Probability – JEE Main MathematicsPractice Questions & PYQs

The probability of answering exactly 8 out of 10 questions correctly depends on how the two incorrect guesses are distributed across the two distinct sets of questions. In the first set of 4 questions, each has a success probability of $\frac{3}{4}$ and a failure probability of $\frac{1}{4}$. In the second set of 6 questions, each has a success probability of $\frac{1}{4}$ and a failure probability of $\frac{3}{4}$. To achieve exactly 8 correct answers, the student must make exactly 2 errors, which can happen in three mutually exclusive scenarios: failing twice in the first set, failing twice in the second set, or failing once in each set.

The first scenario, where both errors occur in the first set, requires the student to answer 2 questions correctly in the first set and all 6 questions correctly in the second set. The probability is $\binom{4}{2} (\frac{3}{4})^2 (\frac{1}{4})^2 \times \binom{6}{6} (\frac{1}{4})^6 (\frac{3}{4})^0 = 6 \times \frac{9}{16} \times \frac{1}{16} \times \frac{1}{4^6} = \frac{54}{4^{10}}$. The second scenario, where both errors occur in the second set, requires the student to answer all 4 questions correctly in the first set and 4 questions correctly in the second set. The probability is $\binom{4}{4} (\frac{3}{4})^4 (\frac{1}{4})^0 \times \binom{6}{2} (\frac{3}{4})^2 (\frac{1}{4})^4 = 1 \times (\frac{3}{4})^4 \times 15 \times (\frac{3}{4})^2 \times (\frac{1}{4})^4 = \frac{15 \times 729}{4^{10}} = \frac{10935}{4^{10}}$. The third scenario, where one error occurs in each set, requires the student to answer 3 questions correctly in the first set and 5 questions correctly in the second set. The probability is $\binom{4}{1} (\frac{3}{4})^3 (\frac{1}{4})^1 \times \binom{6}{1} (\frac{3}{4})^1 (\frac{1}{4})^5 = 4 \times (\frac{3}{4})^3 \times (\frac{1}{4}) \times 6 \times (\frac{3}{4}) \times (\frac{1}{4})^5 = \frac{24 \times 81}{4^{10}} = \frac{1944}{4^{10}}$.

Summing these probabilities, we get $\frac{54 + 10935 + 1944}{4^{10}} = \frac{12933}{4^{10}}$. Factoring this expression to match the form $\frac{27 k}{4^{10}}$, we find that $12933 = 27 \times 479$. Comparing this to the given expression, $k$ is equal to 479.

Binomial distributions are defined by the number of trials $n$ and the probability of success $p$, where the mean is given by $np$ and the variance by $np(1-p)$. By dividing the variance by the mean, we can isolate the probability of failure as $1-p = \frac{4/3}{4} = \frac{1}{3}$, which implies $p = \frac{2}{3}$. Substituting this success probability into the mean equation $n \times \frac{2}{3} = 4$ reveals that the total number of trials is $n=6$.

To determine the cumulative probability $P(X \leq 2)$, we sum the individual probabilities for zero, one, and two successes using the formula $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$. Since the probability of failure is $\frac{1}{3}$ and the success probability is $\frac{2}{3}$, these probabilities are calculated as $P(X=0) = \binom{6}{0} (\frac{2}{3})^0 (\frac{1}{3})^6 = \frac{1}{729}$, $P(X=1) = \binom{6}{1} (\frac{2}{3})^1 (\frac{1}{3})^5 = \frac{12}{729}$, and $P(X=2) = \binom{6}{2} (\frac{2}{3})^2 (\frac{1}{3})^4 = \frac{60}{729}$. Combining these values gives a sum of $P(X \leq 2) = \frac{1+12+60}{729} = \frac{73}{729}$.

The final step is to calculate $54 P(X \leq 2)$, which is $54 \times \frac{73}{729}$. Because $729$ can be expressed as $27 \times 27$ and $54$ as $2 \times 27$, the calculation simplifies to $2 \times \frac{73}{27}$, resulting in $\frac{146}{27}$.

A 6-digit number formed using only digits 1 and 8 is divisible by 21 if and only if it is divisible by both 3 and 7. Each digit can be represented as $1 + 7x_i$, where $x_i \in \{0, 1\}$ is 0 if the digit is 1 and 1 if the digit is 8. Expressing the 6-digit number as $\sum_{i=0}^5 (1 + 7x_i) 10^i$, we obtain $111111 + 7 \sum_{i=0}^5 x_i 10^i$. Because 111111 is a multiple of 21, the entire number is divisible by 21 if and only if the term $\sum_{i=0}^5 x_i 10^i$ is divisible by 3.

