Probability – JEE Main MathematicsPractice Questions & PYQs

The variance of a discrete random variable is determined by the formula Var(x) = E\[x^2]$ - (E[x])^2$, where$E[x^2]$is the expectation of the squared number of defective items and$E[x]$is the mean number. With 3 defective and 7 good oranges, choosing 2 at random results in$\binom{10}{2} = 45$total combinations, and$x$ represents the number of defective oranges drawn, which can be 0, 1, or 2.

We calculate the probability for each value of $x$ as $P(x=0) = \frac{\binom{7}{2}}{45} = \frac{21}{45}$, $P(x=1) = \frac{\binom{3}{1}\binom{7}{1}}{45} = \frac{21}{45}$, and $P(x=2) = \frac{\binom{3}{2}}{45} = \frac{3}{45}$. From these, the expectation $E[x] = 0(\frac{21}{45}) + 1(\frac{21}{45}) + 2(\frac{3}{45}) = \frac{27}{45} = \frac{3}{5}$ and the expectation of the squares E\[x^2]$ = 0^2(\frac{21}{45}) + 1^2(\frac{21}{45}) + 2^2(\frac{3}{45}) = \frac{33}{45} = \frac{11}{15}$. Substituting these into the variance formula yields:

$Var(x) = \frac{11}{15} - \left(\frac{3}{5}\right)^2 = \frac{11}{15} - \frac{9}{25} = \frac{28}{75}$

When tossing a coin three times, there are eight equally likely sequences. The random variable $X$, which counts the number of times a tail follows a head, can be evaluated across these outcomes: HHH, TTT, THH, and TTH each contain zero occurrences of a tail following a head, while HHT, HTH, THT, and HTT each contain one such occurrence. Since four outcomes result in $X=0$ and four result in $X=1$, the probability distribution is defined by $P(X=0) = \frac{1}{2}$ and $P(X=1) = \frac{1}{2}$.

The mean $\mu$ is the expected value of $X$, calculated as $\mu = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2}$. The variance $\sigma^2$ is determined using the formula \sigma^2 = E\[X^2]$ - \mu^2$, where the second moment is$E$[X^2]$ = 0^2 \cdot \frac{1}{2} + 1^2 \cdot \frac{1}{2} = \frac{1}{2}$. Substituting these values, the variance is$\sigma^2 = \frac{1}{2} - (\frac{1}{2})^2 = \frac{1}{4}$. The final expression evaluates to$64(\mu + \sigma^2) = 64(\frac{1}{2} + \frac{1}{4}) = 64 \cdot \frac{3}{4} = 48$.

The most straightforward way to determine the probability that two randomly chosen squares share no side is to use the complement rule, which involves subtracting the number of pairs that do share a side from the total possible combinations. Selecting any two squares from a grid of 16 results in a total of $\frac{16 \times 15}{2} = 120$ possible pairs. Two squares are considered to have a side in common if they are adjacent either horizontally or vertically. In a row of 4 squares, there are 3 adjacent horizontal pairs, and since the board has 4 rows, we account for 12 such pairs. The same logic applies to the 4 columns, where we also find 12 adjacent vertical pairs. By combining these, we find 24 pairs that share a side. Subtracting these from the total, we are left with 96 pairs that share no side. Expressing this as a fraction gives $\frac{96}{120}$, which simplifies to $\frac{4}{5}$.

The powers of $i$ cycle through a pattern of four values: $\{1, i, -1, -i\}$. Since $k_1$ and $k_2$ are chosen randomly, we are essentially selecting a remainder for each when divided by 4, meaning there are $4 \times 4 = 16$ equally likely combinations of $(i^{k_1}, i^{k_2})$. A sum of zero arises only when one value is the negative of the other, which occurs for the specific pairs $(1, -1)$, $(-1, 1)$, $(i, -i)$, and $(-i, i)$. Since there are 4 such combinations out of the 16 total possibilities, the probability of the sum being zero is $\frac{4}{16} = \frac{1}{4}$, which implies the probability of the sum being non-zero is $1 - \frac{1}{4} = \frac{3}{4}$.

