Probability Questions in English

(B) The probability of getting at least one head in $n$ tosses is given by $1 - P(\text{no heads})$.
Since the probability of getting no heads in $n$ tosses is $(1/2)^n$,the probability of getting at least one head is $1 - (1/2)^n$.
We are given that this probability must be $\ge 0.9$.
So,$1 - (1/2)^n \ge 0.9$.
Rearranging the inequality,we get $1 - 0.9 \ge (1/2)^n$,which simplifies to $0.1 \ge (1/2)^n$.
This can be written as $1/10 \ge 1/2^n$,or $2^n \ge 10$.
Testing integer values for $n$:
For $n = 3$,$2^3 = 8$,which is not $\ge 10$.
For $n = 4$,$2^4 = 16$,which is $\ge 10$.
Therefore,the minimum number of tosses required is $4$.

(C) The second ball drawn can be white in two mutually exclusive cases:
Case $(i)$: The first ball is white and the second ball is white.
Probability $= \frac{4}{7} \times \frac{3}{6} = \frac{12}{42} = \frac{2}{7}$.
Case $(ii)$: The first ball is black and the second ball is white.
Probability $= \frac{3}{7} \times \frac{4}{6} = \frac{12}{42} = \frac{2}{7}$.
Therefore,the total probability that the second ball is white is the sum of the probabilities of these two cases:
Total Probability $= \frac{2}{7} + \frac{2}{7} = \frac{4}{7}$.

(C) The probability of $A$ drawing two cards of the same suit with replacement:
There are $4$ suits,each with $13$ cards.
The probability of picking a card of a specific suit is $\frac{13}{52} = \frac{1}{4}$.
Since the cards are drawn with replacement,the probability of drawing two cards of the same suit is $4 \times (\frac{1}{4} \times \frac{1}{4}) = 4 \times \frac{1}{16} = \frac{1}{4}$.
The probability of $B$ throwing a total of $6$ with a pair of dice:
The total outcomes when throwing two dice is $6 \times 6 = 36$.
The favorable outcomes for a sum of $6$ are $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$,which are $5$ outcomes.
So,$P(B) = \frac{5}{36}$.
Since $A$ and $B$ are independent events,the combined probability is:
$P(A \cap B) = P(A) \times P(B) = \frac{1}{4} \times \frac{5}{36} = \frac{5}{144}$.

(A) Total number of balls = $2 + 4 + 3 = 9$.
We need to find the probability of drawing the balls in the specific sequence: $B, B, W, W, W, W, R, R, R$.
The probability is calculated as follows:
$P = P(B_1) \times P(B_2|B_1) \times P(W_1|B_1, B_2) \times P(W_2|B_1, B_2, W_1) \times P(W_3|B_1, B_2, W_1, W_2) \times P(W_4|B_1, B_2, W_1, W_2, W_3) \times P(R_1|B_1, B_2, W_1, W_2, W_3, W_4) \times P(R_2|...) \times P(R_3|...)$
$P = \frac{2}{9} \times \frac{1}{8} \times \frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} \times \frac{3}{3} \times \frac{2}{2} \times \frac{1}{1}$
$P = \frac{2 \times 1 \times 4 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1}{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{288}{362880} = \frac{1}{1260}$.

(B) The total number of possible outcomes when a dice is rolled three times is $6^3 = 216$.
Let the three numbers obtained be $x_1, x_2, x_3$. We require $x_1 < x_2 < x_3$.
This is equivalent to choosing $3$ distinct numbers out of ${1, 2, 3, 4, 5, 6}$,because for any set of $3$ distinct numbers,there is only one way to arrange them in increasing order.
The number of ways to choose $3$ distinct numbers from $6$ is given by the combination formula $\binom{6}{3}$.
$\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
Thus,the number of favourable outcomes is $20$.
The required probability is $\frac{20}{216} = \frac{5}{54}$.