Since $10 \equiv 1 \pmod 3$, the divisibility condition simplifies to $\sum_{i=0}^5 x_i \equiv 0 \pmod 3$. We need the sum of the $x_i$ variables, which represent the choices of digit 8, to be 0, 3, or 6. The total number of valid ways to select these digits is given by the sum of combinations:
$\binom{6}{0} + \binom{6}{3} + \binom{6}{6} = 1 + 20 + 1 = 22$
There are $2^6 = 64$ total possible 6-digit numbers formed with these two digits. The probability $p$ of choosing a multiple of 21 is $\frac{22}{64} = \frac{11}{32}$. Multiplying this by 96, we find $96 \times \frac{11}{32} = 33$.

Determining the probability in this arrangement relies on partitioning the selection process into independent intervals created by the fixed values $x_2=7$ and $x_4=11$. Because the set must be strictly increasing, $x_1$ must be chosen from the 6 integers less than 7, $x_3$ must be chosen from the 3 integers between 7 and 11, and $x_5$ must be chosen from the 7 integers greater than 11.

This identifies three independent selection groups, making the number of favorable outcomes $^6C_1 \times ^3C_1 \times ^7C_1$. Dividing this by the total sample space size, $^{18}C_5$, results in the probability expression $P(E) = \frac{6 \times 3 \times 7}{^{18}C_5}$. Evaluating these combinations leads to a final probability of $\frac{1}{68}$.

In binomial experiments, the probability $P(X=k) = \binom{n}{k} p^k q^{n-k}$ determines the chance of observing exactly $k$ successes, where $p$ and $q$ represent the probabilities of success and failure, respectively. With 5 coin tosses, the condition that the probability of 4 heads equals the probability of 5 heads is defined by the equation $\binom{5}{4} p^4 q = \binom{5}{5} p^5$. This simplifies to the relationship $5q = p$. Because the sum of $p$ and $q$ is 1, this condition forces $6q = 1$, which establishes $q = 1/6$ and $p = 5/6$.

The probability of obtaining at most two heads involves summing the probabilities for the cases of 0, 1, and 2 heads. The total probability is given by:
$\binom{5}{0} q^5 + \binom{5}{1} p q^4 + \binom{5}{2} p^2 q^3$
Substituting the determined values of $p$ and $q$ into the expression leads to $(1/6)^5 + 5(5/6)(1/6)^4 + 10(5/6)^2(1/6)^3$. This results in the sum $(1 + 25 + 250) / 6^5$, simplifying to $276 / 6^5$. Reducing this by dividing both the numerator and denominator by 6 yields $46 / 6^4$.

Calculating the probability of an event requires determining the ratio of favorable outcomes to the total number of possible outcomes, which for a randomly selected 3-digit number is 900. To find the numbers containing at least two odd digits, we analyze the disjoint cases where there are exactly three odd digits and where there are exactly two odd digits. For a 3-digit number to consist entirely of odd digits, we select from the five available odd numbers $\{1, 3, 5, 7, 9\}$ for each position, resulting in $5 \times 5 \times 5 = 125$ possibilities.

When there are exactly two odd digits, we must account for the digit zero separately because it cannot appear in the hundreds place. If the even digit is zero, it must occupy either the tens or the units place, with odd digits filling the remaining two positions, which results in $5 \times 1 \times 5 + 5 \times 5 \times 1 = 50$ distinct numbers. If the even digit is non-zero, we choose from $\{2, 4, 6, 8\}$ and can place this digit in any of the three positions, while the other two slots are filled by odd digits; this calculation gives $3 \times 4 \times 5 \times 5 = 300$ possibilities. Summing these scenarios, we find there are $125 + 50 + 300 = 475$ favorable outcomes. The final probability is $\frac{475}{900}$, which simplifies to $\frac{19}{36}$.

The mean of a random variable following a binomial distribution is defined as $np$, where $n$ is the total number of trials and $p$ represents the probability of success on a single trial. Here, the die is thrown twice, so $n = 2$, and a success occurs when rolling a perfect square, which corresponds to the numbers 1 and 4 on a standard six-faced die. Let $\alpha$ be the probability of rolling a prime number, $\beta$ be the probability of rolling a composite number, and $\gamma$ be the probability of rolling 1. Given the conditions $3\alpha = 6\beta = 2\gamma = k$, we can express the probabilities as $\alpha = \frac{k}{3}$, $\beta = \frac{k}{6}$, and $\gamma = \frac{k}{2}$. Applying the probability constraint $3\alpha + 2\beta + \gamma = 1$, we substitute these expressions to find $k + \frac{k}{3} + \frac{k}{2} = 1$. Simplifying this sum leads to $\frac{11k}{6} = 1$, resulting in a value of $k = \frac{6}{11}$.

