Probability – JEE Main MathematicsPractice Questions & PYQs

A system of linear equations possesses a unique solution only when the determinant of its coefficient matrix is non-zero. For the system defined by the variables $x$, $y$, and $z$, the coefficient matrix leads to a determinant $\Delta = (N-2)(N-3)$. The condition for a unique solution requires $\Delta \neq 0$, which implies that $N$ cannot be 2 or 3.

Considering that $N$ represents the outcome of a fair six-sided die, the possible values are 1, 2, 3, 4, 5, and 6. Excluding the values 2 and 3, the system retains a unique solution for $N$ values of 1, 4, 5, and 6, providing four favorable outcomes and establishing $k=4$. Summing these valid outcomes of 1, 4, 5, and 6 with the value of $k$ results in 20.

When two standard dice are rolled, the possible sums $N$ range from 2 to 12. The inequality $2^N < N!$ is tested by observing specific integer values. For $N=2$, the inequality $4 < 2$ is false, and for $N=3$, the inequality $8 < 6$ is also false. Beginning at $N=4$, the factorial grows faster than the power of two, as $16 < 24$, meaning the condition holds for all $N \ge 4$.

To determine the probability, we consider the complementary case where $N < 4$. There is only one outcome for $N=2$, namely $(1,1)$, and two outcomes for $N=3$, specifically $(1,2)$ and $(2,1)$. Out of the 36 equally likely outcomes for two dice, the probability of obtaining a sum less than 4 is $\frac{1}{36} + \frac{2}{36} = \frac{3}{36}$, which simplifies to $\frac{1}{12}$.

The probability that the inequality holds is therefore $1 - \frac{1}{12} = \frac{11}{12}$. By identifying the coprime integers $m=11$ and $n=12$, we calculate the final expression as $4(11) - 3(12)$, which results in $44 - 36 = 8$.

Probability represents the ratio of favorable occurrences to the entire sample space. Rolling three six-sided dice results in $6 \times 6 \times 6 = 216$ total possible outcomes. For the numbers on the dice to be unique, three distinct values must be chosen from the set of six and arranged in specific orders. This arrangement is determined by $\binom{6}{3} \times 3!$, which counts the number of ways to select and order these values.

This calculation results in $20 \times 6 = 120$ favorable outcomes. Dividing the favorable outcomes by the total sample space produces the fraction $\frac{120}{216}$, which simplifies to $\frac{5}{9}$. With $p=5$ and $q=9$ being co-prime, the value of $q-p$ is calculated as $9 - 5 = 4$.

Bayes' Theorem provides the standard framework for this problem, as it allows us to calculate the probability of a specific cause given an observed effect. By working backward from the occurrence of a defective bolt to the probability of its origin, we can determine which machine was responsible.

To begin, calculate the overall probability of selecting a defective bolt by summing the individual contributions from each machine. Machine A produces $0.2 \times 0.03 = 0.006$ of the total defects, Machine B produces $0.3 \times 0.04 = 0.012$, and Machine C produces $0.5 \times 0.02 = 0.01$. Combining these values shows that the total probability of pulling a defective bolt from the factory output is $0.006 + 0.012 + 0.01 = 0.028$.

The final step is to isolate Machine C's contribution to these defects relative to the total probability calculated above. Dividing the defect rate of Machine C by the total defect rate yields:

$\frac{0.5 \times 0.02}{0.2 \times 0.03 + 0.3 \times 0.04 + 0.5 \times 0.02} = \frac{0.01}{0.028}$

Simplifying this fraction leads to $\frac{10}{28}$, which reduces further to $\frac{5}{14}$.

This problem relies on the law of total probability, where the overall likelihood of an event is the sum of conditional probabilities weighted by the probability of each distinct condition. Here, the number of balls drawn, denoted by $n$, corresponds to the outcome of a fair die roll ranging from 1 to 6, each occurring with a probability of $\frac{1}{6}$. For any specific roll $n$, the probability that all drawn balls are white is calculated by dividing the number of ways to choose $n$ white balls from 6 available by the total number of ways to choose any $n$ balls from the entire set of 10.

