Probability – JEE Main MathematicsPractice Questions & PYQs

In any discrete probability distribution, two fundamental constraints must hold: the sum of all individual probabilities must equal one, and the mean is defined as the sum of each outcome multiplied by its corresponding probability. Applying the first constraint, the probabilities $\frac{1}{5} + a + \frac{1}{3} + \frac{1}{5} + b = 1$ simplify to $a + b = \frac{4}{15}$. Using the given mean of 2.3, the calculation $\sum x P(x) = -2(\frac{1}{5}) - 1(a) + 3(\frac{1}{3}) + 4(\frac{1}{5}) + 6(b) = 2.3$ yields the second relationship $6b - a = 0.9$. Solving this system of two linear equations confirms that $a = \frac{1}{10}$ and $b = \frac{1}{6}$.

With all probability values determined, the variance is found using the computational formula \sigma^2 = E\[X^2]$ - \mu^2$. Calculating the expected value of the squared outcomes,$E$[X^2]$ = \sum x_i^2 P(x_i)$, results in$4(\frac{1}{5}) + 1(\frac{1}{10}) + 9(\frac{1}{3}) + 16(\frac{1}{5}) + 36(\frac{1}{6})$, which sums to$0.8 + 0.1 + 3 + 3.2 + 6 = 13.1$. Subtracting the square of the mean,$2.3^2 = 5.29$, from this result provides a variance of$7.81$. Consequently, the value of$100 \sigma^2$ is 781.

Bayes' theorem provides the framework for this calculation, as it determines the likelihood of a specific category given the observed condition of the person. Initially, we identify the probability of a person belonging to one of the three groups: the smoker and non-vegetarian group, the smoker and vegetarian group, or the non-smoker and vegetarian group. Based on the counts provided, these group probabilities are $\frac{160}{400}$, $\frac{100}{400}$, and $\frac{140}{400}$.

Each group has a unique probability of contracting the chest disorder, specifically $35\%$, $20\%$, and $10\%$. The probability of a person being in the first group and having the disorder is determined by multiplying $\frac{160}{400}$ by $35\%$, which is $0.14$. The overall probability of anyone in the group having the disorder is the sum of the individual probability products for all three categories: $(0.4 \times 0.35) + (0.25 \times 0.20) + (0.35 \times 0.10)$, which results in $0.14 + 0.05 + 0.035 = 0.225$. The required conditional probability is the ratio of the first group's product to this total, $\frac{0.14}{0.225}$, which simplifies to $\frac{140}{225}$ or $\frac{28}{45}$.

A number is divisible by 2 if and only if its last digit is even, implying that for the set of digits 3, 3, 4, 4, 4, 5, and 5, any valid number must conclude with 4. The total number of unique seven-digit arrangements possible is determined by the formula for permutations of a multiset, which is
$\frac{7!}{3!2!2!} = 210$
By fixing one of the 4s in the last position to satisfy the divisibility condition, we are left to arrange the remaining six digits, which are two 3s, two 4s, and two 5s. The number of such favorable arrangements is
$\frac{6!}{2!2!2!} = 90$
Dividing these favorable outcomes by the total count of possible numbers results in a probability of $\frac{90}{210}$, which reduces to $\frac{3}{7}$.

The probability that exactly one of two independent events occurs is determined by summing the probability that event A happens while B does not and the probability that event B happens while A does not. For independent events A and B, the probability of only A occurring is $P(A)(1 - P(B))$, while the probability of only B occurring is $(1 - P(A))P(B)$. Substituting the given values $P(A) = p$ and $P(B) = 2p$ into this framework gives the equation $p(1 - 2p) + (1 - p)(2p) = \frac{5}{9}$.

Expanding these terms leads to $p - 2p^2 + 2p - 2p^2 = \frac{5}{9}$, which simplifies to $3p - 4p^2 = \frac{5}{9}$. Multiplying the equation by $9$ results in $27p - 36p^2 = 5$, which rearranges into the standard quadratic form $36p^2 - 27p + 5 = 0$. Factoring this quadratic equation yields $(3p - 1)(12p - 5) = 0$, providing two potential solutions for $p$: $\frac{1}{3}$ and $\frac{5}{12}$. Comparing these two values, the largest value is $\frac{5}{12}$.