The variance of the number of successes in a hypergeometric distribution is calculated using the formula $Var(X) = n \frac{K}{N} \left(1 - \frac{K}{N}\right) \frac{N - n}{N - 1}$, where $N = 10$ is the total population, $K = 3$ is the number of defective pens, and $n = 2$ is the sample size. Substituting these parameters into the formula gives:

$Var(X) = 2 \times \frac{3}{10} \times \frac{7}{10} \times \frac{8}{9}$

Multiplying the terms yields $Var(X) = \frac{42}{100} \times \frac{8}{9} = \frac{336}{900}$, which simplifies to $\frac{28}{75}$ after dividing the numerator and denominator by 12.

Bayes' Theorem dictates that the probability of an event given observed data is the ratio of the conditional probability weighted by the prior to the total probability. With the prior probabilities of losing a spade or a non-spade card being $P(L_S) = \frac{1}{4}$ and $P(L_N) = \frac{3}{4}$, and conditional probabilities for drawing $n$ spades from the remaining 51 cards being $P(E|L_S) = \frac{\binom{12}{n}}{\binom{51}{n}}$ and $P(E|L_N) = \frac{\binom{13}{n}}{\binom{51}{n}}$, the given probability $P(L_S|E) = \frac{11}{50}$ leads to the equation:
$\frac{11}{50} = \frac{\binom{12}{n} \cdot \frac{1}{4}}{\binom{12}{n} \cdot \frac{1}{4} + \binom{13}{n} \cdot \frac{3}{4}} = \frac{\binom{12}{n}}{\binom{12}{n} + 3\binom{13}{n}}$

Substituting the identity $\binom{13}{n} = \binom{12}{n} \cdot \frac{13}{13-n}$ and simplifying by dividing the numerator and denominator by $\binom{12}{n}$:
$\frac{11}{50} = \frac{1}{1 + 3\left(\frac{13}{13-n}\right)} = \frac{1}{1 + \frac{39}{13-n}} = \frac{13-n}{13-n+39} = \frac{13-n}{52-n}$

Equating the cross-products $11(52-n) = 50(13-n)$ yields $572 - 11n = 650 - 50n$, which simplifies to $39n = 78$ and results in $n = 2$.

The fundamental requirement for any probability distribution is that the sum of probabilities over all possible outcomes equals 1. By assigning a probability $q$ to the events $X=2$ and $X=3$, we establish the equation $2p + 2q = 1$, which simplifies to $p + q = \frac{1}{2}$. To utilize the given condition $E(X^2) = 2E(X)$, we calculate the expected values by summing the product of each outcome and its probability. This gives the expected value $E(X) = 0(p) + 1(p) + 2(q) + 3(q) = p + 5q$, and the expected square value $E(X^2) = 0^2(p) + 1^2(p) + 2^2(q) + 3^2(q) = p + 13q$.

Equating the expectation expression as $p + 13q = 2(p + 5q)$ simplifies to $p + 13q = 2p + 10q$, which yields the relationship $p = 3q$. Substituting this identity into the initial sum equation results in $3q + q = \frac{1}{2}$, leading to $4q = \frac{1}{2}$ and thus $q = \frac{1}{8}$. Consequently, $p$ is equal to $3(\frac{1}{8})$ or $\frac{3}{8}$. Evaluating the final expression $8p - 1$ with this value of $p$ produces $8(\frac{3}{8}) - 1$, which simplifies to 2.

Since each of the four positions in a $2 \times 2$ matrix can be filled by either 0 or 1, there are $2^4 = 16$ total possible matrices. A matrix is invertible only when its determinant is non-zero, so we focus on the condition for non-invertibility where the determinant $ad - bc$ equals zero, which simplifies to $ad = bc$. By evaluating all combinations of entries, there are exactly 10 cases where this condition is met, meaning the remaining 6 matrices must be invertible. Dividing the number of favorable outcomes by the total number of matrices gives a probability of $\frac{6}{16}$, which reduces to $\frac{3}{8}$.

The probability of selecting three distinct numbers that form an increasing geometric progression is calculated by finding the ratio of favorable outcomes to the total number of possible combinations. Selecting three numbers from the set of 40 integers yields a total of ${}^{40}C_3$ possible combinations, which is calculated as:
$\frac{40 \times 39 \times 38}{3 \times 2 \times 1} = 9880$

Identifying the favorable outcomes requires finding sets of three distinct numbers that form an increasing geometric progression. By considering integer common ratios $r$, we find 10 triplets for $r=2$, 4 triplets for $r=3$, 2 triplets for $r=4$, 1 triplet for $r=5$, and 1 triplet for $r=6$, totaling 18 valid triplets.