Mathematically, we aggregate these individual scenarios by summing the ratios for each $n$ from 1 to 6 and multiplying the entire sum by the probability of rolling the die. The expression becomes:
$P = \frac{1}{6} \left( \frac{^6C_1}{^{10}C_1} + \frac{^6C_2}{^{10}C_2} + \frac{^6C_3}{^{10}C_3} + \frac{^6C_4}{^{10}C_4} + \frac{^6C_5}{^{10}C_5} + \frac{^6C_6}{^{10}C_6} \right)$
Computing the values for each combination gives:
$P = \frac{1}{6} \left( \frac{126+70+35+15+5+1}{210} \right) = \frac{42}{210} = \frac{1}{5}$

When rolling a die $n$ times, outcomes follow a binomial distribution where the probability of getting an odd number is $\frac{1}{2}$. Since the probability of rolling an odd number seven times equals the probability of rolling it nine times, the binomial coefficients ${}^n C_7$ and ${}^n C_9$ must be identical, which occurs when $n = 7 + 9 = 16$. The probability of getting an even number exactly twice is ${}^{16} C_2 \times (\frac{1}{2})^{16}$, and because ${}^{16} C_2 = 120$, this simplifies to $\frac{120}{2^{16}}$. Reducing this fraction to $\frac{60}{2^{15}}$ allows for a direct comparison with the expression $\frac{k}{2^{15}}$, which reveals $k = 60$.

The product of two positive integers with a fixed sum is maximized when the two integers are as close as possible. Given the sum is 66, the maximum value $M$ occurs at $x = 33$, yielding $M = 33 \times 33 = 1089$. The sample space $S$ is defined by the inequality $x(66-x) \geq \frac{5}{9} M$, which simplifies to $x(66-x) \geq \frac{5}{9} \times 1089$, or $x(66-x) \geq 605$. Rearranging this into a standard quadratic inequality, we get $x^2 - 66x + 605 \leq 0$. Solving the corresponding quadratic equation reveals that the boundaries for $x$ are 11 and 55, meaning $S$ consists of all integers $x$ such that $11 \leq x \leq 55$.

The total number of integers in the sample space $S$ is $55 - 11 + 1 = 45$. The event $A$ comprises the multiples of 3 within this set, specifically the integers $12, 15, 18, \dots, 54$. Using the formula for the number of terms in an arithmetic progression, the count of these favorable outcomes is $n = \frac{54 - 12}{3} + 1 = 15$. The probability $P(A)$ is the ratio of favorable outcomes to the total number of outcomes, which is $P(A) = \frac{15}{45}$, and simplifying this fraction results in $P(A) = \frac{1}{3}$.

We first determine the value of $\lambda$ using Bayes' Theorem, which relates the conditional probability of selecting urn C given that a red ball is drawn to the individual probabilities of drawing a red ball from each urn. Since each of the three urns is selected with a probability of 1/3, the probability of selecting a red ball from urn A is 0.4, from urn B is 0.5, and from urn C is $\frac{\lambda}{\lambda+4}$. By applying the formula for conditional probability, we have:

$0.4 = \frac{\frac{1}{3} \cdot \frac{\lambda}{\lambda+4}}{\frac{1}{3}(0.4 + 0.5 + \frac{\lambda}{\lambda+4})}$

Simplifying the equation by canceling the 1/3 factors gives $0.4 = \frac{\frac{\lambda}{\lambda+4}}{0.9 + \frac{\lambda}{\lambda+4}}$. Multiplying both sides by the denominator results in $0.4(0.9 + \frac{\lambda}{\lambda+4}) = \frac{\lambda}{\lambda+4}$, which further simplifies to $0.36 = 0.6 \frac{\lambda}{\lambda+4}$, leading to the conclusion that $\lambda = 6$.

With the parabola defined by $y^2 = 6x$, we consider an equilateral triangle inscribed with one vertex at the origin $(0, 0)$. Due to the symmetry of the parabola about the x-axis, the other two vertices must be located at $(x, y)$ and $(x, -y)$. The line from the origin to $(x, y)$ forms an angle of $30^{\circ}$ with the x-axis, meaning the slope is $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$. Since the slope is also equal to $\frac{y}{x}$, we can substitute the parabola equation $x = \frac{y^2}{6}$ to obtain the relation $\frac{y}{y^2/6} = \frac{6}{y} = \frac{1}{\sqrt{3}}$, which solves to $y = 6\sqrt{3}$. Substituting this back, we find $x = \frac{(6\sqrt{3})^2}{6} = 18$. The square of the length of the side of the triangle is the square of the distance from the origin $(0, 0)$ to $(18, 6\sqrt{3})$, which is $18^2 + (6\sqrt{3})^2 = 324 + 108 = 432$.

