Probability – JEE Main MathematicsPractice Questions & PYQs

With two integers selected from the set $\{0, 1, 2, \ldots, 10\}$ with replacement, every selection can be viewed as an ordered pair $(x, y)$. Since there are 11 possible choices for each variable, the total number of equally likely outcomes is $11 \times 11 = 121$.

The condition $|x-y| > 5$ implies the absolute difference between the chosen numbers must be at least 6. This can be evaluated by identifying all pairs where $y > x$ and doubling the result due to symmetry. When $x=0$, $y$ can range from 6 to 10, contributing 5 pairs. For $x=1$, $y$ ranges from 7 to 10, contributing 4 pairs; for $x=2$, $y$ ranges from 8 to 10, contributing 3 pairs; for $x=3$, $y$ ranges from 9 to 10, contributing 2 pairs; and for $x=4$, $y$ must be 10, contributing 1 pair. The value $x=5$ cannot satisfy the condition as no valid $y$ exists. Summing these instances yields $5+4+3+2+1 = 15$ pairs where $y > x$. Symmetrically, there are 15 pairs where $x > y$, resulting in 30 total favorable outcomes.

Dividing the number of favorable cases by the total number of possible combinations gives the final probability of $\frac{30}{121}$.

Bayes' Theorem provides the standard framework for updating the probability of a hypothesis-in this case, the initial composition of the bag-after observing specific evidence, which is the outcome of drawing 2 white and 2 black balls. Since the distribution of white and black balls is unknown, we assume all valid compositions are equally likely, each with a probability of 1/5. The possible compositions capable of producing our result are those with at least 2 balls of each color, ranging from 2 white and 6 black up to 6 white and 2 black.

To determine the target probability, we calculate the likelihood of drawing 2 white and 2 black balls from the specific case of 4 white and 4 black balls and divide it by the total probability of obtaining this outcome across all valid initial configurations. For any configuration with $w$ white balls and $b$ black balls, the probability of drawing 2 white and 2 black balls is $\frac{\binom{w}{2} \times \binom{b}{2}}{\binom{8}{4}}$. Substituting the values for the scenario where the bag contains 4 white and 4 black balls, we obtain the numerator as $\frac{6 \times 6}{70} = \frac{36}{70}$. Summing the individual likelihoods for all possible configurations results in a total probability of $\frac{15 + 30 + 36 + 30 + 15}{70} = \frac{126}{70}$, and dividing the numerator by this sum yields $\frac{36}{126}$, which simplifies to $\frac{2}{7}$.

The probability of independent wins and losses over a fixed number of trials follows the binomial distribution, where the probability of exactly $x$ wins in 10 matches is given by $\binom{10}{x} (\frac{1}{3})^x (\frac{2}{3})^{10-x}$. Since the total number of matches is 10, the variables must satisfy $x+y=10$, which implies $y=10-x$. The condition $|x-y| \leq 2$ can be rewritten by substituting $y$ to get $|x-(10-x)| \leq 2$, which simplifies to $|2x-10| \leq 2$. This inequality is satisfied when $2x-10$ equals $-2, 0,$ or $2$, corresponding to $x$ values of $4, 5,$ and $6$.

To determine $p$, we sum the binomial probabilities for these specific values of $x$:
$p = \binom{10}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^6 + \binom{10}{5} \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^5 + \binom{10}{6} \left(\frac{1}{3}\right)^6 \left(\frac{2}{3}\right)^4$
Factoring out the denominator $3^{10}$ and multiplying by $3^9$ yields:
$3^9 p = \frac{\binom{10}{4} 2^6 + \binom{10}{5} 2^5 + \binom{10}{6} 2^4}{3}$
Using the binomial coefficients $\binom{10}{4} = 210$, $\binom{10}{5} = 252$, and $\binom{10}{6} = 210$, the expression simplifies to:
$\frac{210 \cdot 64 + 252 \cdot 32 + 210 \cdot 16}{3} = \frac{13440 + 8064 + 3360}{3} = \frac{24864}{3} = 8288$

When two events are independent, such as drawing marbles with replacement, the probability of both occurring in sequence is calculated by taking the product of their individual probabilities. First, find the total number of marbles in the box by summing $10 + 30 + 20 + 15 = 75$. Because the first marble is replaced before the second draw, the composition of the box remains constant, making the events independent with the probability of selecting a red marble as $\frac{10}{75}$ and a white marble as $\frac{30}{75}$. Calculating the product of these values gives
$\frac{10}{75} \times \frac{30}{75} = \frac{300}{5625} = \frac{4}{75}$
for the combined event.