This problem relies on Bayes' Theorem, which allows us to update the probability of a hypothesis-that the missing card is not a spade-based on the evidence that two cards drawn from the remaining deck are spades. We must consider two exhaustive scenarios for the missing card: either it is a spade, which occurs with a probability of $\frac{1}{4}$, or it is not a spade, occurring with a probability of $\frac{3}{4}$. If the missing card is a spade, there are 12 spades remaining among the 51 cards, whereas if the missing card is not a spade, all 13 spades remain in the 51 cards.

The conditional probability of drawing two spades depends entirely on which of these two scenarios is true. Given the missing card is a spade, the probability of drawing two spades is $\frac{{}^{12}C_{2}}{{}^{51}C_{2}}$, and if it is not a spade, the probability is $\frac{{}^{13}C_{2}}{{}^{51}C_{2}}$. Applying Bayes' theorem, we find the posterior probability by multiplying the prior probability of each scenario by its respective likelihood and dividing by the total probability of the evidence:

$P(\text{missing not spade} | \text{2 spades}) = \frac{\frac{{}^{13}C_{2}}{{}^{51}C_{2}} \cdot \frac{3}{4}}{\left( \frac{{}^{12}C_{2}}{{}^{51}C_{2}} \cdot \frac{1}{4} \right) + \left( \frac{{}^{13}C_{2}}{{}^{51}C_{2}} \cdot \frac{3}{4} \right)}$

Because the term $\frac{1}{{}^{51}C_{2}}$ appears in every component of the fraction, it cancels out entirely. Substituting the expanded values for the combinations, the expression simplifies to $\frac{13 \cdot 12 \cdot 3}{(12 \cdot 11) + (13 \cdot 12 \cdot 3)}$, which further reduces to $\frac{468}{132 + 468}$. Completing the arithmetic results in $\frac{468}{600}$, which simplifies to $\frac{39}{50}$.

For a quadratic equation $ax^2 + bx + c = 0$ to possess equal roots, its discriminant must be zero, which is defined by the condition $b^2 - 4ac = 0$, or $b^2 = 4ac$. Since the coefficients $a$, $b$, and $c$ are determined by three independent rolls of a standard die, each ranging from 1 to 6, there are $6 \times 6 \times 6 = 216$ total possible outcomes.

The condition $ac = \frac{b^2}{4}$ requires $b$ to be an even number to ensure that $ac$ results in an integer, so we evaluate the scenarios where $b$ is 2, 4, or 6. When $b=2$, $ac=1$, which yields the single combination $(a=1, c=1)$. For $b=4$, $ac=4$, allowing for three valid combinations: $(1,4)$, $(4,1)$, and $(2,2)$. Finally, when $b=6$, $ac=9$, which provides the unique combination $(3,3)$. Aggregating these possibilities gives a total of $1 + 3 + 1 = 5$ favorable outcomes, resulting in a probability of $\frac{5}{216}$.

For independent events $E_1, E_2,$ and $E_3$, let $x, y,$ and $z$ represent the probabilities of each event occurring, respectively. The probability that none of these events occur is $p = (1-x)(1-y)(1-z)$, while the probability that only $E_1$ occurs is $\alpha = x(1-y)(1-z)$. Dividing these two yields the ratio $\frac{\alpha}{p} = \frac{x}{1-x}$, which can be rearranged to find the probability of $E_1$ in terms of $\alpha$ and $p$ as $x = \frac{\alpha}{\alpha+p}$. Applying the same logic for $E_3$ with probability $\gamma$ gives $z = \frac{\gamma}{\gamma+p}$. The required ratio is therefore $\frac{x}{z} = \frac{\alpha}{\alpha+p} \cdot \frac{\gamma+p}{\gamma} = \frac{\alpha}{\gamma} \cdot \frac{\gamma+p}{\alpha+p} = \frac{1 + \frac{p}{\gamma}}{1 + \frac{p}{\alpha}}$.