The variance of the difference between two independent random variables, $\alpha$ and $\beta$, is the sum of their individual variances, defined by the property $\text{Var}(\alpha - \beta) = \text{Var}(\alpha) + \text{Var}(\beta)$. Because $\alpha$ and $\beta$ are independent outcomes of a standard six-sided die, their probability distributions are identical. For a single die roll, the expected value is $E[\alpha] = \frac{1+2+3+4+5+6}{6} = 3.5$, and the mean of the squared outcomes is E\[\alpha^2]$ = \frac{1^{2+}2^{2+}3^{2+}4^{2+}5^{2+}6^2}{6} = \frac{91}{6}$.

Subtracting the square of the mean from the mean of the squared outcomes yields the variance for a single die:
$\text{Var}(\alpha) = \frac{91}{6} - (3.5)^2 = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12}$
Summing these values, the variance of $\alpha - \beta$ is $\frac{35}{12} + \frac{35}{12} = \frac{35}{6}$. In the resulting fraction $\frac{p}{q} = \frac{35}{6}$, the integers $p=35$ and $q=6$ are co-prime. The number $p$ can be expressed as the product of its prime factors, $5^1 \times 7^1$, so the sum of its positive divisors is calculated as $(5^{0+}5^1)(7^{0+}7^1) = (6)(8) = 48$.

To determine the probability that a randomly chosen matrix with entries from $\{0, 1, 2\}$ is invertible, we focus on the condition that its determinant must be non-zero. Since calculating the total number of non-invertible matrices is more straightforward, we calculate the probability of the complement event, where the determinant is zero, and subtract this from one. A 2x2 matrix has four entries, and with three possible choices for each entry, the total number of matrices in the sample space is $3^4 = 81$.

The determinant $ad - bc$ is zero if and only if $ad = bc$. We analyze this by considering all possible product values $k = ad = bc$ for $k \in \{0, 1, 2, 4\}$. If $k=0$, the product $ad=0$ occurs when at least one entry is zero, leaving $3^2 - 2^2 = 5$ possibilities for the pair $(a, d)$, and similarly for $(b, c)$, resulting in $5 \times 5 = 25$ matrices. When $k=1$, the only pair is $(1, 1)$, giving $1 \times 1 = 1$ matrix. For $k=2$, the pairs are $(1, 2)$ and $(2, 1)$, which leads to $2 \times 2 = 4$ matrices. Finally, when $k=4$, the only pair is $(2, 2)$, resulting in $1 \times 1 = 1$ matrix.

Summing these cases, we find that there are $25 + 1 + 4 + 1 = 31$ non-invertible matrices. Subtracting these from the total, we have $81 - 31 = 50$ invertible matrices. Thus, the probability of choosing an invertible matrix is $\frac{50}{81}$.

Probability theory distinguishes between impossible events and events with zero probability, particularly in continuous sample spaces where an outcome can have a measure of zero without being impossible. Consider the sample space $\Omega = [0, 1]$ and the event $A = \{ \frac{1}{2} \}$; here, $P(A) = 0$ is true, yet $A$ is not the empty set $\phi$. Similarly, for the complement $\bar{A}$, we find $P(\bar{A}) = 1$ even though $\bar{A}$ excludes the point $\frac{1}{2}$ and consequently $\bar{A} \neq \Omega$, which demonstrates that both initial statements are false.

Calculating the probability of a target sum over multiple independent trials involves identifying every unique combination of die rolls that achieves that total and summing their individual probabilities. For a die with faces marked 2, 4, 8, 16, 32, and 32, the probability of rolling a specific value $n$ is $\frac{1}{n}$, meaning the probabilities for the relevant outcomes are $P(8) = \frac{1}{8}$, $P(16) = \frac{1}{16}$, and $P(32) = \frac{1}{32} + \frac{1}{32} = \frac{2}{32}$, which simplifies to $\frac{1}{16}$.

To achieve a sum of 48 across three rolls, the possible combinations are $(32, 8, 8)$ and $(16, 16, 16)$. For the combination $(32, 8, 8)$, the sum can be reached in three distinct sequences: $(32, 8, 8)$, $(8, 32, 8)$, and $(8, 8, 32)$. The probability for any one of these sequences is $\frac{2}{32} \times \frac{1}{8} \times \frac{1}{8} = \frac{2}{2048} = \frac{1}{1024}$. Multiplying by the three possible arrangements yields $3 \times \frac{1}{1024} = \frac{3}{2^{10}}$.