Determining the variance for the number of rotten apples drawn from a mix requires constructing a probability distribution based on selecting two items from a total pool of eighteen. The sample space size is defined by the number of ways to choose two apples from eighteen, which is $\binom{18}{2}$ and results in 153 possible outcomes. Calculating individual probabilities for obtaining $x$ rotten apples yields $P(X=0) = \frac{105}{153}$, $P(X=1) = \frac{45}{153}$, and $P(X=2) = \frac{3}{153}$.

Expected values are derived from these probabilities by summing the products of each outcome and its associated probability:
$E(X) = 0 \times \frac{105}{153} + 1 \times \frac{45}{153} + 2 \times \frac{3}{153} = \frac{51}{153} = \frac{1}{3}$
The second moment, $E(X^2)$, follows from summing the squared outcomes weighted by their probabilities:
$E(X^2) = 0^2 \times \frac{105}{153} + 1^2 \times \frac{45}{153} + 2^2 \times \frac{3}{153} = \frac{57}{153}$
Applying the variance formula $\operatorname{Var}(X) = E(X^2) - (E(X))^2$ combines these terms to provide the final measure:
$\operatorname{Var}(X) = \frac{57}{153} - \left(\frac{1}{3}\right)^2 = \frac{57}{153} - \frac{17}{153} = \frac{40}{153}$

To determine the variance of this discrete probability distribution, we first identify the unknown values $a$ and $b$ by utilizing the properties of the probability mass function. Since the sum of all probabilities must equal one, we have $4a + 6b = 1$, and since the mean $\sum X P(X)$ equals $\frac{46}{9}$, we derive the equation $4a + 20b = \frac{23}{9}$ from the given values. Solving this linear system yields $b = \frac{1}{9}$ and $a = \frac{1}{12}$.

With these variables established, the variance is found using the formula $\sigma^2 = E(X^2) - [E(X)]^2$. Calculating the expected value of the squared variable gives $E(X^2) = 0^2(a) + 2^2(2a) + 4^2(a + b) + 6^2(2b) + 8^2(3b)$, which simplifies to $24a + 280b$ and evaluates to $\frac{298}{9}$. Finally, subtracting the square of the mean produces $\frac{298}{9} - (\frac{46}{9})^2$, which results in a variance of $\frac{566}{81}$.

The requirement that each roll must be greater than the previous one implies that the three dice must show three distinct numbers arranged in strictly increasing order. Since any unique set of three distinct numbers chosen from the six available faces can be arranged in only one strictly increasing order, we simply need to determine the number of ways to select three values from the six, expressed as ${}^{6}C_{3}$.

With $20$ favorable outcomes and a total sample space of $6^3 = 216$ possible outcomes for three rolls, the probability is calculated as the ratio $\frac{20}{216}$. Reducing this fraction yields the final result of $\frac{5}{54}$.

When calculating the probability of sequential events without replacement, we must adjust the composition of the available items after each draw. The first step involves drawing 4 white balls from an initial set of 15 balls, which consists of 6 white and 9 black, occurring with a probability of $\frac{\binom{6}{4}}{\binom{15}{4}}$.

For the second draw, the urn contains 11 balls, as the 4 white balls were removed from the initial total. This remaining set consists of 9 black balls and 2 white balls. Since the second draw requires selecting 4 black balls, we calculate the probability based on these 9 black balls out of the 11 total remaining, resulting in a probability of $\frac{\binom{9}{4}}{\binom{11}{4}}$.

