Set Based probability Questions in English

(C) The total number of outcomes when $4$ dice are rolled is $6^4 = 1296$.
To find the number of outcomes where at least one die shows $2$,we subtract the number of outcomes where $2$ does not appear on any die from the total outcomes.
The number of outcomes where $2$ does not appear on any die is $5^4 = 625$.
Therefore,the number of outcomes with at least one $2$ is $1296 - 625 = 671$.

(A) The total number of cards is $n(S) = 52$.
Number of diamond cards is $n(A) = 13$,so $P(A) = \frac{13}{52} = \frac{1}{4}$.
Number of ace cards is $n(B) = 4$,so $P(B) = \frac{4}{52} = \frac{1}{13}$.
The number of ace of diamond cards is $n(A \cap B) = 1$,so $P(A \cap B) = \frac{1}{52}$.
Now,check for independence: $P(A) \times P(B) = \frac{1}{4} \times \frac{1}{13} = \frac{1}{52}$.
Since $P(A \cap B) = P(A) \times P(B)$,the events $A$ and $B$ are independent.

(C) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
We need the probability of getting a sum greater than $7$,which means the sum can be $8, 9, 10, 11,$ or $12$.
The favorable 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 $= 5 + 4 + 3 + 2 + 1 = 15$.
Required probability $= \frac{15}{36} = \frac{5}{12}$.

(A) Total number of balls in the bag $= 3 \text{ (black)} + 4 \text{ (white)} = 7 \text{ balls}$.
Number of favorable outcomes (white balls) $= 4$.
The probability of drawing a white ball is given by the ratio of the number of favorable outcomes to the total number of outcomes.
$\text{Probability} = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{4}{7}$.

(D) The probability of getting a head is $P(H) = \frac{1}{2}$ and the probability of not getting a head is $P(T) = \frac{1}{2}$.
$A$ wins if he gets a head on his $1^{st}$ turn,or if $A$ and $B$ both fail and $A$ gets a head on his $2^{nd}$ turn,and so on.
This is a geometric series: $P(A \text{ wins}) = P(H) + P(T)P(T)P(H) + P(T)P(T)P(T)P(T)P(H) + \dots$
$P(A \text{ wins}) = \frac{1}{2} + (\frac{1}{2})^2 \cdot \frac{1}{2} + (\frac{1}{2})^4 \cdot \frac{1}{2} + \dots$
$P(A \text{ wins}) = \frac{1}{2} + (\frac{1}{2})^3 + (\frac{1}{2})^5 + \dots$
This is an infinite geometric series with first term $a = \frac{1}{2}$ and common ratio $r = (\frac{1}{2})^2 = \frac{1}{4}$.
The sum is $S = \frac{a}{1-r} = \frac{1/2}{1 - 1/4} = \frac{1/2}{3/4} = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}$.

(C) When two balanced dice are tossed,the total number of possible outcomes is $6 \times 6 = 36$.
Let $E$ be the event that the sum of the integers on the upper faces is $9$.
The possible outcomes for the sum to be $9$ are $(3, 6), (4, 5), (5, 4), \text{ and } (6, 3)$.
Thus,the number of favorable outcomes is $4$.
The probability of the event $E$ is given by $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{36} = \frac{1}{9}$.

(A) Total number of cards in a deck = $52$.
Number of aces in a deck = $4$.
Probability of drawing an ace = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{52} = \frac{1}{13}$.
Thus,the correct option is $A$.

(C) The word $PROBABILITY$ contains $11$ letters in total: $P, R, O, B, A, B, I, L, I, T, Y$.
The vowels in the word are $O, A, I, I$.
There are $4$ vowels in total.
The probability of selecting a vowel is given by the ratio of the number of vowels to the total number of letters.
$\text{Probability} = \frac{\text{Number of vowels}}{\text{Total number of letters}} = \frac{4}{11}$.