The only other way to reach a sum of 48 is by rolling $(16, 16, 16)$. Since all three dice must be 16, there is only one arrangement, and the probability is $\frac{1}{16} \times \frac{1}{16} \times \frac{1}{16} = \frac{1}{4096}$, which is $\frac{1}{2^{12}}$. Summing these two distinct possibilities requires a common denominator, where $\frac{3}{2^{10}}$ is equivalent to $\frac{12}{2^{12}}$. Adding $\frac{12}{2^{12}}$ and $\frac{1}{2^{12}}$ results in a total probability of $\frac{13}{2^{12}}$.

The probability that a specific event originated from a particular source is determined by Bayes' Theorem, which compares the likelihood of the event occurring in each scenario to the total probability across all possibilities. Given two bags with different ball compositions, we can calculate the probability of drawing one red and one black ball from each bag and then apply the theorem to solve for the unknown quantity of white balls in Bag B.

Bag A contains a total of 6 balls. The number of ways to choose 1 red ball and 1 black ball is ${}^3C_1 \times {}^1C_1 = 3$, and the total ways to choose any 2 balls is ${}^6C_2 = 15$, resulting in a probability of $\frac{3}{15} = \frac{1}{5}$. Bag B contains $n+5$ balls. The number of ways to choose 1 red and 1 black ball is ${}^2C_1 \times {}^3C_1 = 6$, and the total ways to choose any 2 balls is $\frac{(n+5)(n+4)}{2}$. Thus, the probability of selecting 1 red and 1 black ball from Bag B is $\frac{12}{(n+5)(n+4)}$.

Applying Bayes' Theorem, the conditional probability of selecting Bag A given the observed outcome is expressed by the following equation:
$\frac{\frac{1}{2} \cdot \frac{1}{5}}{\frac{1}{2} \cdot \frac{1}{5} + \frac{1}{2} \cdot \frac{12}{(n+5)(n+4)}} = \frac{6}{11}$
Simplifying by canceling the common factor of $\frac{1}{2}$ leads to the equation $\frac{1/5}{1/5 + \frac{12}{(n+5)(n+4)}} = \frac{6}{11}$. Cross-multiplying the terms gives $\frac{11}{5} = \frac{6}{5} + \frac{72}{(n+5)(n+4)}$. Subtracting $\frac{6}{5}$ from both sides results in $1 = \frac{72}{(n+5)(n+4)}$, which implies $(n+5)(n+4) = 72$. Since $9 \times 8 = 72$, it must be that $n+5 = 9$, which means $n=4$.

A binomial distribution is fundamentally defined by its mean, expressed as $np$, and its variance, given by $npq$. When the sum and product of these two parameters are known, they function as the roots of a quadratic equation. This relationship is structured as x^2 - (\text{\sum})x + (\text{product}) = 0, where the solutions correspond to the mean and the variance.

Plugging in the provided values, the equation becomes $x^2 - 82.5x + 1350 = 0$. Solving this quadratic produces 60 and 22.5 as the roots. Because the variance is always less than the mean in a binomial distribution, the mean is 60 and the variance is 22.5.

The probability $q$ is determined by the ratio of variance to mean, which gives $q = \frac{22.5}{60} = \frac{3}{8}$. Since $p$ is the complement, the probability $p$ is calculated as $p = 1 - \frac{3}{8} = \frac{5}{8}$. Substituting these values into the mean formula $np = 60$ results in $n \cdot \frac{5}{8} = 60$. Solving this linear equation confirms that the number of trials $n = 96$.

The sum of probabilities for any valid random variable distribution must be unity, leading to the equation $k + 2k + 4k + 6k + 8k = 1$, which confirms that $21k = 1$ and therefore $k = \frac{1}{21}$. By applying the conditional probability definition $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$, we evaluate the probability of the event $1 < X < 4$ given the condition $X \le 2$. The denominator $P(X \le 2)$ includes the outcomes where $X$ is 0, 1, or 2, summing to $k + 2k + 4k = 7k$, while the intersection of $1 < X < 4$ and $X \le 2$ restricts the outcomes to just $X = 2$, which has a probability of $4k$. Dividing the intersection probability by the conditional probability results in $\frac{4k}{7k}$, which simplifies to $\frac{4}{7}$.

The binomial probability mass function $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$ governs the random variable $X$ with $n=7$. Substituting the condition $P(X=3) = 5 P(X=4)$ into this formula gives $\binom{7}{3} p^3 (1-p)^4 = 5 \binom{7}{4} p^4 (1-p)^3$. Because $\binom{7}{3}$ is equal to $\binom{7}{4}$, these coefficients cancel out, and dividing both sides by $p^3 (1-p)^3$ yields $(1-p) = 5p$. Solving this linear equation leads to $6p = 1$, or $p = \frac{1}{6}$, which implies the probability of failure is $q = 1 - p = \frac{5}{6}$.