The probability is therefore $\frac{18}{9880}$, which simplifies to $\frac{9}{4940}$ after dividing both the numerator and the denominator by 2. Given that this fraction represents $\frac{m}{n}$ where $\gcd(m, n)=1$, we have $m=9$ and $n=4940$, so the sum $m+n$ equals 4949.

Bayes' Theorem provides the standard framework for updating the probability of a hypothesis-in this case, the coin being unbiased-after observing specific evidence, which is the coin landing on heads. Because the bag contains 20 coins where 19 are unbiased and 1 is double-headed, the initial likelihood of selecting an unbiased coin is $\frac{19}{20}$, while the likelihood of selecting the biased coin is $\frac{1}{20}$. An unbiased coin produces heads with a probability of $\frac{1}{2}$, whereas the double-headed coin produces heads with a probability of $1$.

Calculating the specific probability of having chosen the unbiased coin given the outcome of a head requires dividing the probability of picking an unbiased coin and getting a head by the total probability of getting a head from any coin. The probability of picking an unbiased coin and flipping a head is $\frac{19}{20} \times \frac{1}{2} = \frac{19}{40}$. The total probability of obtaining a head, which accounts for both the unbiased and biased scenarios, is $\frac{19}{20} \times \frac{1}{2} + \frac{1}{20} \times 1 = \frac{19}{40} + \frac{2}{40} = \frac{21}{40}$. Dividing the first value by the total leads to the ratio $\frac{19}{21}$, identifying $m=19$ and $n=21$. Finally, evaluating $n^2 - m^2$ results in $21^2 - 19^2$, which simplifies to $(21 - 19)(21 + 19) = 2 \times 40 = 80$.

The probability is defined as the ratio of favorable committee combinations to the total number of ways to choose 12 individuals from the group of 16, which is $\binom{16}{12} = 1820$. Favorable outcomes are identified by summing the mutually exclusive cases that satisfy the constraints: 3 engineers, 1 doctor, and 8 professors ($\binom{4}{3} \times \binom{2}{1} \times \binom{10}{8} = 360$); 3 engineers, 2 doctors, and 7 professors ($\binom{4}{3} \times \binom{2}{2} \times \binom{10}{7} = 480$); 4 engineers, 1 doctor, and 7 professors ($\binom{4}{4} \times \binom{2}{1} \times \binom{10}{7} = 240$); and 4 engineers, 2 doctors, and 6 professors ($\binom{4}{4} \times \binom{2}{2} \times \binom{10}{6} = 210$). The total number of favorable combinations equals $360 + 480 + 240 + 210 = 1290$, resulting in a final probability of $\frac{1290}{1820} = \frac{129}{182}$.

This scenario represents an infinite geometric series because the game cycle of alternating dice throws repeats until one player successfully achieves their target sum. First, determine the probability of each event: for A, obtaining a sum of 5 involves four outcomes-(1,4), (2,3), (3,2), and (4,1)-yielding $P(A) = \frac{4}{36}$ and a failure probability of $P(\bar{A}) = \frac{32}{36}$. For B, obtaining a sum of 8 involves five outcomes-(2,6), (3,5), (4,4), (5,3), and (6,2)-yielding $P(B) = \frac{5}{36}$ and a failure probability of $P(\bar{B}) = \frac{31}{36}$.

A wins if he succeeds on his first turn, or if both players miss and he succeeds on his second turn, continuing infinitely, which forms the series $P(A) + P(\bar{A})P(\bar{B})P(A) + (P(\bar{A})P(\bar{B}))^2P(A) + \dots$. Applying the sum formula for an infinite geometric progression, $\frac{a}{1-r}$, where $a = \frac{4}{36}$ and the common ratio $r = P(\bar{A})P(\bar{B}) = \frac{32}{36} \times \frac{31}{36} = \frac{992}{1296}$, the result is $\frac{4/36}{1 - 992/1296}$. This simplifies to $\frac{4/36}{304/1296}$, which calculates to $\frac{144}{304}$ and reduces to $\frac{9}{19}$.