The probability of obtaining a perfect square is the sum of the probability of rolling 1 and the probability of rolling 4, which is denoted as $P(1) + P(4) = \gamma + \frac{k}{6}$. Substituting the known values, we have $p = \frac{k}{2} + \frac{k}{6} = \frac{4k}{6} = \frac{2k}{3}$. Using our calculated value for $k$, this probability becomes $p = \frac{2}{3} \cdot \frac{6}{11} = \frac{4}{11}$. By substituting $n = 2$ and $p = \frac{4}{11}$ into the mean formula, we determine the mean of $X$ to be $2 \cdot \frac{4}{11} = \frac{8}{11}$.

The condition for a valid probability distribution requires that every individual probability $P(X)$ must be non-negative, which imposes the constraints $1-d \ge 0$, $1+2d \ge 0$, $1-4d \ge 0$, and $1+3d \ge 0$. These inequalities simplify to $d \le 1$, $d \ge -1/2$, $d \le 1/4$, and $d \ge -1/3$, establishing the permissible range for $d$ as $[-1/3, 1/4]$ and identifying the minimum value as $d = -1/3$. The mean of the distribution is defined by $E(X) = \sum x P(x) = 0\left(\frac{1-d}{4}\right) + 1\left(\frac{1+2d}{4}\right) + 2\left(\frac{1-4d}{4}\right) + 3\left(\frac{1+3d}{4}\right)$. Simplifying this expression yields $E(X) = \frac{1+2d + 2-8d + 3+9d}{4} = \frac{6+3d}{4}$. Substituting the minimum value $d = -1/3$ results in $E(X) = \frac{6+3(-1/3)}{4} = \frac{5}{4}$, and sixty times this mean is $60 \times \frac{5}{4} = 75$.

Understanding probabilities involving multiple events often hinges on the definitions of conditional probability and set operations like unions and complements. By applying the definition $P(B|A) = \frac{P(A \cap B)}{P(A)}$, we can derive $P(A) = \frac{1/9}{2/5} = \frac{5}{18}$, and similarly, using $P(A|B) = \frac{P(A \cap B)}{P(B)}$, we find $P(B) = \frac{1/9}{1/7} = \frac{7}{9}$.

For the first statement, we evaluate $P(A' \cup B)$ by recognizing that it represents the complement of $A$ occurring while $B$ does not, which is calculated as $1 - P(A \cap B')$. Since $P(A \cap B') = P(A) - P(A \cap B) = \frac{5}{18} - \frac{2}{18} = \frac{3}{18} = \frac{1}{6}$, the value of the expression becomes $1 - \frac{1}{6} = \frac{5}{6}$, confirming that statement (S1) is true.

To evaluate the second statement, we apply De Morgan's Law, which relates the intersection of complements to the complement of the union, $P(A' \cap B') = 1 - P(A \cup B)$. Calculating the union first yields $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{5}{18} + \frac{14}{18} - \frac{2}{18} = \frac{17}{18}$, which leads to $P(A' \cap B') = 1 - \frac{17}{18} = \frac{1}{18}$, confirming that (S2) is also true.

Determining the individual probabilities of events $E_1$ and $E_2$ is the necessary first step toward validating the relationships presented in the options. Using the definition of conditional probability $P(E_1|E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)}$, we substitute the known values to find $P(E_2) = \frac{1/8}{1/2} = \frac{1}{4}$. Similarly, applying the definition $P(E_2|E_1) = \frac{P(E_1 \cap E_2)}{P(E_1)}$, we solve for $P(E_1)$ to get $\frac{1/8}{3/4} = \frac{1}{6}$.

Calculating the product $P(E_1) \cdot P(E_2)$ gives $\frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}$, which serves as the benchmark value for our comparison. Focusing on the expression $P(E_1 \cap E_2')$, which denotes the event that $E_1$ occurs but $E_2$ does not, we express this as the difference between the probability of $E_1$ and the probability of the intersection, or $P(E_1) - P(E_1 \cap E_2)$. Performing the subtraction $\frac{1}{6} - \frac{1}{8}$ results in $\frac{4}{24} - \frac{3}{24} = \frac{1}{24}$. Because this result precisely matches the product we calculated, the equality holds true.

