letters: $S, O, C, I, E, T, Y$. The total number of ways to arrange these $7$ letters is $7! = 5040$.
There are $3$ vowels in the word: $O, I, E$. To find the number of arrangements where these $3$ vowels come together,we treat the group $(O, I, E)$ as a single unit.
Now,we have $5$ units to arrange: $(OIE), S, C, T, Y$. These $5$ units can be arranged in $5!$ ways.
The $3$ vowels within the group can be arranged among themselves in $3!$ ways.
Therefore,the total number of favorable arrangements $= 5! \times 3! = 120 \times 6 = 720$.
The required probability $= \frac{\text{Favorable arrangements}}{\text{Total arrangements}} = \frac{5! \times 3!}{7!} = \frac{120 \times 6}{5040} = \frac{720}{5040} = \frac{1}{7}$.

(A) The word $'UNIVERSITY'$ contains $10$ letters, where the letter $I$ appears $2$ times.
Total number of permutations $= \frac{10!}{2!}$.
To find the number of arrangements where two $I$s are together, we treat the two $I$s as a single unit. Now, we have $9$ units (the block $(II)$ and the other $8$ letters), which can be arranged in $9!$ ways.
Number of arrangements where two $I$s are together $= 9!$.
Number of arrangements where two $I$s are not together $= \text{Total arrangements} - \text{Arrangements where } I\text{s are together}$.
$= \frac{10!}{2} - 9! = \frac{10 \times 9!}{2} - 9! = 9!(5 - 1) = 4 \times 9!$.
Required probability $= \frac{4 \times 9!}{\frac{10!}{2}} = \frac{4 \times 9! \times 2}{10!} = \frac{8 \times 9!}{10 \times 9!} = \frac{8}{10} = \frac{4}{5}$.

(C) The word $PENCIL$ contains $6$ distinct letters. The total number of ways to arrange these $6$ letters is $6! = 720$.
To find the number of arrangements where $N$ is next to $E$,we treat $(NE)$ as a single unit. Now,we have $5$ units to arrange: $(NE), P, C, I, L$. These $5$ units can be arranged in $5! = 120$ ways.
Within the unit $(NE)$,the letters $N$ and $E$ can be arranged in $2! = 2$ ways (i.e.,$NE$ or $EN$).
Therefore,the total number of favorable arrangements is $5! \times 2! = 120 \times 2 = 240$.
The probability is $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{240}{720} = \frac{1}{3}$.

(A) When $2$ dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
Let $A$ be the event of getting a sum of $4$.
The possible outcomes for the sum $4$ are $(1, 3), (3, 1), (2, 2)$.
Thus,the number of favourable outcomes is $n(A) = 3$.
The number of unfavourable outcomes is $36 - 3 = 33$.
The odds in favour of an event are given by the ratio of favourable outcomes to unfavourable outcomes.
Therefore,the odds in favour of getting a sum of $4 = \frac{3}{33} = \frac{1}{11}$,which is expressed as $1:11$.

(B) When $2$ dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
Let $A$ be the event of getting a sum of $5$.
The favorable outcomes are $(1, 4), (4, 1), (2, 3), (3, 2)$.
Number of favorable outcomes $n(A) = 4$.
Number of unfavorable outcomes = Total outcomes - Favorable outcomes = $36 - 4 = 32$.
The odds in favour of an event are defined as the ratio of the number of favorable outcomes to the number of unfavorable outcomes.
$\therefore \text{Odds in favour of sum } 5 = \frac{n(A)}{n(A^c)} = \frac{4}{32} = \frac{1}{8}$ or $1:8$.

(C) When $2$ dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
Let $A$ be the event of getting a sum of $6$.
The favorable outcomes for event $A$ are: $(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)$.
Thus,the number of favorable outcomes $n(A) = 5$.
The number of unfavorable outcomes is the total outcomes minus the favorable outcomes: $36 - 5 = 31$.
The odds against an event are defined as the ratio of unfavorable outcomes to favorable outcomes.
Therefore,the odds against getting the sum $6$ are $31:5$.