Determining the overall probability of drawing a white ball from the second bag requires applying the law of total probability, as the outcome depends entirely on whether the ball transferred from the first bag is white or black. Given that Bag 1 contains 4 white and 5 black balls, transferring a white ball occurs with a probability of $4/9$, while transferring a black ball occurs with a probability of $5/9$.

When a white ball is moved to Bag 2, the container holds $n+1$ white balls out of a total of $n+4$, resulting in a conditional probability of $\frac{n+1}{n+4}$ for drawing a white ball. Conversely, transferring a black ball leaves Bag 2 with $n$ white balls out of the same total of $n+4$, leading to a conditional probability of $\frac{n}{n+4}$ for drawing a white ball. The combined probability is represented by the following equation:

$\frac{4}{9} \cdot \frac{n+1}{n+4} + \frac{5}{9} \cdot \frac{n}{n+4} = \frac{29}{45}$

Simplifying the left side of the equation yields $\frac{9n+4}{9n+36} = \frac{29}{45}$. Solving this expression for the variable $n$ reveals that $n = 6$.

The foundation of any probability distribution is that the sum of all possible probabilities must equal one. We set up the infinite series $\sum_{x=0}^{\infty } k(x+1) 3^{-x} = 1$ to isolate the constant $k$. By denoting the series as $S$ and computing $S - \frac{1}{3}S$, we produce the expression $k + \frac{k}{3} + \frac{k}{9} + \dots = \frac{2}{3}S$. Because the geometric series inside the sum converges to $k(\frac{1}{1 - 1/3}) = \frac{3k}{2}$, we equate $\frac{2}{3}S = \frac{3k}{2}$. Given $S=1$, this simplifies to $\frac{2}{3} = \frac{3k}{2}$, which leads directly to $k = \frac{4}{9}$.

Calculating the probability for the tail end of the distribution, $P(X \geq 3)$, is most efficient using the complement rule $1 - (P(X=0) + P(X=1) + P(X=2))$. Applying the formula $P(X=x) = k(x+1) 3^{-x}$ for the first three terms, we find $P(X=0) = k$, $P(X=1) = \frac{2k}{3}$, and $P(X=2) = \frac{3k}{9} = \frac{k}{3}$. Summing these probabilities results in $k + \frac{2k}{3} + \frac{k}{3} = 2k$. Substituting $k = \frac{4}{9}$ into this expression gives $2(\frac{4}{9}) = \frac{8}{9}$, and subtracting this from the total probability of one leaves us with $\frac{1}{9}$.

Conditional probability allows us to determine the likelihood of an initial event occurring when we already have information about a subsequent outcome. In a bag containing 4 white and 6 black balls, the probability that the first selection is black given that the second is also black is calculated by dividing the probability of both balls being black by the total probability that the second ball is black.

The probability that both the first and second balls are black is determined by the sequence of draws, resulting in $\frac{6}{10} \times \frac{5}{9} = \frac{30}{90}$. The second ball can be black in two mutually exclusive ways: either the first ball was also black, yielding a probability of $\frac{6}{10} \times \frac{5}{9} = \frac{30}{90}$, or the first ball was white, yielding $\frac{4}{10} \times \frac{6}{9} = \frac{24}{90}$. Summing these scenarios provides the total probability that the second ball is black, which is $\frac{30}{90} + \frac{24}{90} = \frac{54}{90}$. Dividing the joint probability by this total probability yields $\frac{30}{54}$, which simplifies to $\frac{5}{9}$. Given that this fraction represents $\frac{m}{n}$ with $\gcd(m, n) = 1$, we identify $m=5$ and $n=9$, leading to a sum $m+n$ of $14$.

The probability of independent events occurring together is the product of their individual probabilities, and we combine these products for all mutually exclusive scenarios that result in the target outcome. For two dice where the first has the distribution $\{1, 1, 2, 2, 3, 4\}$ and the second has $\{1, 2, 2, 3, 3, 4\}$, we must identify every pairing that sums to 4 or 5 and calculate the probability of each.

