Probability – JEE Main MathematicsPractice Questions & PYQs

Bayes' Theorem provides the standard framework for updating the probability of a hypothesis, such as the manufacturing origin of a product, given new evidence that the item meets a specific quality standard. To find the overall probability of selecting a standard quality motorcycle, combine the weighted contributions from both plants, which is calculated as $(0.6 \times 0.8) + (0.4 \times 0.9) = 0.48 + 0.36 = 0.84$.

The probability $p$ that a standard quality motorcycle came from plant B is defined by the ratio of the contribution from plant B to the total probability: $p = \frac{0.36}{0.84} = \frac{3}{7}$. Computing the final requested value involves multiplying this probability by 126, leading to $126 \times \frac{3}{7} = 54$.

The probability of rolling a two on any individual throw of a fair die is $p = 1/6$, while the probability of rolling any other number is $q = 5/6$. Success on an even-numbered throw implies the first two appears on the second, fourth, sixth, or subsequent even turns, which creates an infinite geometric series. The first term is the probability of failing once and then succeeding, calculated as $a = q \cdot p = \frac{5}{6} \cdot \frac{1}{6} = \frac{5}{36}$, while the common ratio for successive even turns is $r = q^2 = \frac{25}{36}$. The total probability $P$ is found using the infinite geometric series sum formula $P = \frac{a}{1 - r} = \frac{5/36}{1 - 25/36}$ which simplifies to $\frac{5/36}{11/36}$ and results \in a final value of $\frac{5}{11}$.

Bayes' Theorem provides the framework for determining the conditional probability of an outcome given a specific source, in this case identifying which bag a coin originated from after observing its type. Since there are three bags chosen at random, each has an equal prior probability of selection. The probability of drawing a one-rupee coin from bag X is $\frac{5}{9}$, from bag Y is $\frac{4}{9}$, and from bag Z is $\frac{3}{9}$.

To find the probability that the coin came from bag Y, we calculate the ratio of the likelihood of drawing the coin from bag Y to the total probability of drawing a one-rupee coin from any of the three bags. This is expressed by the following calculation:
$P(Y|1) = \frac{\frac{4}{9}}{\frac{5}{9} + \frac{4}{9} + \frac{3}{9}} = \frac{4}{12} = \frac{1}{3}$

The Principle of Inclusion-Exclusion states that for any three sets, the total number of elements in their union is equal to the sum of the sizes of the individual sets, minus the sum of the sizes of their pairwise intersections, plus the size of their triple intersection. Representing the students who passed Mathematics, Physics, and Chemistry as sets $M$, $P$, and $C$ respectively, we have a total of 40 students passing at least one subject. Letting $x$ represent the number of students who passed all three subjects, the formula is $40 = 20 + 25 + 16 - (n(M \cap P) + n(P \cap C) + n(M \cap C)) + x$. Simplifying this equation, we find that the sum of the pairwise intersections must be $n(M \cap P) + n(P \cap C) + n(M \cap C) = 21 + x$.

To determine the maximum value for $x$, we must adhere to the provided constraints: $n(M \cap P) \le 11$, $n(P \cap C) \le 15$, and $n(M \cap C) \le 15$. Additionally, the triple intersection $x$ must be less than or equal to each pairwise intersection, meaning $n(M \cap P) \ge x$, $n(P \cap C) \ge x$, and $n(M \cap C) \ge x$. If we test the value $x = 11$, then $n(M \cap P)$ is constrained to be exactly 11, while the other two pairwise intersections must also be at least 11. This leads to a minimum sum for the pairwise intersections of $11 + 11 + 11 = 33$. However, according to our simplified equation, if $x = 11$, the sum of these intersections must be $21 + 11 = 32$. Because the required sum of 32 is less than the mathematical minimum of 33, $x = 11$ is impossible.

Testing the value $x = 10$ yields a required pairwise sum of $21 + 10 = 31$. This is entirely feasible because the lower bound for the sum is $10 + 10 + 10 = 30$ and the upper bound is $11 + 15 + 15 = 41$. We can achieve a sum of 31 by choosing, for instance, $n(M \cap P) = 11$, $n(P \cap C) = 10$, and $n(M \cap C) = 10$. Since these values satisfy all given constraints, the maximum number of students who passed all three subjects is 10.