(A) The total number of pages is $100$,so the sample space is $\{1, 2, 3, \dots, 100\}$.
We need to find the page numbers whose digits sum to $11$.
For a two-digit number $xy$,the sum is $x + y = 11$. The possible pairs $(x, y)$ are $(2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2)$.
These correspond to the page numbers: $29, 38, 47, 56, 65, 74, 83, 92$.
There are $8$ such pages.
Note that for the page number $100$,the sum of digits is $1 + 0 + 0 = 1 \neq 11$.
Thus,the number of favourable outcomes is $8$.
The probability is $\frac{8}{100} = \frac{2}{25}$.

(C) The sample space for two children in a family is $S = \{BB, BG, GB, GG\}$,where $B$ denotes a boy and $G$ denotes a girl.
Total number of outcomes $n(S) = 4$.
The event of having both boys is $E = \{BB\}$.
Number of favorable outcomes $n(E) = 1$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{1}{4}$.

(C) The total number of outcomes when a dice is thrown twice is $6 \times 6 = 36$.
Let $E_1$ be the event of getting $1$ in the first throw. The outcomes are $(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$.
Let $E_2$ be the event of not getting $1$ in the second throw. The outcomes for the second throw can be $2, 3, 4, 5, 6$.
We want the probability of getting $1$ in the first throw $AND$ not getting $1$ in the second throw.
The favorable outcomes are $(1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$,which are $5$ outcomes.
The required probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{36}$.

(B) The sample space for tossing a coin is $S_1 = \{H, T\}$,so $P(\text{Head}) = \frac{1}{2}$.
The sample space for rolling a dice is $S_2 = \{1, 2, 3, 4, 5, 6\}$,so $P(6) = \frac{1}{6}$.
Since these are independent events,the probability of both occurring is $P(\text{Head} \cap 6) = P(\text{Head}) \times P(6) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$.

(B) The sample space $S$ for tossing a coin twice is $S = \{(H, H), (H, T), (T, H), (T, T)\}$.
Total number of outcomes $n(S) = 4$.
Let $E$ be the event of getting a head both times,so $E = \{(H, H)\}$.
Number of favorable outcomes $n(E) = 1$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{1}{4}$.

(B) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
The sum $2$ can be obtained in $1$ way: $(1, 1)$.
The sum $8$ can be obtained in $5$ ways: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$.
The sum $12$ can be obtained in $1$ way: $(6, 6)$.
The total number of favorable outcomes is $1 + 5 + 1 = 7$.
Therefore,the required probability is $\frac{7}{36}$.

(B) Let $A$ be the event of getting $4, 5,$ or $6$ in the first throw. The number of favorable outcomes is $3$. Total outcomes = $6$.
$P(A) = \frac{3}{6} = \frac{1}{2}$.
Let $B$ be the event of getting $1, 2, 3,$ or $4$ in the second throw. The number of favorable outcomes is $4$. Total outcomes = $6$.
$P(B) = \frac{4}{6} = \frac{2}{3}$.
Since the two throws are independent events,the probability of both occurring is $P(A \cap B) = P(A) \times P(B)$.
$P(A \cap B) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$.

(A) Let $A$ be the event that the card drawn from the first pack is an ace of hearts,and $B$ be the event that the card drawn from the second pack is an ace of hearts.
The probability of drawing an ace of hearts from one pack is $P(A) = \frac{1}{52}$.
The probability of $NOT$ drawing an ace of hearts from one pack is $P(A') = 1 - \frac{1}{52} = \frac{51}{52}$.
The probability that neither card is an ace of hearts is $P(A' \cap B') = P(A') \times P(B') = \frac{51}{52} \times \frac{51}{52} = \frac{2601}{2704}$.
The probability that at least one of them is an ace of hearts is $1 - P(A' \cap B') = 1 - \frac{2601}{2704} = \frac{2704 - 2601}{2704} = \frac{103}{2704}$.

(C) Total number of items $= 6 + 10 = 16$.
Let $N$ be the event that the item is a nail and $R$ be the event that the item is rusted.
Number of nails $n(N) = 6$.
Number of rusted items $n(R) = \frac{6}{2} + \frac{10}{2} = 3 + 5 = 8$.
Number of rusted nails $n(N \cap R) = \frac{6}{2} = 3$.
We need to find the probability of $N \cup R$.
Using the formula $P(N \cup R) = P(N) + P(R) - P(N \cap R)$:
$P(N \cup R) = \frac{6}{16} + \frac{8}{16} - \frac{3}{16} = \frac{11}{16}$.

