Probability – JEE Main MathematicsPractice Questions & PYQs

The total number of one-one functions from a set of four elements to a set of five elements is determined by the permutation ${}^5P_4$, which equals 120. Favorable cases occur for distinct mappings $(f(a), f(b), f(c), f(d))$ from the set $\{1, 2, 3, 4, 5\}$ that satisfy the condition $f(a)+2 f(b) -f(c)=f(d)$. Since 6 such arrangements exist that adhere to the one-one requirement, the probability is found by taking the ratio $\frac{6}{120}$, which simplifies to $\frac{1}{20}$.

The probability of any subset $A$ within the sample space $S$ is calculated by summing the probabilities of its individual elements $E_n$, which simplifies to the sum of the indices of the elements in $A$ divided by 36. For the condition $P(A) \geq \frac{4}{5}$ to hold, the sum of the indices of the elements in the subset $A$ must be at least $36 \times \frac{4}{5}$, which equals 28.8; thus, the sum must be at least 29. Because calculating subsets with a sum of 29 or greater is cumbersome, it is more efficient to consider the complement set $A^c = S \setminus A$, where the sum of the indices of the elements in $A^c$ must be less than or equal to $36 - 29 = 7$.

By identifying all subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ that sum to 7 or less, we can determine the number of valid sets $A$. The empty set results in a sum of 0, which is one possible combination. For a sum of 1, there is only one set, $\{1\}$. For a sum of 2, there is one set, $\{2\}$. For a sum of 3, there are two sets, $\{3\}$ and $\{1, 2\}$. For a sum of 4, there are two sets, $\{4\}$ and $\{1, 3\}$. For a sum of 5, there are three sets: $\{5\}$, $\{1, 4\}$, and $\{2, 3\}$. For a sum of 6, there are four sets: $\{6\}$, $\{1, 5\}$, $\{2, 4\}$, and $\{1, 2, 3\}$. Finally, for a sum of 7, there are five sets: $\{7\}$, $\{1, 6\}$, $\{2, 5\}$, $\{3, 4\}$, and $\{1, 2, 4\}$. Summing these combinations yields $1 + 1 + 1 + 2 + 2 + 3 + 4 + 5 = 19$ possible subsets $A$ that satisfy the probability requirement.

In a binomial distribution where the mean is $np = \alpha$ and the variance is $npq = \frac{\alpha}{3}$, the probability of failure $q$ is determined by dividing the variance by the mean, resulting in $q = \frac{1}{3}$. Consequently, the probability of success is $p = 1 - q = \frac{2}{3}$. Using the condition $P(X=1) = \binom{n}{1} p^1 q^{n-1} = \frac{4}{243}$, we substitute these values to obtain $n \cdot \frac{2}{3} \cdot (\frac{1}{3})^{n-1} = \frac{4}{243}$. This simplifies to $n \cdot \frac{2}{3^n} = \frac{4}{243}$, which confirms that the number of trials is $n = 6$. Finally, calculating $P(X=4) + P(X=5)$ involves evaluating $\binom{6}{4} (\frac{2}{3})^4 (\frac{1}{3})^2 + \binom{6}{5} (\frac{2}{3})^5 (\frac{1}{3})^1$, which yields $\frac{240}{729} + \frac{192}{729} = \frac{432}{729}$, reducing to $\frac{16}{27}$.

Two numbers are coprime if their highest common factor is 1, which means the randomly chosen number $n$ from the set $\{1, 2, \dots, 2022\}$ must not be divisible by the prime factors of 2022. Since the prime factorization of 2022 is $2 \times 3 \times 337$, we seek the count of numbers in the set that are not divisible by 2, 3, or 337.

Calculating the multiples of 2 and 3 yields 1011 and 674 integers, respectively. Applying the principle of inclusion-exclusion, we subtract the multiples of 6 to avoid double-counting, as there are 337 numbers divisible by 6. This calculation, $1011 + 674 - 337$, results in 1348 numbers divisible by either 2 or 3.