(B) Total cards = $52$. Number of aces = $4$. Number of non-ace cards = $48$.
We want the first $10$ cards to be non-aces and the $11$th card to be an ace.
The probability of picking $10$ non-ace cards in a row is given by the product of probabilities for each draw:
$P = \frac{48}{52} \times \frac{47}{51} \times \frac{46}{50} \times \frac{45}{49} \times \frac{44}{48} \times \frac{43}{47} \times \frac{42}{46} \times \frac{41}{45} \times \frac{40}{44} \times \frac{39}{43} \times \frac{4}{42}$.
Canceling the common terms in the numerator and denominator:
$P = \frac{48 \times 47 \times 46 \times 45 \times 44 \times 43 \times 42 \times 41 \times 40 \times 39}{52 \times 51 \times 50 \times 49 \times 48 \times 47 \times 46 \times 45 \times 44 \times 43} \times \frac{4}{42}$.
$P = \frac{42 \times 41 \times 40 \times 39}{52 \times 51 \times 50 \times 49} \times \frac{4}{42} = \frac{41 \times 40 \times 39}{52 \times 51 \times 50 \times 49} \times 4$.
Simplifying the fraction,we get $\frac{164}{4165}$.

(D) Let $P(T)$ be the probability of a test occurring in a class, $P(T) = 1/5$.
Let $P(T')$ be the probability of no test occurring, $P(T') = 1 - 1/5 = 4/5$.
The student is absent for $2$ classes. The student misses at least one test if a test occurs in the first class, the second class, or both.
It is easier to calculate the complement: the probability that the student misses no tests.
The student misses no tests if no test occurs in either of the $2$ classes.
$P(\text{misses no tests}) = P(T') \times P(T') = (4/5) \times (4/5) = 16/25$.
Therefore, the probability that the student misses at least one test is $1 - P(\text{misses no tests}) = 1 - 16/25 = 9/25$.

(C) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
We need to find the favorable outcomes for getting a sum of $3$,$5$,or $11$:
- For a sum of $3$: The possible outcomes are $(1, 2)$ and $(2, 1)$. (Total $2$ cases)
- For a sum of $5$: The possible outcomes are $(1, 4), (2, 3), (3, 2),$ and $(4, 1)$. (Total $4$ cases)
- For a sum of $11$: The possible outcomes are $(5, 6)$ and $(6, 5)$. (Total $2$ cases)
Total favorable outcomes $= 2 + 4 + 2 = 8$.
The required probability is $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{8}{36} = \frac{2}{9}$.

(D) Let $P(O)$ be the probability of getting an odd number and $P(E)$ be the probability of getting an even number.
Since the die is biased such that an even number is twice as likely as an odd number, we have $P(E) = 2P(O)$.
Since the sum of probabilities is $1$, $P(E) + P(O) = 1$, which gives $3P(O) = 1$, so $P(O) = \frac{1}{3}$ and $P(E) = \frac{2}{3}$.
The sum of two numbers is even if both are even or both are odd.
$P(\text{Sum is even}) = P(E, E) + P(O, O)$.
$P(E, E) = P(E) \times P(E) = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$.
$P(O, O) = P(O) \times P(O) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$.
Therefore, $P(\text{Sum is even}) = \frac{4}{9} + \frac{1}{9} = \frac{5}{9}$.

(D) Let $T$ be the event that India wins the toss and $V$ be the event that India wins the match.
Given:
$P(T) = 3/4$
$P(T^c) = 1 - 3/4 = 1/4$
$P(V|T) = 4/5$
$P(V|T^c) = 1/2$
Using the Law of Total Probability:
$P(V) = P(T) \times P(V|T) + P(T^c) \times P(V|T^c)$
$P(V) = (3/4 \times 4/5) + (1/4 \times 1/2)$
$P(V) = 12/20 + 1/8 = 3/5 + 1/8$
$P(V) = (24 + 5) / 40 = 29/40$
Therefore,the chance of India's victory is $29/40$.

(B) Total number of cards in a pack $= 52$.
Number of kings in a pack $= 4$.
Number of queens in a pack $= 4$.
Since the events of drawing a king and drawing a queen are mutually exclusive,the probability of drawing either a king or a queen is the sum of their individual probabilities.
Probability of drawing a king $P(K) = \frac{4}{52} = \frac{1}{13}$.
Probability of drawing a queen $P(Q) = \frac{4}{52} = \frac{1}{13}$.
Probability of drawing a king or a queen $P(K \cup Q) = P(K) + P(Q) = \frac{1}{13} + \frac{1}{13} = \frac{2}{13}$.