Bayes' Theorem provides the framework for determining the probability that the chosen ball came from bag A given that it is white. Since each bag is equally likely to be selected, the prior probability is $\frac{1}{2}$ for both, with bag A having a $\frac{3}{10}$ probability of yielding a white ball and bag B having a $\frac{3}{5}$ probability. The total probability of selecting a white ball is the sum of these joint probabilities:
$\frac{1}{2} \times \frac{3}{10} + \frac{1}{2} \times \frac{3}{5} = \frac{3}{20} + \frac{6}{20} = \frac{9}{20}$
The required probability is the ratio of the chance of selecting bag A and drawing a white ball to this total probability, which is $\frac{3/20}{9/20}$, simplifying to $\frac{1}{3}$.

A quadratic equation fails to have real roots when its discriminant is negative. For the equation $64 x^{2+}5 N x+1=0$, the discriminant is determined by $(5 N)^2 - 4(64)(1)$, which simplifies to $25 N^2 - 256$. Requiring this value to be less than zero leads to $25 N^2 < 256$, or $N^2 < 10.24$. Since $N$ represents the number of tosses performed until a head appears, it must be a positive integer, leaving $N=1, 2,$ or $3$ as the only possibilities.

The probability of getting a head on any single toss is $\frac{1}{4}$, while the probability of getting a tail is $\frac{3}{4}$. For $N=1$, the first toss must be a head, yielding a probability of $\frac{1}{4}$. For $N=2$, we need a tail followed by a head, resulting in $\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$. For $N=3$, we require two tails followed by a head, giving $\left(\frac{3}{4}\right)^2 \times \frac{1}{4} = \frac{9}{64}$. Adding these individual probabilities together, we get $\frac{16}{64} + \frac{12}{64} + \frac{9}{64}$, which equals $\frac{37}{64}$. Because $37$ and $64$ share no common factors, the difference $q-p$ is calculated as $64 - 37$, which results in $27$.

In a binomial distribution $\mathrm{B}(n, p)$, the mean is $np$ and the variance is $npq$, where $q = 1 - p$. Given the sum $np(1+q) = 5$ and the product $n^2p^2q = 6$, squaring the sum and dividing the result by the product leads to the equation $\frac{(1+q)^2}{q} = \frac{25}{6}$.

This simplifies to $6q^2 - 13q + 6 = 0$, which factors into $(3q - 2)(2q - 3) = 0$. Since $q$ represents a probability value, $q = 2/3$ is the valid root, which implies $p = 1/3$. Inserting these parameters into the mean equation $n(1/3) + n(1/3)(2/3) = 5$ yields $n(5/9) = 5$, confirming $n = 9$. Calculating the final expression $6(n + p - q)$ results in $6(9 + 1/3 - 2/3)$, which simplifies to $6(9 - 1/3)$ or $54 - 2$, equal to 52.

Calculating the probability that at least three players pick the correct T-shirt is most effectively achieved by determining the complement event, where zero, one, or two players pick correctly, and subtracting the sum of these probabilities from one. The probability of any specific number of matches relates to the number of derangements, denoted as $D_n \approx \frac{n!}{e}$, which represents permutations where no item is in its original position. For fifteen players, the probability that zero players match is $\frac{D_{15}}{15!}$, which is $\frac{1}{e}$. For exactly one player to match, the calculation is $\frac{{}^{15}C_1 \cdot D_{14}}{15!}$, simplifying to $\frac{1}{e}$. For exactly two players to match, the calculation is $\frac{{}^{15}C_2 \cdot D_{13}}{15!}$, which simplifies to $\frac{1}{2e}$. Summing these individual probabilities results in $\frac{1}{e} + \frac{1}{e} + \frac{1}{2e} = \frac{5}{2e}$. Subtracting this sum from the total probability of one yields the result $1 - \frac{5}{2e}$.

Calculating the probability that the bag contains at least 5 black balls given that two were drawn requires evaluating all possible bag compositions, as the bag could contain anywhere from 2 to 6 black balls. Each potential composition provides a number of ways to draw two black balls, defined by the combination ${}^kC_2$ where $k$ is the number of black balls. The total number of ways to draw two black balls across all possible scenarios is the sum of these combinations: ${}^2C_2 + {}^3C_2 + {}^4C_2 + {}^5C_2 + {}^6C_2$, which evaluates to $1 + 3 + 6 + 10 + 15 = 35$. The favorable scenarios, where the bag contains 5 or 6 black balls, are represented by the sum ${}^5C_2 + {}^6C_2$, which equals $10 + 15 = 25$. Dividing these successful outcomes by the total sample space yields $\frac{25}{35}$, which simplifies to $\frac{5}{7}$.