Next, identify the multiples of 337 that were not included in this count. The full set of multiples of 337 is $\{337, 674, 1011, 1348, 1685, 2022\}$. Among these, only 337 and 1685 are not already accounted for as multiples of 2 or 3. Adding these 2 values to the previous 1348 yields a total of 1350 numbers that are not coprime to 2022. Subtracting this from 2022 leaves 672 numbers that share no common factors with 2022. The required probability is therefore $\frac{672}{2022}$, which simplifies to $\frac{112}{337}$.

A system of linear equations possesses a unique solution if and only if the determinant of its coefficient matrix is non-zero, whereas it has no solution if the determinant is zero but the augmented matrix remains inconsistent. The determinant of the coefficient matrix for this system is calculated as follows:
$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{vmatrix} = \lambda - 5$
For a unique solution, we require $\Delta \neq 0$, which means $\lambda \neq 5$. Since $\lambda$ is the outcome of a fair die roll, there are 5 favorable values out of 6, resulting in a probability $p = \frac{5}{6}$.

For the system to have no solution, we must have $\Delta = 0$, which restricts $\lambda$ to the value 5. In this case, we check for inconsistency using the determinant of the matrix formed by replacing the third column with the constants, denoted as $\Delta_3$:
$\Delta_3 = \begin{vmatrix} 1 & 1 & 5 \\ 1 & 2 & \mu \\ 1 & 3 & 1 \end{vmatrix} = 6 - 2\mu$
The system is inconsistent if $\Delta_3 \neq 0$, which simplifies to $6 - 2\mu \neq 0$, or $\mu \neq 3$. Among the 6 possible values for $\mu$, 5 satisfy this condition. Because the rolls are independent, the probability $q$ of having no solution is the product of the probability that $\lambda = 5$ (which is $\frac{1}{6}$) and the probability that $\mu \neq 3$ (which is $\frac{5}{6}$), resulting in $q = \frac{5}{36}$.

The number of tosses required to observe the first six follows a geometric distribution, which is defined by a memoryless property. This property ensures that if we have already failed to roll a six in the first two attempts, the probability of requiring additional tosses to achieve the first success is independent of the past outcomes. Because the condition $X > 2$ implies that the first two trials resulted in failure, the conditional probability $P(X \geq 5 | X > 2)$ is equivalent to the probability of failing two more times in succession. Since each toss is independent with a failure probability of $\frac{5}{6}$, we simply calculate $(\frac{5}{6})^2$, which results in $\frac{25}{36}$.

Rolling a die multiple times creates a sequence of independent trials that follow the binomial distribution, where the probability of success, such as rolling an odd number, is consistently $\frac{1}{2}$. When the probability of rolling an odd number exactly twice equals the probability of rolling an even number exactly three times across $n$ rolls, the geometric terms $(\frac{1}{2})^n$ cancel out, reducing the equality to the binomial coefficients ${}^{n}C_{2} = {}^{n}C_{3}$. Based on the symmetry of combinations, this condition is satisfied when the sum of the counts equals $n$, meaning the die is rolled $n=5$ times.

Determining the likelihood of obtaining an odd number an odd number of times requires summing the probabilities for exactly one, three, or five occurrences. Using the binomial formula, this sum becomes ${}^{5}C_{1}\left(\frac{1}{2}\right)^{5} + {}^{5}C_{3}\left(\frac{1}{2}\right)^{5} + {}^{5}C_{5}\left(\frac{1}{2}\right)^{5}$. Substituting the combination values $5 + 10 + 1$ and multiplying by the total outcomes of $2^{5}$, or $32$, results in $\frac{16}{32}$, which simplifies to $\frac{1}{2}$.