The given conditions $p = \frac{\alpha\beta}{\alpha-2\beta}$ and $p = \frac{2\beta\gamma}{\beta-3\gamma}$ provide a system to solve for $\beta$ in terms of $p, \alpha,$ and $\gamma$. Solving the first equation for $\beta$ results in $\beta = \frac{\alpha p}{\alpha+2p}$, and solving the second yields $\beta = \frac{3p\gamma}{p-2\gamma}$. Setting these expressions equal to each other, we have $\frac{\alpha p}{\alpha+2p} = \frac{3p\gamma}{p-2\gamma}$. Dividing both sides by $p$ and inverting both fractions leads to the relation $1 + 2\frac{p}{\alpha} = \frac{1}{3}\frac{p}{\gamma} - \frac{2}{3}$. Multiplying the entire equation by 3 gives $3 + 6\frac{p}{\alpha} = \frac{p}{\gamma} - 2$, which simplifies to $\frac{p}{\gamma} = 5 + 6\frac{p}{\alpha}$. Substituting this into our expression for the ratio, we find $\frac{x}{z} = \frac{1 + (5 + 6\frac{p}{\alpha})}{1 + \frac{p}{\alpha}} = \frac{6 + 6\frac{p}{\alpha}}{1 + \frac{p}{\alpha}} = 6$.

When two individuals independently toss three coins, the probability that they match counts of heads depends on the binomial distribution for a single person, where the probabilities of obtaining 0, 1, 2, or 3 heads are $\frac{1}{8}$, $\frac{3}{8}$, $\frac{3}{8}$, and $\frac{1}{8}$ respectively. Because the tossing events for each person are independent, the probability of both achieving the same outcome of $r$ heads is the square of the individual probability $P(r)^2$.

Matching the counts requires summing the joint probabilities of all compatible outcomes, which occurs when both people get zero heads, both get one head, both get two heads, or both get three heads. This calculation is structured as:
$\left(\frac{1}{8}\right)^2 + \left(\frac{3}{8}\right)^2 + \left(\frac{3}{8}\right)^2 + \left(\frac{1}{8}\right)^2$
Summing these values results in $\frac{1}{64} + \frac{9}{64} + \frac{9}{64} + \frac{1}{64}$, which combines to $\frac{20}{64}$ and simplifies to $\frac{5}{16}$.

Assigning distinct balls to boxes follows a binomial probability model where each ball acts independently with a specific probability of landing in a chosen box. Since there are 4 options for each of the 9 balls, the total number of ways to distribute them is $4^9$. To ensure box $B_3$ contains exactly 3 balls, we choose 3 balls out of 9 in $\binom{9}{3}$ ways, while the remaining 6 balls must be placed into any of the other 3 boxes, resulting in $3^6$ possible configurations.

The probability of this outcome is the ratio of these favorable arrangements to the total, represented as $\frac{\binom{9}{3} \cdot 3^6}{4^9}$. Because $\binom{9}{3} = 84$, we can rewrite the expression to isolate the factor $(\frac{3}{4})^9$ by adjusting the powers of 3:

$P = \frac{84 \cdot 3^6}{4^9} = \frac{84}{3^3} \cdot \frac{3^6 \cdot 3^3}{4^9} = \frac{84}{27} \cdot \left(\frac{3}{4}\right)^9 = \frac{28}{9} \cdot \left(\frac{3}{4}\right)^9$

Comparing this result to the form $k(\frac{3}{4})^9$ gives $k = \frac{28}{9}$, which is approximately 3.11. Given that the absolute difference $|3.11 - 3| = 0.11$ is less than 1, the value of $k$ falls within the range $|x-3| < 1$.

In problems involving the arrangement of all letters of a word, the probability that a specific letter occupies a particular position is determined by the total number of positions available. Because each of the 11 letters in the word EXAMINATION has an equal likelihood of being placed in any of the 11 slots, the probability of the letter M appearing at the fourth position is uniform across all positions.