The mean of this distribution is $np = 7 \times \frac{1}{6} = \frac{7}{6}$, and the variance is $npq = 7 \times \frac{1}{6} \times \frac{5}{6} = \frac{35}{36}$. Adding these two values together results in $\frac{7}{6} + \frac{35}{36} = \frac{42}{36} + \frac{35}{36} = \frac{77}{36}$.

A $2 \times 2$ matrix is singular if its determinant equals zero, which occurs when the product of the diagonal entries equals the product of the off-diagonal entries. With each of the four entries independently chosen from the first 10 prime numbers, there are $10^4$ total possible matrices in the sample space. The singular matrices arise either when all entries are identical, which accounts for 10 cases, or when exactly two distinct primes are utilized in specific configurations, which contributes another 180 cases. Combining these possibilities yields 190 singular matrices in total, and evaluating the probability relative to the sample space results in $\frac{190}{10^4}$, which simplifies to $\frac{19}{10^3}$.

A relation on the set $\{x, y\}$ is a subset of the Cartesian product of the set with itself, meaning there are $2^{2 \times 2} = 16$ total possible relations. For a relation to be both symmetric and transitive, it must satisfy specific conditions that limit the inclusion of ordered pairs. The favorable relations are the empty set, the singleton sets $\{(x, x)\}$ and $\{(y, y)\}$, the identity relation $\{(x, x), (y, y)\}$, and the universal relation $\{(x, x), (y, y), (x, y), (y, x)\}$. Since there are 5 favorable outcomes out of the total, the probability is $\frac{5}{16}$.

The variance of a discrete random variable $\sigma^2$ is determined by the formula $\sigma^2 = E(X^2) - (E(X))^2$, where $E(X)$ is the expected value of the outcomes and $E(X^2)$ is the expected value of the square of the outcomes. Drawing three balls from a bag containing 4 white and 6 black balls results in a total of $\binom{10}{3} = 120$ possible combinations. The probability distribution for $X$, representing the count of white balls drawn, is calculated as follows: $P(X=0) = \binom{6}{3} / 120 = 1/6$, $P(X=1) = (\binom{4}{1} \times \binom{6}{2}) / 120 = 1/2$, $P(X=2) = (\binom{4}{2} \times \binom{6}{1}) / 120 = 3/10$, and $P(X=3) = \binom{4}{3} / 120 = 1/30$.

The mean value of the random variable is found by summing the product of each outcome and its probability: $E(X) = 0(1/6) + 1(1/2) + 2(3/10) + 3(1/30) = 0 + 0.5 + 0.6 + 0.1 = 1.2$. To find $E(X^2)$, we weight the square of each outcome by its respective probability, which yields $E(X^2) = 0^2(1/6) + 1^2(1/2) + 2^2(3/10) + 3^2(1/30) = 0 + 0.5 + 1.2 + 0.3 = 2.0$. Substituting these values into the variance equation results in $\sigma^2 = 2.0 - (1.2)^2 = 2.0 - 1.44 = 0.56$. Consequently, the value of $100 \sigma^2$ is 56.

For any set of mutually exclusive events, each individual probability must satisfy the condition $0 \le P(E_i) \le 1$, and their collective sum must satisfy $0 < \sum P(E_i) \le 1$. Applying these constraints, the condition $0 \le \frac{2+3p}{6} \le 1$ restricts $p$ to the interval $[-\frac{2}{3}, \frac{4}{3}]$. Similarly, evaluating $0 \le \frac{2-p}{8} \le 1$ and $0 \le \frac{1-p}{2} \le 1$ yields the intervals $[-6, 2]$ and $[-1, 1]$ respectively. Intersecting these individual constraints identifies that $p$ must lie within $[-\frac{2}{3}, 1]$. Considering the sum condition, we evaluate $0 < \frac{2+3p}{6} + \frac{2-p}{8} + \frac{1-p}{2} \le 1$, which simplifies to $0 < \frac{26-3p}{24} \le 1$. This inequality implies $p \ge \frac{2}{3}$, and when combined with our previous intersection, defines the valid range for $p$ as $[\frac{2}{3}, 1]$. Given that the minimum and maximum values of $p$ are $p_2 = \frac{2}{3}$ and $p_1 = 1$, their sum is $p_1+p_2 = \frac{5}{3}$.