The mean of a random variable is the weighted average calculated by summing the products of each possible outcome and its respective probability, where in this case the experiment stops at the first head or after three tails. Since the head is three times as likely as a tail, the probabilities are $P(H) = \frac{3}{4}$ and $P(T) = \frac{1}{4}$. The experiment ends at $X = 1$ if the first toss is a head, with probability $\frac{3}{4}$. It ends at $X = 2$ if the sequence is TH, occurring with probability $\frac{3}{16}$. It ends at $X = 3$ if the sequence is either TTH or TTT, yielding a combined probability of $\frac{3}{64} + \frac{1}{64} = \frac{1}{16}$. The expected value is calculated by summing the products of each outcome and its probability:

$E[X] = 1 \left( \frac{3}{4} \right) + 2 \left( \frac{3}{16} \right) + 3 \left( \frac{1}{16} \right) = \frac{12}{16} + \frac{6}{16} + \frac{3}{16} = \frac{21}{16}$

Analyzing the sample space of 36 outcomes involves partitioning the possibilities according to the defined constraints. The intersection $(A \cup B) \cap C$ expands using the distributive law to $(A \cap C) \cup (B \cap C)$. Since event $B$ describes a first-even and second-odd roll, while event $C$ describes a first-odd and second-even roll, the two events are mutually exclusive, which means the set $B \cap C$ contributes zero outcomes. The calculation therefore focuses entirely on finding the pairs that belong to both $A$ and $C$.

In event $A$, the first die is less than the second, and in $C$, the first die is odd while the second is even. Setting the first die to 1 allows the second to be 2, 4, or 6 (3 outcomes); setting it to 3 allows the second to be 4 or 6 (2 outcomes); and setting it to 5 restricts the second to 6 (1 outcome). Summing these specific instances results in a total of 6 favourable cases for the event $(A \cup B) \cap C$. This confirms that the count for this intersection is 6, while the independent counts for events $A$, $B$, and $C$ are 15, 9, and 9 respectively.

For the product of five dice outcomes to be positive, no outcome can be zero, and the total number of negative signs must be even. Given that the die has six sides, the probability of rolling a positive number $\{1, 2, 3\}$ is $1/2$, a negative number $\{-2, -1\}$ is $1/3$, and a zero is $1/6$. Since any zero would force the product to be zero, we must exclude it and focus only on the non-zero outcomes.

To achieve a positive product, we must have either zero, two, or four negative results among the five rolls. The probability of obtaining five positive outcomes is ${}^{5}C_{5} \left(\frac{1}{2}\right)^5$. For the case with exactly two negative outcomes, we calculate ${}^{5}C_{2} \left(\frac{1}{3}\right)^2 \left(\frac{1}{2}\right)^3$, and for the case with exactly four negative outcomes, we calculate ${}^{5}C_{4} \left(\frac{1}{3}\right)^4 \left(\frac{1}{2}\right)^1$. Summing these probabilities results in $\frac{1}{32} + \frac{10}{72} + \frac{5}{162}$, which simplifies to $\frac{521}{2592}$.

When we repeat an independent trial a fixed number of times, we rely on the binomial distribution to determine probabilities. In this scenario, there are 5 trials, and getting a sum of 5 with two dice constitutes a success. Since there are 4 pairs that sum to 5 out of 36 possible outcomes, the probability of success is $\frac{1}{9}$, meaning the probability of failure is $\frac{8}{9}$.

To calculate the probability of at least 4 successes, we sum the probabilities of getting exactly 4 successes and exactly 5 successes. Applying the binomial probability formula $\binom{n}{r} p^r q^{n-r}$, the probability for 4 successes is $\binom{5}{4} (\frac{1}{9})^4 (\frac{8}{9})$, which simplifies to $\frac{40}{9^5}$. The probability of getting 5 successes is $\binom{5}{5} (\frac{1}{9})^5$, which is $\frac{1}{9^5}$. Adding these results in a total probability of $\frac{41}{9^5}$. To express this with a denominator of $3^{11}$, we rewrite $9^5$ as $(3^2)^5$, which equals $3^{10}$. Multiplying the numerator and denominator by 3 gives the following:

$\frac{41}{3^{10}} = \frac{123}{3^{11}}$

This calculation confirms that $k$ equals 123.