Because these two dependent events must occur in succession to satisfy the condition, we multiply their individual probabilities to determine the total likelihood of the sequence:

$\frac{\binom{6}{4}}{\binom{15}{4}} \times \frac{\binom{9}{4}}{\binom{11}{4}} = \frac{3}{715}$

The condition for a quadratic equation $ax^{2+}bx+c=0$ to have real roots is that its discriminant $D=b^{2-}4ac$ must be greater than or equal to zero. Given that $a, b,$ and $c$ are outcomes from independent rolls of a tetrahedral die, each variable belongs to the set $\{1, 2, 3, 4\}$, creating a total of $4^3=64$ possible combinations.

To determine the number of favorable outcomes, we evaluate $b^2 \ge 4ac$ for each possible value of $b$. When $b=1$, $b^2=1$, and since $4ac$ must be at least 4, no solutions exist. When $b=2$, $b^2=4$, requiring $ac \le 1$, which is only possible when $(a, c) = (1, 1)$. For $b=3$, $b^2=9$, implying $ac \le 2.25$, which is satisfied by the three pairs $(1, 1), (1, 2),$ and $(2, 1)$. Finally, when $b=4$, $b^2=16$, requiring $ac \le 4$, which is satisfied by $(1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (1, 4), (4, 1),$ and $(2, 2)$, yielding eight valid outcomes.

Combining these results, there are $1+3+8=12$ favorable outcomes in total. The probability is $\frac{12}{64}$, which simplifies to $\frac{3}{16}$. Since $\gcd(3, 16)=1$, we have $m=3$ and $n=16$, leading to $m+n=19$.

Selecting a sample without replacement from a finite population where items are categorized as either defective or non-defective follows a hypergeometric distribution. In this specific scenario, the population size is $N=12$ with $K=3$ defectives, and the sample size is $n=5$. Because the selection process is finite and without replacement, the expected value of the number of defectives $X$ is given by $E[X] = n \frac{K}{N}$, which evaluates to $5 \times \frac{3}{12} = 1.25$.

The variance of a hypergeometric random variable incorporates a finite population correction factor, defined by the formula $\sigma^2 = n \frac{K}{N} \left(1 - \frac{K}{N}\right) \left(\frac{N-n}{N-1}\right)$. Applying the given values to this expression leads to $\sigma^2 = 5 \times \frac{3}{12} \times \frac{9}{12} \times \frac{7}{11}$. Reducing this product results in $\sigma^2 = \frac{105}{176}$. Given that $\frac{m}{n} = \frac{105}{176}$ with $\gcd(105, 176) = 1$, the calculation $n-m$ yields $176 - 105 = 71$.

A quadratic equation $ax^2 + bx + c = 0$ yields repeated roots only when the discriminant $b^2 - 4ac = 0$, which requires $b^2 = 4ac$. Since there are 8 choices for each of the three variables, the total number of possible combinations is $8^3 = 512$. By systematically checking values of $b$, we find the favorable triples: when $b = 2$, we have $ac = 1$, yielding (1, 2, 1); when $b = 4$, we have $ac = 4$, yielding (1, 4, 4), (4, 4, 1), and (2, 4, 2); when $b = 6$, we have $ac = 9$, yielding (3, 6, 3); and when $b = 8$, we have $ac = 16$, yielding (2, 8, 8), (8, 8, 2), and (4, 8, 4). There are 8 such successful combinations, so the probability is $\frac{8}{512}$, which simplifies to $\frac{1}{64}$.

The event where Ajay appears in the exam is composed of two mutually exclusive scenarios: one where both Ajay and Vijay appear, and another where Ajay appears while Vijay does not. Expressing this relationship as $P(A) = P(A \cap V) + P(A \cap \overline{V})$ allows us to isolate the specific probability we need to find. Because we know the probability of Ajay not appearing is $\frac{2}{7}$, we calculate his probability of appearing as $P(A) = 1 - \frac{2}{7} = \frac{5}{7}$. Given that the probability of both appearing is $\frac{1}{5}$, we subtract this from the total probability of Ajay appearing, giving $P(A \cap \overline{V}) = \frac{5}{7} - \frac{1}{5}$. Solving this arithmetic yields $\frac{25 - 7}{35}$, which simplifies to $\frac{18}{35}$.