The requirement that $2^{n} - 2$ is a multiple of 3 is equivalent to the modular congruence $2^{n} \equiv 2 \pmod 3$. Because 2 is congruent to -1 modulo 3, this expression simplifies to $(-1)^{n} \equiv -1 \pmod 3$, a condition that holds true if and only if $n$ is an odd integer.

There are 90 possible two-digit numbers in the range from 10 to 99. Within this set, half of the numbers are even and half are odd, resulting in 45 odd integers. The probability is calculated by dividing these 45 favorable outcomes by the 90 total possibilities, resulting in $\frac{45}{90}$, which simplifies to $\frac{1}{2}$.

An electric instrument operates only if both independent units function. Given the first unit functions with a probability of 0.9 and the second with 0.8, the probability of the instrument working is $0.9 \times 0.8 = 0.72$. The total probability of the instrument failing is the complement of this successful state, determined as $1 - 0.72 = 0.28$.

The specific failure case where the first unit fails while the second unit functions depends on the probability of the first failing (0.1) and the second functioning (0.8). Their intersection is $0.1 \times 0.8 = 0.08$. The conditional probability $P$ is the ratio of this intersection to the total probability of failure, which is $P = 0.08 / 0.28 = 8 / 28$. Multiplying this value by 98, expressed as $98 \times (8 / 28)$, results in 28.

The mean of a discrete random variable is found by calculating the weighted average of all possible outcomes, which requires evaluating the infinite series $\sum_{j=1}^{\infty } j \cdot (\frac{1}{3})^j$. By treating this as an arithmetico-geometric series that follows the pattern $\sum_{j=1}^{\infty } j x^j = \frac{x}{(1-x)^2}$ for $|x| < 1$, we substitute $x = \frac{1}{3}$ to obtain a mean value of $\frac{1/3}{(2/3)^2} = \frac{3}{4}$.

To determine the probability that the random variable is both positive and even, we sum the probabilities for all even integers $j=2, 4, 6, \dots$. This results in a geometric series $\sum_{k=1}^{\infty } (\frac{1}{3^2})^k = \frac{1}{9} + \frac{1}{81} + \dots$, where the first term is $a = \frac{1}{9}$ and the common ratio is $r = \frac{1}{9}$. Applying the infinite geometric sum formula $\frac{a}{1-r}$ yields $\frac{1/9}{1 - 1/9} = \frac{1/9}{8/9}$, which simplifies to $\frac{1}{8}$.

Each element of the set $\{1, 2, 3, 4, 5\}$ can exist in one of four mutually exclusive states in relation to two randomly selected subsets: it can be in both, in only the first, in only the second, or in neither. Because each of the five elements has these four independent choices, the total number of ways to form the two subsets is $4^{5}$, which is equivalent to $2^{10}$.

For the intersection to contain exactly two elements, we must choose 2 specific elements from the 5 to be in the intersection, which can be done in ${}^{5}C_{2}$ ways. The remaining 3 elements must then be assigned to any of the 3 other possible states-in only the first, in only the second, or in neither-which provides $3^{3}$ independent possibilities. Multiplying these counts gives ${}^{5}C_{2} \times 3^{3} = 270$ favorable outcomes, and dividing this by the total sample space results in a probability of $\frac{270}{2^{10}}$, which simplifies to $\frac{135}{2^{9}}$.

The probability of getting at least one head in a series of coin tosses is most efficiently determined by subtracting the probability of getting no heads at all from the total probability of $1$. Since a fair coin has a probability of $0.5$ for tails, the likelihood of landing on tails across all $n$ tosses is $(\frac{1}{2})^n$, leading to the inequality $1 - (\frac{1}{2})^n \ge 0.9$. This simplifies to $(\frac{1}{2})^n \le 0.1$, which is equivalent to $2^n \ge 10$. Since $2^3 = 8$ falls short and $2^4 = 16$ satisfies the requirement, the minimum integer $n$ is $4$.