Bayes' Theorem provides the framework to determine the probability of a cause, specifically being a smoker, given that an effect has been observed, which is a diagnosis of lung cancer. Since 25 percent of the population are smokers, the probability of selecting a smoker is $P(S) = \frac{1}{4}$, making the probability of selecting a non-smoker $P(NS) = \frac{3}{4}$. Because a smoker is 27 times more likely to develop lung cancer than a non-smoker, we can define the probability of cancer for a non-smoker as $x$ and for a smoker as $27x$.

Applying these variables to the formula for conditional probability yields:

$P(S|C) = \frac{P(C|S)P(S)}{P(C|S)P(S) + P(C|NS)P(NS)}$

Substituting the values into the equation, we get:

$P(S|C) = \frac{27x \times \frac{1}{4}}{(27x \times \frac{1}{4}) + (x \times \frac{3}{4})}$

The variable $x$ cancels out from the numerator and denominator, which simplifies the expression to $\frac{27/4}{27/4 + 3/4}$, resulting in $\frac{27}{30}$ or $\frac{9}{10}$. Matching this to the form $\frac{k}{10}$ confirms that $k = 9$.

Choosing two real numbers from the interval $[0, 60]$ creates a sample space represented by a square with a side length of 60 and a total area of 3600. The condition $|x - y| \le a$ defines the region between the lines $y = x - a$ and $y = x + a$, and it is more straightforward to calculate the complement, which is the region where $|x - y| > a$. This excluded area consists of two isosceles right triangles at the corners of the square, each with legs of length $(60-a)$, yielding a total excluded area of $(60-a)^2$.

The probability of this complement is $\frac{(60-a)^2}{3600}$, which means the probability of the given event is $1 - \frac{(60-a)^2}{3600} = \frac{11}{36}$. Rearranging this expression gives $\frac{(60-a)^2}{3600} = \frac{25}{36}$, which simplifies further to $(60-a)^2 = 2500$. Taking the square root of both sides results in $60 - a = 50$, giving a final value of $a = 10$.

Calculating the probability of the event $B$ begins with identifying the distribution of the sample space. Given that the total probability must equal $1$ and each term is half its predecessor, the probabilities follow a geometric progression where $P(w_n) = \frac{P(w_1)}{2^{n-1}}$. Solving the infinite sum $\sum_{n=1}^{\infty } P(w_1) (\frac{1}{2})^{n-1} = 1$ leads to $2P(w_1) = 1$, which gives $P(w_1) = \frac{1}{2}$. Thus, the general probability for any outcome is defined by $P(w_n) = \frac{1}{2^n}$.

The set $A$ consists of integers generated by $2k+3\ell$ for $k, \ell \in \mathbb{N}$, creating the sequence $\{5, 7, 8, 9, 10, \dots\}$. This set contains all natural numbers with the exception of $\{1, 2, 3, 4, 6\}$. Calculating the probability of the complement by subtracting the sum of these excluded indices from $1$ provides the value for $P(B)$. The sum of probabilities for the excluded indices is $\frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \frac{1}{2^6}$, which simplifies to $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64}$. Combining these terms as $\frac{32 + 16 + 8 + 4 + 1}{64}$ results in $\frac{61}{64}$, and subtracting this from $1$ yields $\frac{3}{64}$.

This scenario models a geometric distribution where the process stops upon the first success, defined as rolling a number less than $n$ with probability $p = \frac{n-1}{n}$. The expected number of trials for such a distribution is the reciprocal of the probability of success, yielding an expected mean value of $E[X] = \frac{n}{n-1}$. Equating this expression to the given mean of $\frac{n}{9}$ results in the following relation:
$\frac{n}{n-1} = \frac{n}{9}$
Dividing both sides by $n$, which is valid since $n > 1$, simplifies the equation to $\frac{1}{n-1} = \frac{1}{9}$, which confirms that $n = 10$.