When two positive integers have a fixed sum, their product is greatest when they are equal, which in this case means $x = y = 12$, yielding a maximum product of $12 \times 12 = 144$. The problem requires the product of these integers to be at least $\frac{3}{4}$ of this maximum, so we seek pairs where $xy \ge 108$.

To identify all such pairs where $x+y=24$, we substitute $y = 24 - x$ into the inequality, resulting in $x(24 - x) \ge 108$, which simplifies to $x^2 - 24x + 108 \le 0$. Solving the quadratic equation $x^2 - 24x + 108 = 0$ gives roots of 6 and 18, meaning the inequality holds for all integers $x$ in the range $6 \le x \le 18$. Counting the integers in this inclusive range provides 13 valid values for $x$, each corresponding to a unique pair $(x, y)$.

Given that there are 23 possible pairs of positive integers that sum to 24, ranging from $(1, 23)$ to $(23, 1)$, the probability of selecting a pair that satisfies the condition is $\frac{13}{23}$. Since $\gcd(13, 23) = 1$, we identify $m = 13$ and $n = 23$, which makes the difference $n - m$ equal to 10.

Bayes' Theorem provides the standard framework for calculating the conditional probability of an event given that an outcome has already occurred. Here, the probability of selecting any specific urn is $\frac{1}{3}$, and the likelihood of drawing a black ball from Urn A, Urn B, and Urn C is $\frac{5}{12}$, $\frac{7}{12}$, and $\frac{6}{12}$ respectively. We determine the probability of the ball coming from Urn A by taking the ratio of the probability of picking Urn A and drawing a black ball to the total probability of drawing a black ball from any of the three urns. Calculating this gives the expression $\frac{\frac{1}{3} \times \frac{5}{12}}{\frac{1}{3} \times \left( \frac{5}{12} + \frac{7}{12} + \frac{6}{12} \right)}$, which simplifies to $\frac{5}{5 + 7 + 6}$ and yields a final result of $\frac{5}{18}$.

A quadratic equation $ax^2 + bx + c = 0$ possesses two distinct real roots if its discriminant is strictly greater than zero, which corresponds to the condition $b^2 - 4ac > 0$, or equivalently $b^2 > 4ac$. Given that $a, b,$ and $c$ are integers chosen from the set $\{1, 2, 3, 4, 5, 6\}$, there are $6 \times 6 \times 6 = 216$ total possible combinations. To determine the probability, we must count the favorable outcomes by examining each possible value of $b$ from 1 to 6. For $b=1$ and $b=2$, $b^2$ is less than or equal to $4ac$ for all values of $a$ and $c$, meaning no cases satisfy the condition.

For $b=3$, we require $9 > 4ac$, which implies $ac < 2.25$. This condition is met by the pairs $(a, c)$ being $(1, 1), (1, 2),$ and $(2, 1)$, resulting in 3 favorable cases. When $b=4$, the condition $16 > 4ac$ simplifies to $ac < 4$, which is satisfied by $(1, 1), (1, 2), (2, 1), (1, 3),$ and $(3, 1)$, yielding 5 cases.

For $b=5$, the condition $25 > 4ac$ requires $ac < 6.25$. The pairs fulfilling this are $(1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (1, 4), (4, 1), (1, 5), (5, 1), (1, 6), (6, 1), (2, 3), (3, 2),$ and $(2, 2)$, providing 14 successful outcomes. Finally, for $b=6$, the requirement $36 > 4ac$ means $ac < 9$, which includes $(1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (1, 4), (4, 1), (1, 5), (5, 1), (1, 6), (6, 1), (2, 3), (3, 2), (2, 4), (4, 2),$ and $(2, 2)$, adding another 16 cases. Summing these counts gives $3 + 5 + 14 + 16 = 38$. Since the probability $p$ is the ratio of favorable cases to total outcomes, $p = \frac{38}{216}$, and consequently, $216p$ equals 38.