To obtain a sum of 4, the possible outcomes are $(1, 3)$, $(2, 2)$, and $(3, 1)$, resulting in probabilities of $\frac{2}{6} \times \frac{2}{6} = \frac{4}{36}$, $\frac{2}{6} \times \frac{2}{6} = \frac{4}{36}$, and $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$, respectively. For a sum of 5, the combinations are $(1, 4)$, $(2, 3)$, $(3, 2)$, and $(4, 1)$, which yield probabilities of $\frac{2}{6} \times \frac{1}{6} = \frac{2}{36}$, $\frac{2}{6} \times \frac{2}{6} = \frac{4}{36}$, $\frac{1}{6} \times \frac{2}{6} = \frac{2}{36}$, and $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$. Summing these individual values gives a total probability of $\frac{4+4+1+2+4+2+1}{36} = \frac{18}{36}$, which simplifies to $\frac{1}{2}$.

Solving probability problems involving two events often relies on the definition of conditional probability, $P(A|B) = \frac{P(A \cap B)}{P(B)}$, combined with set-theoretic identities such as De Morgan's laws. The quadratic equation $12x^2 - 7x + 1 = 0$ factors into $(4x - 1)(3x - 1) = 0$, yielding roots of 1/4 and 1/3. Using these roots as $P(A|B)$ and $P(B|A)$, their product 1/12 is equal to $\frac{P(A \cap B)^2}{P(A)P(B)}$, which implies $P(A)P(B) = \frac{0.01}{1/12} = 0.12$. The \sum of the roots 7/12 is equivalent to $P(A \cap B) \frac{P(A) + P(B)}{P(A)P(B)}$, allowing us to determine $P(A) + P(B) = \frac{7}{12} × \frac{0.12}{0.1} = 0.7$.

Defining the union $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.7 - 0.1 = 0.6$, De Morgan's laws simplify the target expression. The numerator $P(\overline{A} \cup \overline{B})$ is equivalent to $P(\overline{A \cap B}) = 1 - 0.1 = 0.9$, while the denominator $P(\overline{A} \cap \overline{B})$ is equivalent to $P(\overline{A \cup B}) = 1 - 0.6 = 0.4$. Dividing these results gives the final value of 9/4.

The relative ordering of the vowels A and E in the word GARDEN is dictated by symmetry, as these two letters must appear in one of only two arrangements: alphabetical order or non-alphabetical order. Because each permutation of the letters is equally likely, these two outcomes occur with equal probability, meaning the chance of the vowels appearing in alphabetical order is $\frac{1}{2}$. This is confirmed by the ratio of the number of arrangements with the vowels in a fixed relative position, $\frac{6!}{2!}$, to the total number of permutations $6!$, yielding $P(v) = \frac{\frac{6!}{2!}}{6!} = \frac{1}{2}$. To find the probability that the vowels are not in alphabetical order, simply subtract this value from the total probability, resulting in $1 - \frac{1}{2} = \frac{1}{2}$.

The conditional probability of an event $B$ given a condition $A \cup \overline{B}$ is calculated using the ratio $\frac{P(B \cap (A \cup \overline{B}))}{P(A \cup \overline{B})}$. The intersection in the numerator simplifies to $P(B \cap A)$ because the intersection of $B$ with its own complement is empty. Given that $P(A) = P(A \cap B) + P(A \cap \overline{B})$, one can determine $P(A \cap B)$ by calculating $0.7 - 0.5 = 0.2$.

For the denominator, $P(A \cup \overline{B})$ is determined by the addition rule $P(A) + P(\overline{B}) - P(A \cap \overline{B})$. Substituting $P(\overline{B}) = 1 - 0.4 = 0.6$, the calculation becomes $0.7 + 0.6 - 0.5$, which equals $0.8$. Dividing the intersection probability $0.2$ by the total probability of the condition $0.8$ results in $\frac{1}{4}$.

Finding the probability that a randomly chosen integer from 1 to 50 is a multiple of 4, 6, or 7 relies on the Inclusion-Exclusion Principle to accurately account for numbers shared between these groups. First, determine the total count of multiples for each condition within the range: there are 12 multiples of 4, 8 multiples of 6, and 7 multiples of 7. Because some numbers fall into more than one category, identify these intersections by finding the least common multiple for each pair. There are 4 multiples of 12, which is the least common multiple of 4 and 6; 1 multiple of 42, which is the least common multiple of 6 and 7; and 1 multiple of 28, which is the least common multiple of 4 and 7. Since no integer in this range is a multiple of 84, the least common multiple of 4, 6, and 7, the intersection of all three sets is 0. Using the formula for the union of these events, $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$, the calculation becomes the sum of the individual probabilities minus the sum of the overlapping probabilities, resulting in 21/50.