(A) The total number of outcomes when a coin is tossed $4$ times is $2^4 = 16$.
Let $E$ be the event of getting at least one tail.
The complement event $E'$ is the event of getting no tails,which means getting all heads.
The only outcome for $E'$ is $(H, H, H, H)$,so the number of outcomes for $E'$ is $1$.
The probability of $E'$ is $P(E') = \frac{1}{16}$.
Therefore,the probability of getting at least one tail is $P(E) = 1 - P(E') = 1 - \frac{1}{16} = \frac{15}{16}$.

(B) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
Let $E$ be the event that the sum of the numbers appearing is more than $10$.
The possible outcomes for the sum being more than $10$ are $(5, 6), (6, 5), (6, 6)$.
There are $3$ such outcomes.
Therefore,the probability $P(E) = \frac{3}{36} = \frac{1}{12}$.

(B) The total number of possible outcomes when throwing $2$ dice is $6 \times 6 = 36$.
The outcomes resulting in a sum of $5$ are $(1, 4), (2, 3), (3, 2), (4, 1)$,which are $4$ cases.
The outcomes resulting in a sum of $6$ are $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$,which are $5$ cases.
The total number of favourable outcomes is $4 + 5 = 9$.
Therefore,the required probability is $\frac{9}{36} = \frac{1}{4}$.

(B) sure event is an event that is certain to occur.
By definition,the probability of a sample space $S$ is $P(S) = 1$.
Therefore,the probability of a sure event is $1$.

(B) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
An odd sum is obtained if one die shows an odd number and the other shows an even number.
The possible pairs 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 find there are $18$ favorable outcomes.
The probability is $\frac{18}{36} = \frac{1}{2}$.

(C) Total number of tickets $= 10,000$.
Number of tickets divisible by $20$ is given by $\frac{10,000}{20} = 500$.
Probability of drawing a ticket divisible by $20 = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{500}{10,000} = \frac{1}{20}$.

(B) When two dice are thrown,the total number of outcomes is $6 \times 6 = 36$.
Let $A$ be the event of getting a multiple of $2$ on the first die: $A = \{2, 4, 6\}$,so $n(A) = 3$.
Let $B$ be the event of getting a multiple of $3$ on the second die: $B = \{3, 6\}$,so $n(B) = 2$.
The outcomes where the first die is a multiple of $2$ and the second is a multiple of $3$ are: $(2,3), (2,6), (4,3), (4,6), (6,3), (6,6)$. There are $3 \times 2 = 6$ such outcomes.
The outcomes where the first die is a multiple of $3$ and the second is a multiple of $2$ are: $(3,2), (3,4), (3,6), (6,2), (6,4), (6,6)$. There are $2 \times 3 = 6$ such outcomes.
The outcome $(6,6)$ is common to both sets.
Using the principle of inclusion-exclusion,the total number of favourable outcomes is $6 + 6 - 1 = 11$.
Therefore,the required probability is $\frac{11}{36}$.

(D) Let $A, B,$ and $C$ be the events that the three students solve the problem respectively.
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 are $P(A') = 1 - \frac{1}{3} = \frac{2}{3}$,$P(B') = 1 - \frac{1}{4} = \frac{3}{4}$,and $P(C') = 1 - \frac{1}{5} = \frac{4}{5}$.
The probability that the problem is not solved by any of them is $P(A' \cap B' \cap C') = P(A') \times P(B') \times P(C') = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{2}{5}$.
Therefore,the probability that the problem is solved is $1 - P(A' \cap B' \cap C') = 1 - \frac{2}{5} = \frac{3}{5}$.