The expected value of a random variable representing the number of successes in a sample is determined by multiplying the sample size by the probability of drawing a specific type of item. With three balls drawn from a total of nine, where five are blue and four are yellow, the probability of selecting a blue ball is $5/9$ and a yellow ball is $4/9$. This results in mean values of $\overline{X} = 3 \times \frac{5}{9} = \frac{5}{3}$ and $\overline{Y} = 3 \times \frac{4}{9} = \frac{4}{3}$. Substituting these values into the required expression gives $7\left(\frac{5}{3}\right) + 4\left(\frac{4}{3}\right) = \frac{35}{3} + \frac{16}{3} = \frac{51}{3} = 17$.

The probability of combined independent events is calculated by multiplying individual probabilities, while the total likelihood for a set of outcomes requires accounting for every possible arrangement. With the coin biased such that a head is twice as likely as a tail, the probability of obtaining a tail is $\frac{1}{3}$ and the probability of a head is $\frac{2}{3}$. One specific sequence consisting of two tails and one head has a probability of $\frac{1}{3} \times \frac{1}{3} \times \frac{2}{3} = \frac{2}{27}$. Because there are $3$ distinct sequences that yield this combination-TTH, THT, and HTT-the overall probability is found by multiplying this single sequence probability by $3$:
$3 \times \frac{2}{27} = \frac{2}{9}$

The number of defective items in a sample drawn without replacement from a finite population follows a hypergeometric distribution. Calculating the variance requires determining the probability distribution for the random variable $X$, finding the expected value and the expected value of its square, and then using the relation \sigma^2 = E\[X^2]$ - (E[X])^2$.

In a lot of 10 items containing 3 defective ones, drawing a sample of 5 allows for 0, 1, 2, or 3 defective items with corresponding probabilities of $1/12$, $5/12$, $5/12$, and $1/12$. The expected value is $E[X] = 0(1/12) + 1(5/12) + 2(5/12) + 3(1/12)$, which equals $18/12$. The expectation of the square is E\[X^2]$ = 0^2(1/12) + 1^2(5/12) + 2^2(5/12) + 3^2(1/12)$, which simplifies to$34/12$. Using these values in the variance formula,$\sigma^2 = 34/12 - (18/12)^2$, results in$7/12$. Multiplying this variance by 96 gives$96 \times \frac{7}{12}$, which equals 56.

The probability of this event relies on determining the ratio of favorable distribution patterns to the total possible arrangements. Distributing three letters into five addresses allows for $5^3 = 125$ total outcomes since each letter has five independent choices. Focusing on the condition of using exactly two addresses, we select the two target addresses using ${}^5C_2 = 10$, and for these targets, we find $2^3 - 2 = 6$ valid distribution patterns by excluding scenarios where only one address is used. Combining these results gives 60 favorable outcomes, leading to a final probability of $60/125 = 12/25$.

The number of tosses required to obtain the first six on a fair die follows a geometric distribution with a success probability of $p = \frac{1}{6}$ and a failure probability of $q = \frac{5}{6}$. The probability that the first success occurs on exactly the $k$-th toss is $P(X=k) = q^{k-1}p$, and the probability that the first success occurs at or after the $k$-th toss is given by $P(X \ge k) = q^{k-1}$. Based on these definitions, we find that $a = P(X=3) = q^2 p$ and $b = P(X \ge 3) = q^2$.

To determine the conditional probability $c = P(X \ge 6 | X > 3)$, we utilize the ratio $\frac{P(X \ge 6 \cap X > 3)}{P(X > 3)}$. Since the event $X \ge 6$ is a subset of $X > 3$, the numerator simplifies to $P(X \ge 6) = q^5$, and the denominator is $P(X > 3) = P(X \ge 4) = q^3$. Consequently, $c = \frac{q^5}{q^3} = q^2$. Substituting these expressions into the final term, we obtain $\frac{b+c}{a} = \frac{q^2 + q^2}{q^2 p} = \frac{2q^2}{q^2 p} = \frac{2}{p}$. Given that $p = \frac{1}{6}$, the final value is $\frac{2}{1/6} = 12$.