(C) Total number of cards in a pack $= 52$.
Number of kings in a pack $= 4$.
Number of queens in a pack $= 4$.
Since the cards are drawn one by one without replacement:
Probability of drawing a king first $= P(K_1) = \frac{4}{52} = \frac{1}{13}$.
After drawing one king,the remaining number of cards is $51$.
Probability of drawing a queen second given that the first was a king $= P(Q_2|K_1) = \frac{4}{51}$.
Required probability $= P(K_1) \times P(Q_2|K_1) = \frac{4}{52} \times \frac{4}{51} = \frac{1}{13} \times \frac{4}{51} = \frac{4}{663}$.

(D) Let $A, B, C$ be the events of getting $I, II$ and $III$ division respectively,and $D$ be the event of failing the examination.
Since these events are mutually exclusive and exhaustive,the sum of their probabilities must be $1$.
$P(A) = \frac{1}{10} = 0.1$
$P(B) = \frac{3}{5} = 0.6$
$P(C) = \frac{1}{4} = 0.25$
$P(A) + P(B) + P(C) + P(D) = 1$
$0.1 + 0.6 + 0.25 + P(D) = 1$
$0.95 + P(D) = 1$
$P(D) = 1 - 0.95 = 0.05$
$P(D) = \frac{5}{100} = \frac{1}{20}$
Since $\frac{1}{20}$ is not among the given options,the correct answer is "None of these".

(B) Let $p$ be the probability of getting a number greater than $4$ (i.e.,$5$ or $6$) in a single toss.
$p = \frac{2}{6} = \frac{1}{3}$.
Let $q$ be the probability of failure (i.e.,getting $1, 2, 3,$ or $4$).
$q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.
We need the probability that the success occurs on an even number of tosses (i.e.,$2^{nd}, 4^{th}, 6^{th}, \dots$ toss).
The probability is given by the sum of the infinite geometric series:
$P = P(FS) + P(FFFS) + P(FFFFFS) + \dots$
$P = qp + q^3p + q^5p + \dots$
This is a geometric progression with first term $a = qp$ and common ratio $r = q^2$.
The sum of an infinite geometric series is $S = \frac{a}{1-r}$.
$P = \frac{qp}{1 - q^2} = \frac{(\frac{2}{3})(\frac{1}{3})}{1 - (\frac{2}{3})^2} = \frac{\frac{2}{9}}{1 - \frac{4}{9}} = \frac{\frac{2}{9}}{\frac{5}{9}} = \frac{2}{5}$.

(A) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
Let $E$ be the event that at least one die shows the digit $6$.
The outcomes where at least one die shows $6$ are:
$(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1)$.
Counting these outcomes,we find there are $11$ favorable outcomes.
Therefore,the probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{11}{36}$.

(B) Total number of balls $= 30$.
Let $A$ be the event of getting a multiple of $5$. The multiples of $5$ between $1$ and $30$ are ${5, 10, 15, 20, 25, 30}$. So,$n(A) = 6$.
Let $B$ be the event of getting a multiple of $7$. The multiples of $7$ between $1$ and $30$ are ${7, 14, 21, 28}$. So,$n(B) = 4$.
Since there are no common multiples of $5$ and $7$ in the range $1$ to $30$,$n(A \cap B) = 0$.
The number of favorable outcomes is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 6 + 4 - 0 = 10$.
The probability $P(A \cup B) = \frac{n(A \cup B)}{\text{Total outcomes}} = \frac{10}{30} = \frac{1}{3}$.

(C) Let $p$ be the probability of killing the bird,so $p = 3/4$.
Let $q$ be the probability of missing the bird,so $q = 1 - p = 1 - 3/4 = 1/4$.
The person tries $n = 5$ times.
The probability that he does not kill the bird at all means he misses the bird in all $5$ attempts.
This is given by the binomial probability formula for $X=0$ successes: $P(X=0) = q^n = (1/4)^5$.
Calculating this,we get $P(X=0) = 1^5 / 4^5 = 1 / 1024$.
Therefore,the probability that he may not kill the bird is $1/1024$.

(A) The outcome of each coin toss is an independent event.
The probability of getting a head on any single toss of a fair coin is always $\frac{1}{2}$,regardless of the results of previous tosses.
Therefore,the probability of getting a head on the $5^{th}$ toss is $\frac{1}{2}$.