The total number of ways to select two distinct squares from a chessboard is defined by the combination ${}^{64} C_{2}$, which equals $32 \times 63$ total pairings. Favorable outcomes occur only when two squares share a common edge, either horizontally or vertically. Since each row of eight squares contains seven adjacent pairs and there are eight rows, there are 56 horizontal pairs; similarly, each of the eight columns provides seven vertical pairs, leading to 56 vertical pairings for a total of $2 \times 7 \times 8$ successful outcomes. Comparing these favorable configurations to the total possibilities yields the fraction $\frac{2 \times 7 \times 8}{32 \times 63}$, which simplifies to $\frac{1}{18}$.

Probability is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes within a sample space. When rolling two dice, where each face is numbered $\{1, 2, 3, 5, 7, 11\}$, the total number of unique combinations is determined by multiplying the possible values of the first die by those of the second, which results in $6 \times 6 = 36$ total outcomes.

To identify the specific outcomes where the sum of the two top faces is less than or equal to 8, we can test each possible value for the first die systematically. If the first die shows 1, the second die can be 1, 2, 3, 5, or 7, providing 5 valid pairs since the maximum sum is $1+7=8$. When the first die is 2, the second die can be 1, 2, 3, or 5, yielding 4 valid pairs because $2+7=9$ exceeds our limit. Starting with 3 on the first die, the second die can be 1, 2, 3, or 5, creating another 4 valid pairs since $3+7=10$ is too high. If the first die is 5, the second die can be 1, 2, or 3, contributing 3 pairs as $5+5=10$ is too high. Finally, if the first die is 7, only 1 is possible for the second die because $7+1=8$. Summing these favorable outcomes gives $5 + 4 + 4 + 3 + 1 = 17$, meaning the probability is $\frac{17}{36}$.

A missile hits its target only if it survives interception and then successfully completes the strike. Since these two conditions must both be met, the probability of a single missile scoring a hit is the product of the probability that it remains unintercepted and the probability that it hits given that it is not intercepted. Given that the probability of interception is $\frac{1}{3}$, the missile has a $\frac{2}{3}$ chance of proceeding unimpeded. Multiplying this by the $\frac{3}{4}$ chance of hitting the target results in a $\frac{1}{2}$ probability for one missile to land successfully. Because the three missiles are fired independently, we calculate the probability of all three hitting by multiplying their individual success probabilities together, leading to $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$.

Forming a $2 \times 2$ matrix with four dice throws creates a sample space of $6^4$ total possibilities. We are looking for the instances where the matrix is non-singular, meaning the determinant $ad - bc \neq 0$, under the strict constraint that all four numbers filling the matrix are unique. By calculating the number of ways to pick and arrange four distinct digits from the set $\{1, 2, 3, 4, 5, 6\}$, we find there are ${}^6C_4 \times 4! = 15 \times 24 = 360$ arrangements that satisfy the distinct-entry requirement.

Among these arrangements, some result in a determinant of zero because $ad = bc$. Identifying the number combinations that satisfy $ad = bc$ with distinct values requires checking for product equality. The combination $\{1, 6\}$ and $\{2, 3\}$ allows for eight arrangements where the product of the diagonals equals the product of the off-diagonals, and the combination $\{2, 6\}$ and $\{3, 4\}$ similarly produces eight such cases. Removing these sixteen singular matrices from the total of 360 distinct-entry arrangements leaves 344 favorable outcomes. The resulting probability is $\frac{344}{1296}$, which simplifies to $\frac{43}{162}$.

The Inclusion-Exclusion Principle provides the standard framework for calculating the probability that at least one of three events $A, B$, and $C$ occurs, defined as $P(A \cup B \cup C) = P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)] + P(A \cap B \cap C)$. The given conditions involve the probability that exactly one of two events occurs, which corresponds to the sum of their individual probabilities minus twice their intersection. This leads to three distinct equations: $P(A) + P(B) - 2P(A \cap B) = 1 - k$, $P(B) + P(C) - 2P(B \cap C) = 1 - 2k$, and $P(C) + P(A) - 2P(C \cap A) = 1 - k$.