(C) fair die has $6$ faces,numbered $1, 2, 3, 4, 5, 6$.
Total number of possible outcomes $n(S) = 6$.
The number of favorable outcomes for getting $5$ is $n(E) = 1$ (since only one face has the number $5$).
The probability $P(E)$ is given by the formula:
$P(E) = \frac{n(E)}{n(S)} = \frac{1}{6}$.

(B) The total number of cards in a pack is $52$.
The probability of drawing a queen of clubs is $P(A) = \frac{1}{52}$.
The probability of drawing a king of hearts is $P(B) = \frac{1}{52}$.
Since these are mutually exclusive events,the probability of getting a queen of clubs or a king of hearts is $P(A \cup B) = P(A) + P(B)$.
Therefore,the required probability is $\frac{1}{52} + \frac{1}{52} = \frac{2}{52} = \frac{1}{26}$.

(C) When three coins are tossed simultaneously,the total number of possible outcomes is $2^3 = 8$.
The sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
We want the probability of getting at least $2$ tails.
The favorable outcomes are $\{HTT, THT, TTH, TTT\}$.
The number of favorable outcomes is $4$.
The probability $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{8} = \frac{1}{2}$.

(B) The sample space $S$ for a single throw of a fair die is $\{1, 2, 3, 4, 5, 6\}$.
The total number of outcomes is $n(S) = 6$.
Let $E$ be the event of getting a number less than $7$.
All outcomes in the sample space are less than $7$,so $E = \{1, 2, 3, 4, 5, 6\}$.
The number of favorable outcomes is $n(E) = 6$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{6}{6} = 1$.
Therefore,the correct option is $B$.

(C) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
Let $E$ be the event of getting a sum less than $11$.
It is easier to find the complement event $E'$,which is getting a sum of $11$ or more.
The outcomes for a sum $\ge 11$ are: $(5, 6), (6, 5), (6, 6)$.
There are $3$ such outcomes.
Therefore,the number of favourable outcomes for $E$ is $36 - 3 = 33$.
The probability $P(E) = \frac{33}{36} = \frac{11}{12}$.

(B) An ordinary or non-leap year has $365$ days.
$365$ days $= 52$ weeks and $1$ day.
This remaining $1$ day can be any of the $7$ days of the week: {Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday}.
For the year to have $53$ Sundays,the remaining $1$ day must be a Sunday.
There is only $1$ favorable outcome out of $7$ possible outcomes.
Therefore,the required probability $= \frac{1}{7}$.

(B) Total number of cards in a pack $= 52$.
Court cards (face cards) include jack,queen,and king of each suit.
There are $4$ suits,so the total number of court cards $= 4 \times 3 = 12$.
The probability of drawing a court card $= \frac{\text{Number of court cards}}{\text{Total number of cards}} = \frac{12}{52}$.
Simplifying the fraction,we get $\frac{12}{52} = \frac{3}{13}$.
Thus,the correct option is $B$.

(C) When two dice are thrown,the total number of outcomes is $6 \times 6 = 36$.
Let $E$ be the event that the sum of the numbers on the two dice is a multiple of $4$.
The possible sums that are multiples of $4$ are $4, 8, 12$.
The outcomes for sum $= 4$ are $(1, 3), (2, 2), (3, 1)$.
The outcomes for sum $= 8$ are $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$.
The outcomes for sum $= 12$ are $(6, 6)$.
The total number of favorable outcomes is $3 + 5 + 1 = 9$.
Therefore,the probability $P(E) = \frac{9}{36} = \frac{1}{4}$.

(A) Total number of outcomes = $5 \text{ (prizes)} + 20 \text{ (blanks)} = 25$.
Number of favorable outcomes (getting a prize) = $5$.
Probability of getting a prize = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{25} = \frac{1}{5}$.

(B) When a fair die is thrown,the sample space is $S = \{1, 2, 3, 4, 5, 6\}$.
The total number of outcomes is $n(S) = 6$.
Let $E$ be the event of getting a number greater than $2$.
The favorable outcomes are $E = \{3, 4, 5, 6\}$.
The number of favorable outcomes is $n(E) = 4$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{4}{6} = \frac{2}{3}$.