To find this value, calculate the total number of arrangements of the 11 letters, which is $\frac{11!}{2! 2! 2!}$ given the repetitions of A, I, and N. When the letter M is fixed in the fourth position, the remaining 10 letters can be arranged in $\frac{10!}{2! 2! 2!}$ ways. The required probability is obtained by dividing the favorable arrangements by the total arrangements:

$\frac{\frac{10!}{2! 2! 2!}}{\frac{11!}{2! 2! 2!}} = \frac{10!}{11!} = \frac{1}{11}$

In a binomial distribution with $n$ independent trials, the probability of obtaining exactly $k$ successes is calculated using the formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$, where $p$ represents the probability of success and $q = 1-p$ is the probability of failure. For this scenario with 5 trials, we can express the given probabilities as $5pq^4 = 0.4096$ and $10p^2q^3 = 0.2048$. By dividing these two equations, the common terms simplify to $\frac{q}{2p} = 2$, which establishes the relationship $q = 4p$.

Since the sum of the probabilities must satisfy $p + q = 1$, substituting $4p$ for $q$ gives $5p = 1$, resulting in a success probability of $p = \frac{1}{5}$ and a failure probability of $q = \frac{4}{5}$. With these values identified, we can determine the probability of exactly 3 successes by evaluating $\binom{5}{3} p^3 q^2$. Substituting our values into this formula gives $10 \cdot (\frac{1}{5})^3 \cdot (\frac{4}{5})^2$, which simplifies to $10 \cdot \frac{1}{125} \cdot \frac{16}{25}$. Multiplying these terms results in $\frac{160}{3125}$, which reduces to $\frac{32}{625}$.

The total probability of all possible outcomes in a distribution must always sum to 1, which allows us to determine the constant $K$ by setting the sum $K + 2K + 2K + 3K + K = 9K = 1$, identifying $K = \frac{1}{9}$.

When calculating the conditional probability $p = P(1 < X < 4 \mid X < 3)$, we must restrict our focus to the intersection where both conditions are met, meaning $1 < X < 3$. Within the given set of outcomes, this intersection occurs exclusively at $X = 2$, so the probability of the numerator is $P(X = 2) = 2K$. The condition $X < 3$ includes outcomes $X = 1$ and $X = 2$, resulting in a denominator of $P(X = 1) + P(X = 2) = K + 2K = 3K$, which simplifies the conditional probability $p$ to $\frac{2K}{3K} = \frac{2}{3}$.

With these values established, we substitute $p = \frac{2}{3}$ and $K = \frac{1}{9}$ into the equation $5p = \lambda K$ to get $5 \times \frac{2}{3} = \lambda \times \frac{1}{9}$, which simplifies to $\frac{10}{3} = \frac{\lambda}{9}$. Multiplying both sides by 9 to isolate the variable results in $\lambda = 30$.

Let $x$, $y$, and $z$ represent the probabilities $P(B_1)$, $P(B_2)$, and $P(B_3)$ respectively. Given the events are independent, the probability that none of them occur is $P = (1-x)(1-y)(1-z)$, while the probabilities that only one event occurs are $\alpha = x(1-y)(1-z)$, $\beta = y(1-x)(1-z)$, and $\gamma = z(1-x)(1-y)$. These definitions yield the useful ratios $\frac{\alpha}{P} = \frac{x}{1-x}$, $\frac{\beta}{P} = \frac{y}{1-y}$, and $\frac{\gamma}{P} = \frac{z}{1-z}$.

Dividing the equation $(\alpha - 2\beta)P = \alpha\beta$ by $P^2$ results in $\frac{\alpha}{P} - 2\frac{\beta}{P} = \frac{\alpha}{P} \cdot \frac{\beta}{P}$. Substituting the simplified ratios into this equation produces $\frac{x}{1-x} - 2\frac{y}{1-y} = \frac{x}{1-x} \cdot \frac{y}{1-y}$, which, after multiplying by $(1-x)(1-y)$, simplifies to $x(1-y) - 2y(1-x) = xy$, or $x = 2y$.

The relation $(\beta - 3\gamma)P = 2\beta\gamma$ behaves identically when divided by $P^2$, leading to $\frac{\beta}{P} - 3\frac{\gamma}{P} = 2\frac{\beta}{P} \cdot \frac{\gamma}{P}$. Substituting the appropriate ratios results in $\frac{y}{1-y} - 3\frac{z}{1-z} = 2\frac{y}{1-y} \cdot \frac{z}{1-z}$, which simplifies to $y = 3z$. Combining these two findings gives $x = 6z$, establishing that the required ratio $\frac{P(B_1)}{P(B_3)}$ is $6$.