(B) Let $A$ be the event of drawing a queen and $B$ be the event of drawing an ace.
The total number of cards in a deck is $52$.
The number of queens in a deck is $4$,so $P(A) = \frac{4}{52}$.
The number of aces in a deck is $4$,so $P(B) = \frac{4}{52}$.
Since a card cannot be both a queen and an ace simultaneously,the events are mutually exclusive,meaning $P(A \cap B) = 0$.
The probability of drawing either a queen or an ace is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{4}{52} + \frac{4}{52} - 0 = \frac{8}{52}$.
Simplifying the fraction,we get $\frac{8}{52} = \frac{2}{13}$.

(C) Let $A$ be the event that the roll number is a multiple of $5$,and $B$ be the event that the roll number is a multiple of $7$.
The set of roll numbers from $1$ to $25$ is $S = \{1, 2, 3, ..., 25\}$,so the total number of outcomes is $n(S) = 25$.
The multiples of $5$ in the range are $A = \{5, 10, 15, 20, 25\}$,so $n(A) = 5$.
The multiples of $7$ in the range are $B = \{7, 14, 21\}$,so $n(B) = 3$.
Since there are no common multiples of $5$ and $7$ in the range $1$ to $25$,$A \cap B = \emptyset$,so $n(A \cap B) = 0$.
The probability of the union of two events is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A) = \frac{5}{25}$,$P(B) = \frac{3}{25}$,and $P(A \cap B) = 0$.
Therefore,the required probability is $P(A \cup B) = \frac{5}{25} + \frac{3}{25} - 0 = \frac{8}{25}$.

(A) Let $A$ be the event that the chosen integer is divisible by $6$,and $B$ be the event that the chosen integer is divisible by $8$.
The total number of outcomes is $n(S) = 200$.
The number of integers divisible by $6$ is $n(A) = \lfloor \frac{200}{6} \rfloor = 33$.
The number of integers divisible by $8$ is $n(B) = \lfloor \frac{200}{8} \rfloor = 25$.
The integers divisible by both $6$ and $8$ are divisible by $\text{lcm}(6, 8) = 24$. The number of such integers is $n(A \cap B) = \lfloor \frac{200}{24} \rfloor = 8$.
Using the inclusion-exclusion principle,the number of integers divisible by $6$ or $8$ is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 33 + 25 - 8 = 50$.
The probability is $P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{50}{200} = \frac{1}{4}$.

(C) When $2$ dice are rolled,the total number of outcomes is $6 \times 6 = 36$.
Let $A$ be the event that the sum is divisible by $3$. The possible sums are $3, 6, 9, 12$.
The outcomes for $A$ are: $(1,2), (2,1), (1,5), (5,1), (2,4), (4,2), (3,3), (3,6), (6,3), (4,5), (5,4), (6,6)$.
Number of outcomes in $A$,$n(A) = 12$.
Let $B$ be the event that the sum is divisible by $4$. The possible sums are $4, 8, 12$.
The outcomes for $B$ are: $(1,3), (3,1), (2,2), (2,6), (6,2), (3,5), (5,3), (4,4), (6,6)$.
Number of outcomes in $B$,$n(B) = 9$.
The intersection $A \cap B$ is the event that the sum is divisible by both $3$ and $4$ (i.e.,divisible by $12$).
The only outcome with a sum of $12$ is $(6,6)$.
So,$n(A \cap B) = 1$.
The probability $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{12}{36} + \frac{9}{36} - \frac{1}{36} = \frac{20}{36} = \frac{5}{9}$.

(A) The total number of outcomes when throwing $2$ dice is $6 \times 6 = 36$.
Event $A$: Getting a total of $11$.
The favorable outcomes are $(5, 6)$ and $(6, 5)$.
Thus,$n(A) = 2$,and $P(A) = \frac{2}{36}$.
Event $B$: Getting an odd number on each die.
The odd numbers on a die are ${1, 3, 5}$.
The favorable outcomes are $(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)$.
Thus,$n(B) = 9$,and $P(B) = \frac{9}{36}$.
Intersection $A \cap B$: Getting a total of $11$ $AND$ an odd number on each die.
Since the sum of two odd numbers is always even,it is impossible to get a total of $11$ (which is odd) if both dice show odd numbers.
Thus,$A \cap B = \emptyset$,and $P(A \cap B) = 0$.
Using the addition rule of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) = \frac{2}{36} + \frac{9}{36} - 0 = \frac{11}{36}$.