(C) Let $A$ and $B$ be the two persons. Each tosses a coin $3$ times.
The probability of getting $r$ heads in $3$ tosses follows a binomial distribution $P(X=r) = \binom{3}{r} (\frac{1}{2})^3$.
For $r = 0, 1, 2, 3$,the probabilities are:
$P(0) = \binom{3}{0} (\frac{1}{8}) = \frac{1}{8}$
$P(1) = \binom{3}{1} (\frac{1}{8}) = \frac{3}{8}$
$P(2) = \binom{3}{2} (\frac{1}{8}) = \frac{3}{8}$
$P(3) = \binom{3}{3} (\frac{1}{8}) = \frac{1}{8}$
Since the events are independent,the probability that both get an equal number of heads is the sum of the probabilities of both getting $0, 1, 2,$ or $3$ heads:
$P = P(0)P(0) + P(1)P(1) + P(2)P(2) + P(3)P(3)$
$P = (\frac{1}{8} \times \frac{1}{8}) + (\frac{3}{8} \times \frac{3}{8}) + (\frac{3}{8} \times \frac{3}{8}) + (\frac{1}{8} \times \frac{1}{8})$
$P = \frac{1}{64} + \frac{9}{64} + \frac{9}{64} + \frac{1}{64} = \frac{20}{64} = \frac{5}{16}$.

(A) Let the two positive numbers be $x$ and $y$. Given $x + y = 100$,where $x, y \in \{1, 2, 3, \dots, 99\}$.
The total number of possible pairs $(x, y)$ is $99$.
We want the product $xy > 1000$.
Substituting $y = 100 - x$,we get $x(100 - x) > 1000$,which simplifies to $100x - x^2 > 1000$,or $x^2 - 100x + 1000 < 0$.
Solving the quadratic equation $x^2 - 100x + 1000 = 0$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{100 \pm \sqrt{10000 - 4000}}{2} = \frac{100 \pm \sqrt{6000}}{2} = 50 \pm 10\sqrt{15}$.
Since $\sqrt{15} \approx 3.87$,$x \approx 50 \pm 38.7$.
So,$x$ must be in the range $(11.3, 88.7)$.
The possible integer values for $x$ are $12, 13, \dots, 88$.
The number of such values is $88 - 12 + 1 = 77$.
The probability is $\frac{77}{99} = \frac{7}{9}$.

(C) Each tetrahedron has $4$ faces,so when $3$ tetrahedrons are tossed,the total number of outcomes is $4^3 = 64$.
We need the sum of the numbers on the upward faces to be $5$.
Let the outcomes be $(x_1, x_2, x_3)$ where $x_i \in \{1, 2, 3, 4\}$.
The possible combinations that sum to $5$ are:
$(1, 1, 3)$ which can occur in $3$ permutations: $(1, 1, 3), (1, 3, 1), (3, 1, 1)$.
$(1, 2, 2)$ which can occur in $3$ permutations: $(1, 2, 2), (2, 1, 2), (2, 2, 1)$.
Total favorable outcomes $= 3 + 3 = 6$.
Therefore,the required probability $= \frac{6}{64} = \frac{3}{32}$.

(B) The last digit of the square of an integer depends only on the last digit of the integer itself.
Let the last digit of the integer be $x$,where $x \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
The squares of these digits are:
$0^2 = 0$
$1^2 = 1$
$2^2 = 4$
$3^2 = 9$
$4^2 = 16$ (last digit $6$)
$5^2 = 25$ (last digit $5$)
$6^2 = 36$ (last digit $6$)
$7^2 = 49$ (last digit $9$)
$8^2 = 64$ (last digit $4$)
$9^2 = 81$ (last digit $1$)
The last digits of the squares are $\{0, 1, 4, 9, 6, 5, 6, 9, 4, 1\}$.
We want the last digit of the square to be $1$ or $5$.
The cases where the last digit is $1$ are $x = 1$ and $x = 9$ (total $2$ cases).
The case where the last digit is $5$ is $x = 5$ (total $1$ case).
Total favorable outcomes = $2 + 1 = 3$.
Total possible outcomes = $10$.
Therefore,the required probability = $\frac{3}{10}$.