Since the balls are drawn with replacement, every draw remains an independent event where all six colors are equally likely to appear. When drawing two balls, there are $6 \times 6 = 36$ total possible outcomes, and a match occurs when both are the same color, which can happen in $6$ specific ways. This leads to the probability $p = \frac{6}{36} = \frac{1}{6}$.

When drawing four balls, the total number of outcomes is $6^4 = 1296$. To calculate $q$, the probability of getting exactly three balls of the same color, we first select one of the $6$ colors to repeat, arrange those three in ${}^{4}C_{3} = 4$ possible positions, and choose one of the remaining $5$ colors for the final ball. This yields $6 \times 4 \times 5 = 120$ favorable outcomes, resulting in $q = \frac{120}{1296} = \frac{5}{54}$.

Comparing these values by expressing $p$ with a common denominator of $54$ gives $p = \frac{9}{54}$. The ratio $p:q$ is therefore $9:5$, and because $9$ and $5$ are coprime, we assign $m = 9$ and $n = 5$. Summing these values gives $m + n = 14$.

In any discrete probability distribution, the sum of all probabilities across the sample space must be 1. For the function $P(X=x) = k(x+1) 3^{-x}$, we set the total sum $k \sum_{x=0}^{\infty } (x+1) 3^{-x} = 1$. By isolating the summation part as $S = 1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \dots$, we determine its value by observing that $S - \frac{1}{3}S = 1 + \frac{1}{3} + \frac{1}{9} + \dots$, which represents an infinite geometric series summing to $\frac{1}{1 - 1/3} = \frac{3}{2}$. With $\frac{2}{3}S = \frac{3}{2}$, we find $S = \frac{9}{4}$, and since $kS = 1$, we must have $k = \frac{4}{9}$. Calculating $P(X \ge 2)$ requires subtracting the probabilities of $x=0$ and $x=1$ from the total probability of 1. Substituting $k$, we get $P(X=0) = \frac{4}{9}$ and $P(X=1) = \frac{4}{9}(2)(\frac{1}{3}) = \frac{8}{27}$, leading to $1 - (\frac{12}{27} + \frac{8}{27}) = 1 - \frac{20}{27}$, which results in $\frac{7}{27}$.

The binomial distribution $B(n, p)$ is characterized by a mean of $np$ and a variance of $npq$, where $q = 1-p$. Given that the difference between the mean and variance is 1, we can express this as $np - npq = 1$. Factoring out $np$, this becomes $np(1-q) = 1$, which simplifies to $np^2 = 1$, allowing us to establish the relationship $n = \frac{1}{p^2}$.

We are also provided with the condition $2P(X=2) = 3P(X=1)$. Using the binomial probability formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$, we can expand this as $2 \binom{n}{2} p^2 q^{n-2} = 3 \binom{n}{1} p^1 q^{n-1}$. Substituting the binomial coefficients, this simplifies to $n(n-1) p^2 q^{n-2} = 3np q^{n-1}$. Dividing both sides by $np q^{n-2}$ (assuming $p, q \neq 0$), we get $(n-1)p = 3q$, or more explicitly, $(n-1)p = 3(1-p)$.

Substituting $n = \frac{1}{p^2}$ into this equation gives $(\frac{1}{p^2} - 1)p = 3(1-p)$, which simplifies further to $\frac{1}{p} - p = 3 - 3p$. Multiplying the entire equation by $p$ results in $1 - p^2 = 3p - 3p^2$, which rearranges into the quadratic equation $2p^2 - 3p + 1 = 0$. Factoring this leads to $(2p-1)(p-1) = 0$, yielding solutions $p=1/2$ or $p=1$. Since $p=1$ would imply a variance of zero and a trivial distribution, we choose $p=1/2$, which gives $n = \frac{1}{(1/2)^2} = 4$.

Finally, we calculate $n^2 P(X > 1)$, which is equivalent to $n^2(1 - P(X=0) - P(X=1))$. With $n=4$ and $p=1/2$, the probability $P(X=0) = \binom{4}{0} (1/2)^4 = 1/16$, and $P(X=1) = \binom{4}{1} (1/2)^4 = 4/16$. Therefore, $P(X > 1) = 1 - 1/16 - 4/16 = 11/16$. Multiplying this by $n^2 = 16$ gives $16 \times (11/16) = 11$.