(B) Let $A$ be the event that the number is divisible by $3$.
$A = \{3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36\}$,so $n(A) = 12$.
Let $B$ be the event that the number is a perfect square.
$B = \{1, 4, 9, 16, 25, 36\}$,so $n(B) = 6$.
The intersection $A \cap B$ contains numbers divisible by $3$ and are perfect squares.
$A \cap B = \{9, 36\}$,so $n(A \cap B) = 2$.
The total number of outcomes is $36$.
Using the addition rule for probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{12}{36} + \frac{6}{36} - \frac{2}{36} = \frac{16}{36} = \frac{4}{9}$.

(C) Total items $= 50 + 150 = 200$.
Number of rusted bolts $= 50 / 2 = 25$.
Number of rusted nuts $= 150 / 2 = 75$.
Total rusted items $= 25 + 75 = 100$.
Let $A$ be the event of choosing a rusted item and $B$ be the event of choosing a bolt.
$P(A) = 100 / 200 = 1/2$.
$P(B) = 50 / 200 = 1/4$.
$P(A \cap B)$ is the probability of choosing a rusted bolt,which is $25 / 200 = 1/8$.
Using the addition rule of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = 100/200 + 50/200 - 25/200 = 125/200 = 5/8$.

(A) Let $A$ be the event of getting a doublet and $B$ be the event of getting a total of $10$.
The sample space $S$ has $6 \times 6 = 36$ outcomes.
The set of doublets is $A = \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}$,so $n(A) = 6$.
The set of outcomes with a total of $10$ is $B = \{(4,6), (5,5), (6,4)\}$,so $n(B) = 3$.
The intersection $A \cap B = \{(5,5)\}$,so $n(A \cap B) = 1$.
The probability of $A \cup B$ is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{6}{36} + \frac{3}{36} - \frac{1}{36} = \frac{8}{36} = \frac{2}{9}$.
The probability that neither a doublet nor a total of $10$ appears is $1 - P(A \cup B) = 1 - \frac{2}{9} = \frac{7}{9}$.

(B) Let $A$ be the event of getting a sum of $9$ and $B$ be the event of getting a sum of $11$.
The sample space for throwing $2$ dice has $6 \times 6 = 36$ outcomes.
The outcomes for event $A$ (sum $= 9$) are: $(3, 6), (4, 5), (5, 4), (6, 3)$. So,$n(A) = 4$.
The outcomes for event $B$ (sum $= 11$) are: $(5, 6), (6, 5)$. So,$n(B) = 2$.
Since $A$ and $B$ are mutually exclusive events,$P(A \cup B) = P(A) + P(B) = \frac{4}{36} + \frac{2}{36} = \frac{6}{36} = \frac{1}{6}$.
The probability that the sum is neither $9$ nor $11$ is $1 - P(A \cup B) = 1 - \frac{1}{6} = \frac{5}{6}$.

(D) Let $A$ be the event of getting a spade card,$B$ be the event of getting an ace card,and $C$ be the event of getting a red card.
The total number of cards is $52$.
$P(A) = \frac{13}{52}$,$P(B) = \frac{4}{52}$,$P(C) = \frac{26}{52}$.
Intersection probabilities:
$P(A \cap B) = \frac{1}{52}$ (the ace of spades).
$P(B \cap C) = \frac{2}{52}$ (the ace of hearts and the ace of diamonds).
$P(C \cap A) = 0$ (a card cannot be both red and a spade).
$P(A \cap B \cap C) = 0$.
Using the inclusion-exclusion principle:
$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)$
$= \frac{13}{52} + \frac{4}{52} + \frac{26}{52} - \frac{1}{52} - \frac{2}{52} - 0 + 0$
$= \frac{13 + 4 + 26 - 1 - 2}{52} = \frac{40}{52} = \frac{10}{13}$.