(B) Let the two integers be $x$ and $y$. Each integer can be either even $(E)$ or odd $(O)$.
There are $4$ possible outcomes for the parity of the two integers: $(E, E), (E, O), (O, E), (O, O)$.
Each outcome is equally likely with a probability of $\frac{1}{4}$.
$1$. If both integers are even $(E, E)$,the product is even.
$2$. If one is even and one is odd $(E, O)$ or $(O, E)$,the product is even.
$3$. If both integers are odd $(O, O)$,the product is odd.
The product is even in $3$ out of the $4$ cases: $(E, E), (E, O), (O, E)$.
Therefore,the required probability is $\frac{3}{4}$.

(D) The probability of a single bit being correct is $1 - p$.
Since the errors in different bits are independent,the probability that all $16$ bits are correct is $(1 - p)^{16}$.
An incorrect number is formed if at least one bit is incorrect.
Therefore,the probability of forming an incorrect number is $1 - P(\text{all bits are correct})$.
Thus,the probability is $1 - (1 - p)^{16}$.

(D) 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 head.
The complement event $E'$ is the event of getting no heads,which means getting $4$ tails.
There is only $1$ outcome where no heads appear: $(T, T, T, T)$.
So,$P(E') = \frac{1}{16}$.
The probability of getting at least one head is $P(E) = 1 - P(E')$.
$P(E) = 1 - \frac{1}{16} = \frac{15}{16}$.

(D) century consists of $100$ years. In any century,there are $25$ leap years and $75$ non-leap years.
Probability of choosing a leap year $P(L) = \frac{25}{100} = \frac{1}{4}$.
Probability of choosing a non-leap year $P(NL) = \frac{75}{100} = \frac{3}{4}$.
$A$ leap year has $366$ days,which is $52$ weeks and $2$ extra days. The probability of having $53$ Sundays in a leap year is $\frac{2}{7}$.
$A$ non-leap year has $365$ days,which is $52$ weeks and $1$ extra day. The probability of having $53$ Sundays in a non-leap year is $\frac{1}{7}$.
Total probability $= P(L) \times P(53 \text{ Sundays in leap year}) + P(NL) \times P(53 \text{ Sundays in non-leap year})$.
Total probability $= (\frac{1}{4} \times \frac{2}{7}) + (\frac{3}{4} \times \frac{1}{7}) = \frac{2}{28} + \frac{3}{28} = \frac{5}{28}$.

(B) Let $P(K)$ be the probability of getting face $K$. Given $P(K) \propto K$,so $P(K) = cK$ for some constant $c$.
Since the sum of all probabilities must be $1$,we have $\sum_{K=1}^{6} P(K) = 1$.
$c(1 + 2 + 3 + 4 + 5 + 6) = 1 \implies c(21) = 1 \implies c = \frac{1}{21}$.
The outcomes that are even numbers are $2, 4,$ and $6$.
The probability of getting an even number is $P(2) + P(4) + P(6) = c(2 + 4 + 6) = 12c$.
Substituting $c = \frac{1}{21}$,we get $\frac{12}{21} = \frac{4}{7}$.

(C) When a single die is thrown,the total number of possible outcomes is $S = \{1, 2, 3, 4, 5, 6\}$,so the total number of outcomes is $n(S) = 6$.
The even numbers on a die are $\{2, 4, 6\}$,so the number of favorable outcomes is $n(E) = 3$.
The probability of getting an even number is given by $P(E) = \frac{n(E)}{n(S)} = \frac{3}{6} = \frac{1}{2}$.

(C) Total number of cards in a pack = $52$.
Number of aces in a pack = $4$.
Probability of drawing the first ace = $\frac{4}{52} = \frac{1}{13}$.
Since the card is drawn without replacement,the remaining number of cards = $51$ and the remaining number of aces = $3$.
Probability of drawing the second ace = $\frac{3}{51} = \frac{1}{17}$.
The probability that both cards are aces = $\frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$.

(D) When three coins are tossed together,the total number of possible outcomes is $2^3 = 8$.
The sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
Let $E$ be the event of getting at least one head.
The complement event $E'$ is the event of getting no heads,which means getting all tails $(TTT)$.
The number of outcomes for $E'$ is $1$.
Therefore,$P(E') = \frac{1}{8}$.
The probability of getting at least one head is $P(E) = 1 - P(E')$.
$P(E) = 1 - \frac{1}{8} = \frac{7}{8}$.