Summing these three expressions yields $2[P(A) + P(B) + P(C)] - 2[P(A \cap B) + P(B \cap C) + P(C \cap A)] = 3 - 4k$. Dividing everything by 2 leaves the combined probability term $[P(A) + P(B) + P(C)] - [P(A \cap B) + P(B \cap C) + P(C \cap A)]$ equal to $\frac{3}{2} - 2k$. By substituting this back into the primary Inclusion-Exclusion formula and including the intersection probability $P(A \cap B \cap C) = k^2$, the union of the events is found to be $P(A \cup B \cup C) = (\frac{3}{2} - 2k) + k^2$. Rewriting this expression as $k^2 - 2k + \frac{3}{2}$ allows it to be completed as $(k - 1)^2 + \frac{1}{2}$. Since $(k - 1)^2$ must be positive for all valid values of $k$, the total probability is always greater than $\frac{1}{2}$.

Because $g: S \to S$ is an onto function on a finite set, it is necessarily a bijection, meaning the total number of such functions is $6!$.

The condition $g(3) = 2g(1)$ can be satisfied in three distinct ways: when $g(1)=1$ and $g(3)=2$, when $g(1)=2$ and $g(3)=4$, or when $g(1)=3$ and $g(3)=6$. In each of these cases, two domain elements are fixed, and the remaining 4 elements must be assigned to the remaining 4 values in the codomain in $4!$ ways. Total favorable outcomes are therefore $3 \times 4!$, and the resulting probability is $\frac{3 \times 4!}{6!}$, which simplifies to $\frac{1}{10}$.

Probability relies on identifying the number of favorable outcomes relative to the total number of possible outcomes. A 4-digit number leaves a remainder of 2 when divided by 5 if its unit digit is either 2 or 7. To find the total number of 4-digit integers containing exactly one 7, we consider all cases where the thousands place is not zero and exactly one digit is 7, which equals 2673 total cases.

When the number ends in 7, the thousands place can be any digit from 1-9 except 7, while the other two digits can be any of the 9 digits that are not 7, giving $8 \times 9 \times 9 = 648$ possibilities. When the number ends in 2, we consider all ways to place a 7 in the remaining three positions while treating the thousands place as if it could be zero initially ($3 \times 9 \times 9 = 243$), and then subtract the cases where the thousands place is 0 to avoid invalid 4-digit numbers ($2 \times 9 = 18$), which totals 225 possibilities. Summing these favorable outcomes gives $648 + 225 = 873$. Dividing the favorable outcomes by the total sample space gives the required probability of $\frac{873}{2673}$, which simplifies to $\frac{97}{297}$.

The divisibility of any integer by 3 relies on the sum of its digits being a multiple of 3. Given the digits $\{0, 1, 2, 3, 4, 5, 6\}$, their total sum is 21, so constructing a 6-digit integer involves excluding one digit such that the remaining sum stays divisible by 3, which restricts the excluded value to the set $\{0, 3, 6\}$.

To determine the total sample space, we consider all 6-digit integers formed without repetition. The first position cannot be zero, offering 6 choices, and the remaining 5 positions are filled using the remaining 6 digits in $6!$ ways, creating a total of $6 \cdot 6!$ valid integers.

Favorable outcomes depend on the excluded digit. Removing 0 results in the set $\{1, 2, 3, 4, 5, 6\}$, yielding $6!$ arrangements. Removing 3 or 6 results in sets containing 0, where the first digit must be non-zero, yielding $5 \cdot 5!$ arrangements for each of these two cases. Summing these favorable outcomes, $6! + 5 \cdot 5! + 5 \cdot 5!$, and dividing by the total sample space, $6 \cdot 6!$, simplifies the probability to $\frac{1}{6} + \frac{5}{36} + \frac{5}{36}$, resulting in a final value of $\frac{4}{9}$.