(C) When two dice are thrown,the total number of outcomes is $6 \times 6 = 36$.
The outcomes where the numbers on the two dice are the same are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$,which are $6$ outcomes.
The number of outcomes where the numbers are different is $36 - 6 = 30$.
The outcomes where the sum is $6$ are $(1,5), (2,4), (3,3), (4,2), (5,1)$.
Since we are given that the numbers must be different,we exclude $(3,3)$.
The favorable outcomes are $(1,5), (2,4), (4,2), (5,1)$,which are $4$ outcomes.
The required probability is $\frac{4}{30} = \frac{2}{15}$.

(A) Let $E_1$ be the event that the man is selected and $E_2$ be the event that the woman is selected.
Given $P(E_1) = 1/4$ and $P(E_2) = 1/3$.
The probability that the man is not selected is $P(\bar{E_1}) = 1 - P(E_1) = 1 - 1/4 = 3/4$.
The probability that the woman is not selected is $P(\bar{E_2}) = 1 - P(E_2) = 1 - 1/3 = 2/3$.
Since the events are independent,the probability that none of them will be selected is $P(\bar{E_1} \cap \bar{E_2}) = P(\bar{E_1}) \times P(\bar{E_2})$.
Therefore,$P(\bar{E_1} \cap \bar{E_2}) = (3/4) \times (2/3) = 6/12 = 1/2$.

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

(B) Total number of cards = $52$.
Number of heart cards = $13$.
Number of king cards = $4$.
Number of cards that are both heart and king (King of Hearts) = $1$.
Using the principle of inclusion-exclusion,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 maximum sum possible is $6 + 6 = 12$.
Since the sum $13$ is impossible to obtain,the number of favorable outcomes is $0$.
Therefore,the probability is $\frac{0}{36} = 0$.

(C) The total number of outcomes when three dice are thrown 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 $P = \frac{4}{216} = \frac{1}{54}$.

(D) Total number of articles $= 10 + 6 = 16$.
Let $E$ be the event that the chosen article is either good or defective.
Since every article in the box is either good or defective,this is a sure event.
The number of favorable outcomes $= 10 + 6 = 16$.
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.
By definition,the probability of an impossible event is $0$.
Conversely,the probability of a sure event is $1$.

(B) When two dice are tossed,the total number of outcomes is $6 \times 6 = 36$.
The possible sums range from $2$ to $12$. The prime numbers in this range are $\{2, 3, 5, 7, 11\}$.
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 $= 1 + 2 + 4 + 6 + 2 = 15$.
Required probability $= \frac{15}{36} = \frac{5}{12}$.

(A) Let the probabilities of solving the problem by the three persons be $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{4}$,and $P(C) = \frac{1}{5}$.
Since they work independently,the probability that they do not solve the problem is given by $P(A') = 1 - \frac{1}{3} = \frac{2}{3}$,$P(B') = 1 - \frac{1}{4} = \frac{3}{4}$,and $P(C') = 1 - \frac{1}{5} = \frac{4}{5}$.
The probability that none of them can 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 $6$.
Therefore,the probability is $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36}$.

(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:
For sum $9$: $(3,6), (4,5), (5,4), (6,3)$ ($4$ outcomes)
For sum $10$: $(4,6), (5,5), (6,4)$ ($3$ outcomes)
For sum $11$: $(5,6), (6,5)$ ($2$ outcomes)
For sum $12$: $(6,6)$ ($1$ outcome)
Total favorable outcomes $= 4 + 3 + 2 + 1 = 10$.
The required probability $= \frac{10}{36} = \frac{5}{18}$.

(B) The word $POSSESSIVE$ contains $10$ letters: $P, O, S, S, E, S, S, I, V, E$.
Total number of outcomes $n(S) = 10$.
The letter $S$ appears $4$ times in the word.
Number of favorable outcomes $n(E) = 4$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{4}{10}$.

(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 $(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), \text{ and } (6, 6, 6)$.
There are $6$ such favorable outcomes.
Therefore,the required probability is $\frac{6}{216} = \frac{1}{36}$.