The probability of obtaining exactly $k$ heads in $n$ tosses of a fair coin is determined by the binomial distribution formula ${}^nC_k (\frac{1}{2})^n$. Since the probability of getting $7$ heads is equal to the probability of getting $9$ heads, their binomial coefficients must be equal, which implies ${}^nC_7 = {}^nC_9$. Given the fundamental symmetry property where ${}^nC_a = {}^nC_b$ only when $n = a + b$, we identify the total number of tosses as $n = 16$.

To find the probability of getting exactly $2$ heads, we calculate ${}^{16}C_2 (\frac{1}{2})^{16}$. Evaluating the binomial coefficient ${}^{16}C_2$ results in $\frac{16 \times 15}{2} = 120$. Applying this to the probability formula, we have $120 \times (\frac{1}{2})^{16} = \frac{120}{2^{16}}$. By decomposing the numerator as $120 = 15 \times 2^3$, the expression simplifies to $\frac{15 \times 2^3}{2^{16}}$, which reduces to $\frac{15}{2^{13}}$.

When a die is rolled twice, a sum of seven is achieved through the specific pairs (1,6), (6,1), (2,5), (5,2), (3,4), and (4,3). Given the biased probabilities where the pair (1,6) has face probabilities of $\frac{1}{6} - x$ and $\frac{1}{6} + x$ while the other pairs consist of faces with probabilities of $\frac{1}{6}$, the total probability is calculated as:
$2 \left[ \left(\frac{1}{6} - x\right)\left(\frac{1}{6} + x\right) + \left(\frac{1}{6}\right)^2 + \left(\frac{1}{6}\right)^2 \right]\$ = \frac{13}{96}$
Evaluating the terms inside the brackets results in 2 \left[ \frac{1}{36} - x^2 + \frac{1}{36} + \frac{1}{36} \right]\ = \frac{13}{96}$, which simplifies to$2 \left[ \frac{3}{36} - x^2 \right]$ = \frac{13}{96}$. Distributing the$2$yields$\frac{6}{36} - 2x^2 = \frac{13}{96}$, which reduces to$\frac{1}{6} - 2x^2 = \frac{13}{96}$. Rearranging to isolate$x^2$gives$2x^2 = \frac{16}{96} - \frac{13}{96} = \frac{3}{96} = \frac{1}{32}$. Consequently,$x^2 = \frac{1}{64}$, confirming that the value of$x$is$\frac{1}{8}$.

The probability of getting exactly $r$ correct answers out of 8 independent trials, where each has an equal chance of success or failure, follows a Binomial distribution defined by $\binom{8}{r} (\frac{1}{2})^8$. To find the smallest value of $n$ such that the probability of guessing at least $n$ correct answers is less than half, we set up the inequality $\sum_{r=n}^{8} \binom{8}{r} (\frac{1}{2})^8 < \frac{1}{2}$.

Multiplying both sides of the inequality by $256$ yields $\sum_{r=n}^{8} \binom{8}{r} < 128$. Since the sum of all binomial coefficients $\sum_{r=0}^{8} \binom{8}{r}$ equals $2^8 = 256$, we can express the required sum as $256 - \sum_{r=0}^{n-1} \binom{8}{r} < 128$, which simplifies to $\sum_{r=0}^{n-1} \binom{8}{r} > 128$. We calculate the values of the individual binomial coefficients: $\binom{8}{0} = 1$, $\binom{8}{1} = 8$, $\binom{8}{2} = 28$, $\binom{8}{3} = 56$, and $\binom{8}{4} = 70$. Adding these sequentially, the sum up to $n-1=3$ is $1 + 8 + 28 + 56 = 93$, which is not greater than $128$, but including the next term $\binom{8}{4} = 70$ brings the total to $163$. Because the sum only exceeds $128$ when $n-1$ reaches $4$, the smallest integer value for $n$ is $5$.