(C) Let $A$ be the event that $5$ appears on at least one die.
The sample space for two dice is $36$. The outcomes where $5$ appears are: $(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)$.
Total number of outcomes in $A$ is $n(A) = 11$.
Let $B$ be the event that the sum is $10$ or greater.
We need to find the conditional probability $P(B|A) = \frac{n(A \cap B)}{n(A)}$.
The outcomes in $A$ where the sum is $10$ or greater are: $(5, 5), (5, 6), (6, 5)$.
Thus,$n(A \cap B) = 3$.
Therefore,the required probability is $P(B|A) = \frac{3}{11}$.

(B) Let the total number of students in the college be $100$.
Since girls constitute $60\%$ of the total students,the number of girls $= 60$ and the number of boys $= 100 - 60 = 40$.
Number of boys studying Mathematics $= 25\% \text{ of } 40 = \frac{25}{100} \times 40 = 10$.
Number of girls studying Mathematics $= 10\% \text{ of } 60 = \frac{10}{100} \times 60 = 6$.
Total number of students studying Mathematics $= 10 + 6 = 16$.
We need to find the probability that the student is a girl,given that the student studies Mathematics.
Using the conditional probability formula,$P(\text{Girl} | \text{Maths}) = \frac{\text{Number of girls studying Maths}}{\text{Total number of students studying Maths}}$.
$P(\text{Girl} | \text{Maths}) = \frac{6}{16} = \frac{3}{8}$.

(C) Total number of possible outcomes when two dice are thrown is $6 \times 6 = 36$.
The favorable outcomes where the first die shows $1$ are: $(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$.
The number of favorable outcomes is $6$.
Therefore,the required probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}$.

(C) The last digit of any number can be any of the $10$ digits: ${0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$.
For the product of four numbers to have a last digit of $1, 3, 5,$ or $7$,each of the four numbers must have a last digit that is odd and not $9$ (since $9 \times 9 = 81$,which ends in $1$,but $9$ is not in the set ${1, 3, 5, 7}$).
Actually,for the product to end in $1, 3, 5,$ or $7$,each individual number must end in one of these digits.
There are $4$ favorable digits ${1, 3, 5, 7}$ out of $10$ possible digits for each number.
The probability for one number is $P = \frac{4}{10} = \frac{2}{5}$.
Since there are four independent numbers,the probability that all four have a last digit in ${1, 3, 5, 7}$ is $\left( \frac{2}{5} \right)^4 = \frac{16}{625}$.

(A) The total number of possible outcomes when a coin is tossed $n$ times is $2^n$.
The number of ways to get heads an odd number of times is given by the sum of binomial coefficients: $\binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \dots$
It is a known property of binomial coefficients that the sum of odd-indexed terms is equal to the sum of even-indexed terms,which is $2^{n-1}$.
Therefore,the number of favorable outcomes is $2^{n-1}$.
The required probability is $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{2^{n-1}}{2^n} = \frac{1}{2}$.

(B) leap year consists of $366$ days,which is equal to $52$ weeks and $2$ extra days.
These $2$ extra days can be any of the following combinations:
(Sunday,Monday),(Monday,Tuesday),(Tuesday,Wednesday),(Wednesday,Thursday),(Thursday,Friday),(Friday,Saturday),(Saturday,Sunday).
There are $7$ possible outcomes for these $2$ days.
Let $A$ be the event of having $53$ Fridays and $B$ be the event of having $53$ Saturdays.
$P(A) = \frac{2}{7}$ (since Friday occurs in (Thursday,Friday) and (Friday,Saturday)).
$P(B) = \frac{2}{7}$ (since Saturday occurs in (Friday,Saturday) and (Saturday,Sunday)).
$P(A \cap B) = \frac{1}{7}$ (since both occur only in (Friday,Saturday)).
Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{2}{7} + \frac{2}{7} - \frac{1}{7} = \frac{3}{7}$.

(D) The total number of two-digit numbers that can be formed using the digits $1, 2, 3, 4, 5$ with repetition allowed is $5 \times 5 = 25$.
$A$ two-digit number is divisible by $4$ if the number formed by its digits is a multiple of $4$. The possible two-digit numbers formed by these digits that are divisible by $4$ are: $12, 24, 32, 44, 52$.
There are $5$ such favourable outcomes.
The probability is given by the ratio of favourable outcomes to the total number of outcomes:
$P = \frac{5}{25} = \frac{1}{5}$.
Since $1/5$ is not among the given options,the correct answer is 'None of these'.