(A) Let $A,$ $B,$ and $C$ be the events that the number is divisible by $2,$ $3,$ and $5$ respectively.
Total number of tickets $n(S) = 200.$
$n(A) = \lfloor \frac{200}{2} \rfloor = 100, n(B) = \lfloor \frac{200}{3} \rfloor = 66, n(C) = \lfloor \frac{200}{5} \rfloor = 40.$
$n(A \cap B) = \lfloor \frac{200}{LCM(2,3)} \rfloor = \lfloor \frac{200}{6} \rfloor = 33.$
$n(B \cap C) = \lfloor \frac{200}{LCM(3,5)} \rfloor = \lfloor \frac{200}{15} \rfloor = 13.$
$n(C \cap A) = \lfloor \frac{200}{LCM(5,2)} \rfloor = \lfloor \frac{200}{10} \rfloor = 20.$
$n(A \cap B \cap C) = \lfloor \frac{200}{LCM(2,3,5)} \rfloor = \lfloor \frac{200}{30} \rfloor = 6.$
Using the Principle of Inclusion-Exclusion:
$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)$
$n(A \cup B \cup C) = 100 + 66 + 40 - 33 - 13 - 20 + 6 = 146.$
Required probability $= \frac{n(A \cup B \cup C)}{n(S)} = \frac{146}{200} = \frac{73}{100}.$

(B) We know that $P(A) = 1 - P(\bar{A})$.
Given $P(\bar{A}) = 0.65$,so $P(A) = 1 - 0.65 = 0.35$.
For mutually exclusive events,$P(A \cap B) = 0$.
The addition rule for probability is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $0.65 = 0.35 + p - 0$.
Solving for $p$: $p = 0.65 - 0.35 = 0.30$.

(A) The probability that neither $A$ nor $B$ occurs is given by $P(A' \cap B') = 1 - P(A \cup B)$.
Since $A$ and $B$ are mutually exclusive events,$P(A \cap B) = 0$.
Therefore,$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.3 - 0 = 0.8$.
Now,the required probability is $1 - P(A \cup B) = 1 - 0.8 = 0.2$.

(C) To receive at least a $B$ grade, the student must receive either an $A$ grade or a $B$ grade.
$P(\text{at least } B \text{ grade}) = P(B \text{ grade}) + P(A \text{ grade})$
Given, $P(A \text{ grade}) = 0.30$ and $P(B \text{ grade}) = 0.38$.
$P(\text{at least } B \text{ grade}) = 0.38 + 0.30 = 0.68$.

(C) Let $A$ be the event that the contractor gets a plumbing contract and $B$ be the event that he gets an electric contract.
Given: $P(A) = 2/3$.
The probability that he will not get an electric contract is $P(\bar{B}) = 5/9$. Therefore,$P(B) = 1 - P(\bar{B}) = 1 - 5/9 = 4/9$.
The probability of getting at least $1$ contract is $P(A \cup B) = 4/5$.
We use the formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values: $4/5 = 2/3 + 4/9 - P(A \cap B)$.
$P(A \cap B) = 2/3 + 4/9 - 4/5$.
Finding a common denominator $(45)$: $P(A \cap B) = (30 + 20 - 36) / 45 = 14/45$.
Thus,the probability that he will get both contracts is $14/45$.

(A) Let $A$ be the event that a spade is drawn and $B$ be the event that an ace is drawn.
Total number of cards in a pack $= 52$.
Number of spades $= 13$.
Number of aces $= 4$.
Number of cards that are both a spade and an ace (the ace of spades) $= 1$.
Probability of winning the bet $= P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$.
Probability of losing the bet $= 1 - P(A \cup B) = 1 - \frac{4}{13} = \frac{9}{13}$.
Odds against winning the bet are defined as the ratio of the probability of losing to the probability of winning.
Odds against $= \frac{9}{13} : \frac{4}{13} = 9 : 4$.