The probability of generating a specific sequence like '1001' is determined by analyzing the alternating parity of the positions in the binary string. Because the chance of generating a 0 or 1 depends strictly on whether the position is odd or even, we must calculate the likelihood for two separate, mutually exclusive cases: one where the pattern starts at an odd index and one where it starts at an even index.

For a pattern beginning at an odd position, the sequence of probabilities for the digits 1, 0, 0, 1 corresponds to odd, even, odd, and even placements, respectively. Given that the probability of 1 at an odd place is $1 - \frac{1}{3} = \frac{2}{3}$, and the probability of 0 at an even place is $\frac{1}{2}$, we calculate the product as $\frac{2}{3} \times \frac{1}{2} \times \frac{1}{3} \times \frac{1}{2} = \frac{1}{18}$. If the pattern begins at an even position, the sequence of probabilities follows an even, odd, even, and odd order, which calculates to $\frac{1}{2} \times \frac{1}{3} \times \frac{1}{2} \times \frac{2}{3} = \frac{1}{18}$. Adding these two independent scenarios provides the total probability of $\frac{1}{18} + \frac{1}{18} = \frac{1}{9}$.

A quadratic expression $Ax^{2+}Bx+C$ remains positive for all real $x$ if its leading coefficient is positive and its discriminant $D = B^2 - 4AC$ is strictly less than zero. For the expression $x^{2}+2(a+4)x-5a+64$, the coefficient of $x^2$ is 1, so we only need to satisfy the condition $D = [2(a+4)]^2 - 4(1)(-5a+64) < 0$. Dividing the inequality by 4 simplifies the calculation to $(a+4)^2 - (-5a+64) < 0$, which expands to $a^2 + 8a + 16 + 5a - 64 < 0$, or $a^2 + 13a - 48 < 0$. Factoring the quadratic gives $(a+16)(a-3) < 0$, meaning the inequality holds for $a \in (-16, 3)$. Within the given range of integers $[-5, 30]$, the valid values for $a$ are $\{-5, -4, -3, -2, -1, 0, 1, 2\}$, providing 8 favorable outcomes. Since there are 36 total integers from -5 to 30, the probability is $\frac{8}{36}$, which reduces to $\frac{2}{9}$.

For three distinct numbers to form an arithmetic progression, the sum of the first and last terms must be even, as it must equal twice the middle term. This parity requirement means we must select either two even numbers or two odd numbers, because the sum of two integers is only even if both are even or both are odd. Within any set of 11 consecutive natural numbers, there are either six even and five odd integers, or five even and six odd integers.

The total number of ways to select three numbers from the set is given by ${}^{11}C_{3}$, which equals 165. For favorable outcomes, selecting two even numbers provides ${}^{6}C_{2} = 15$ possible combinations, while selecting two odd numbers provides ${}^{5}C_{2} = 10$ possible combinations. Each of these pairs automatically fixes the unique third number required to complete an arithmetic progression with a positive common difference. Summing these possibilities gives 25 favorable selections, resulting in a probability of $\frac{25}{165}$, which simplifies to $\frac{5}{33}$.

In a sequential game where players take turns, the probability of winning can be modeled as an infinite geometric series. Because the game ends as soon as either player achieves their target score, A wins only if they succeed on their first turn, third turn, fifth turn, and so on, while avoiding all previous losses by both themselves and player B.

The probability of throwing a total of 6 with two fair dice is $5/36$, while the probability of throwing a total of 7 is $6/36$, which simplifies to $1/6$. Player A wins on the first turn with a probability of $5/36$. If A does not win, the game continues only if A fails to throw a 6, which happens with a probability of $31/36$, and player B subsequently fails to throw a 7, which happens with a probability of $5/6$.

The probability of the game lasting past the first full round of throws is the product of these failure probabilities, $r = (31/36) \times (5/6) = 155/216$. The total probability of A winning is the sum of an infinite geometric series where the first term is $a = 5/36$ and the common ratio is $r = 155/216$. Calculating this sum using the formula $P = \frac{a}{1 - r}$ provides the final probability. Substituting the values yields $P = \frac{5/36}{1 - 155/216} = \frac{5/36}{61/216} = \frac{5}{36} \times \frac{216}{61} = \frac{30}{61}$ which represents the overall chance of player A securing the win.