(B) Total number of cards in a pack is $52$. There is only $1$ ace of heart in the pack.
Let $A$ be the event that the first card is an ace of heart and $B$ be the event that the second card is an ace of heart.
There are two mutually exclusive cases:
$(i)$ The first card is an ace of heart and the second card is not an ace of heart:
$P(E_1) = \frac{1}{52} \times \frac{51}{51} = \frac{1}{52}$
$(ii)$ The first card is not an ace of heart and the second card is an ace of heart:
$P(E_2) = \frac{51}{52} \times \frac{1}{51} = \frac{1}{52}$
Therefore,the required probability is $P(E_1) + P(E_2) = \frac{1}{52} + \frac{1}{52} = \frac{2}{52} = \frac{1}{26}$.

(A) Let $P(A), P(B)$,and $P(C)$ be the probabilities of students $A, B$,and $C$ solving the problem respectively.
$P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4}$.
The problem is solved if at least one of them solves it.
It is easier to calculate the probability that the problem is not solved by anyone.
The probability that $A$ does not solve the problem is $P(A') = 1 - \frac{1}{2} = \frac{1}{2}$.
The probability that $B$ does not solve the problem is $P(B') = 1 - \frac{1}{3} = \frac{2}{3}$.
The probability that $C$ does not solve the problem is $P(C') = 1 - \frac{1}{4} = \frac{3}{4}$.
Since these events are independent,the probability that none of them solve the problem is $P(\text{None}) = P(A') \times P(B') \times P(C') = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{6}{24} = \frac{1}{4}$.
The probability that the problem is solved is $P(\text{Solved}) = 1 - P(\text{None}) = 1 - \frac{1}{4} = \frac{3}{4}$.

(B) The total number of possible outcomes when rolling $2$ dice is $6 \times 6 = 36$.
$A$ doublet occurs when both dice show the same number.
The favourable outcomes are: $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$.
There are $6$ such favourable outcomes.
The probability of getting a doublet is given by the ratio of favourable outcomes to total outcomes:
$\text{Probability} = \frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \frac{6}{36} = \frac{1}{6}.$

(D) Total number of possible outcomes when rolling $2$ dice is $6 \times 6 = 36$.
For the sum to be $7$,the favorable outcomes are $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$. There are $6$ such outcomes.
For the sum to be $12$,the only favorable outcome is $(6, 6)$. There is $1$ such outcome.
Since these are mutually exclusive events,the total probability is the sum of the individual probabilities:
$P(7 \text{ or } 12) = P(7) + P(12) = \frac{6}{36} + \frac{1}{36} = \frac{7}{36}$.

(A) Total number of ways to select $4$ shoes out of $20$ is given by $\binom{20}{4} = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4845$.
To find the probability of at least one pair,we calculate the complement: the probability that no pair is selected.
To select $4$ shoes such that no two form a pair,we must first choose $4$ distinct pairs out of $10$,which can be done in $\binom{10}{4}$ ways. From each of these $4$ pairs,we must choose $1$ shoe,which can be done in $2^4$ ways.
Number of ways to select $4$ shoes with no pair $= \binom{10}{4} \times 2^4 = 210 \times 16 = 3360$.
Probability of no pair $= \frac{3360}{4845} = \frac{224}{323}$.
Probability of at least one pair $= 1 - \frac{224}{323} = \frac{99}{323}$.

(B) Total number of balls in the bag = $3 \text{ (red)} + 7 \text{ (black)} = 10 \text{ balls}$.
Since the first ball taken out is red,it is not replaced.
After removing one red ball,the remaining number of balls in the bag = $10 - 1 = 9$.
The remaining number of red balls in the bag = $3 - 1 = 2$.
Therefore,the probability that the second ball is red,given that the first ball was red,is the number of remaining red balls divided by the total number of remaining balls.
Probability = $\frac{2}{9}$.

(B) leap year contains $366$ days,which is equal to $52$ weeks and $2$ extra days.
There are definitely $52$ Sundays in the $52$ full weeks.
For the remaining $2$ days,the possible combinations 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
There are $7$ total possible outcomes for the remaining $2$ days.
To have $53$ Sundays,one of the remaining $2$ days must be a Sunday.
This occurs in $2$ cases: $(i)$ Sunday and Monday,and $(vii)$ Saturday and Sunday.
Therefore,the required probability is $\frac{2}{7}$.

(B) Let $B_1$ be the event of choosing the first bag and $B_2$ be the event of choosing the second bag. Since the bag is chosen randomly,$P(B_1) = P(B_2) = \frac{1}{2}$.
In the first bag,there are $3$ white and $2$ black balls,so the total number of balls is $5$. The probability of picking a black ball from the first bag is $P(Black|B_1) = \frac{2}{5}$.
In the second bag,there are $2$ white and $4$ black balls,so the total number of balls is $6$. The probability of picking a black ball from the second bag is $P(Black|B_2) = \frac{4}{6} = \frac{2}{3}$.
Using the law of total probability,the probability of picking a black ball is $P(Black) = P(B_1) \times P(Black|B_1) + P(B_2) \times P(Black|B_2)$.
$P(Black) = \frac{1}{2} \times \frac{2}{5} + \frac{1}{2} \times \frac{2}{3} = \frac{1}{5} + \frac{1}{3} = \frac{3+5}{15} = \frac{8}{15}$.

(B) Let $E_1$ be the event of choosing bag $x$ and $E_2$ be the event of choosing bag $y$. Since the bag is chosen at random,$P(E_1) = P(E_2) = 1/2$.
Let $W$ be the event of picking a white ball.
The probability of picking a white ball from bag $x$ is $P(W|E_1) = 3/5$.
The probability of picking a white ball from bag $y$ is $P(W|E_2) = 2/6 = 1/3$.
Using the law of total probability,$P(W) = P(E_1) \cdot P(W|E_1) + P(E_2) \cdot P(W|E_2)$.
$P(W) = (1/2) \cdot (3/5) + (1/2) \cdot (1/3) = 3/10 + 1/6 = (9 + 5) / 30 = 14/30 = 7/15$.

(C) Total number of pens in the first box = $4 + 2 = 6$.
Total number of pens in the second box = $3 + 5 = 8$.
The probability of selecting a white pen from the first box = $\frac{4}{6} = \frac{2}{3}$.
The probability of selecting a white pen from the second box = $\frac{3}{8}$.
Since the selection from each box is an independent event,the probability that both pens are white is the product of their individual probabilities.
Required probability = $\frac{2}{3} \times \frac{3}{8} = \frac{6}{24} = \frac{1}{4}$.

(B) Let $A$ be the first bag and $B$ be the second bag.
For bag $A$: $P(\text{red}_A) = \frac{3}{8}$,$P(\text{black}_A) = \frac{5}{8}$.
For bag $B$: $P(\text{red}_B) = \frac{6}{10}$,$P(\text{black}_B) = \frac{4}{10}$.
The event 'one is red and the other is black' can happen in two mutually exclusive ways:
$1$. Red from bag $A$ and black from bag $B$.
$2$. Black from bag $A$ and red from bag $B$.
Required probability $= P(\text{red}_A) \times P(\text{black}_B) + P(\text{black}_A) \times P(\text{red}_B)$
$= (\frac{3}{8} \times \frac{4}{10}) + (\frac{5}{8} \times \frac{6}{10})$
$= \frac{12}{80} + \frac{30}{80} = \frac{42}{80} = \frac{21}{40}$.

(A) Let $p$ be the probability of hitting the target, so $p = 1/5$.
Let $q$ be the probability of missing the target, so $q = 1 - p = 1 - 1/5 = 4/5$.
For $n = 10$ independent shots, the probability of missing the target in all $10$ shots is $q^{10} = (4/5)^{10}$.
The probability of at least one hit is given by $P(\text{at least one hit}) = 1 - P(\text{no hits})$.
Therefore, $P(\text{at least one hit}) = 1 - (4/5)^{10}$.

(C) When four coins are tossed,the total number of possible outcomes is $2^4 = 16$.
The event of getting 'at least one head' is the complement of the event of getting 'no head' (i.e.,all tails).
The probability of getting no head is $P(\text{no head}) = \frac{1}{16}$.
Therefore,the probability of getting at least one head is $P(\text{at least one head}) = 1 - P(\text{no head})$.
$P(\text{at least one head}) = 1 - \frac{1}{16} = \frac{15}{16}$.