The binomial distribution is characterized by the probability mass function $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$, where $n$ is the number of trials and $p$ is the probability of success. By applying the condition $3 P(X=0) = P(X=1)$ with $n=33$, we set up the equation $3 \binom{33}{0} p^0 (1-p)^{33} = \binom{33}{1} p^1 (1-p)^{32}$. Simplifying this expression gives $3(1-p) = 33p$, which simplifies further to $1-p = 11p$, meaning $p = \frac{1}{12}$ and consequently the ratio $\frac{1-p}{p} = 11$.

For the first term of the target expression, $\frac{P(X=15)}{P(X=18)}$, we substitute the probability mass function to get $\frac{\binom{33}{15} p^{15} (1-p)^{18}}{\binom{33}{18} p^{18} (1-p)^{15}}$. Because $\binom{33}{15} = \binom{33}{18}$, the combination coefficients cancel out, leaving $\frac{(1-p)^3}{p^3} = \left(\frac{1-p}{p}\right)^3$, which equals $11^3$. Similarly, for the second term $\frac{P(X=16)}{P(X=17)}$, the expression becomes $\frac{\binom{33}{16} p^{16} (1-p)^{17}}{\binom{33}{17} p^{17} (1-p)^{16}}$. Since $\binom{33}{16} = \binom{33}{17}$, they cancel, reducing the term to $\frac{1-p}{p}$, which equals 11. Subtracting these two results, $11^3 - 11$, leads to the final calculation of $1331 - 11 = 1320$.

Bayes' theorem provides the framework to determine the probability of an earlier event, such as the color of a transferred ball, based on the outcome of a subsequent event, such as drawing a ball of a specific color from the second bag. Bag I contains ten balls in total, so the probability of transferring a red, black, or white ball is $3/10$, $4/10$, and $3/10$, respectively. After the transfer, Bag II contains ten balls, consisting of the original nine plus the one added.

If a red ball is transferred, Bag II becomes composed of three red, five black, and two white balls, making the probability of drawing a black ball $5/10$. If a black ball is transferred, Bag II contains two red, six black, and two white balls, which results in a $6/10$ probability of drawing a black ball. If a white ball is transferred, Bag II contains two red, five black, and three white balls, leading to a $5/10$ probability of drawing a black ball.

The total probability of drawing a black ball is the sum of these conditional possibilities: $(3/10 \times 5/10) + (4/10 \times 6/10) + (3/10 \times 5/10)$. This simplifies to $15/100 + 24/100 + 15/100$, which equals $54/100$. Applying Bayes' theorem, the probability that the transferred ball was red, given that the ball drawn from Bag II was black, is calculated as the ratio of the probability of the red transfer and black draw event to the total probability of drawing a black ball:
$\frac{15/100}{54/100} = \frac{15}{54} = \frac{5}{18}$

To determine the probability that a randomly chosen qualified candidate is female, we must divide the number of qualified female candidates by the total number of qualified candidates. Given that 60% of all candidates qualified, if we assume a total of 100 candidates, then 60 candidates passed the exam. Based on the condition that the number of females qualifying is twice the number of males qualifying, these 60 qualified candidates are distributed such that 40 are female and 20 are male. The probability of selecting a female candidate from the qualified group is calculated as the ratio $\frac{40}{60}$, which simplifies to $\frac{2}{3}$.

The inclusion-exclusion principle is the starting point for this problem, as it allows us to determine the intersection of two events from their individual probabilities and the probability of their union. By calculating $P(A \cap B) = P(A) + P(B) - P(A \cup B)$, we find $\frac{1}{3} + \frac{1}{5} - \frac{1}{2} = \frac{1}{30}$. Once we have this intersection, we can evaluate conditional probabilities using the definition $P(E|F) = \frac{P(E \cap F)}{P(F)}$.

We need to compute $P(A|B')$ and $P(B|A')$. For the first term, the numerator $P(A \cap B')$ represents the probability of $A$ occurring while $B$ does not, which is $P(A) - P(A \cap B) = \frac{1}{3} - \frac{1}{30} = \frac{9}{30}$. Since the denominator is the complement $P(B') = 1 - P(B) = \frac{4}{5}$, the term becomes $\frac{9/30}{4/5} = \frac{3}{8}$. Similarly, for the second term, the numerator $P(B \cap A')$ is $P(B) - P(A \cap B)$, yielding $\frac{1}{5} - \frac{1}{30} = \frac{5}{30}$. Dividing this by $P(A') = 1 - P(A) = \frac{2}{3}$ gives $\frac{5/30}{2/3} = \frac{1}{4}$. Adding these two values together, $\frac{3}{8} + \frac{1}{4}$ results in $\frac{5}{8}$.

In a binomial distribution, the mean and variance are defined as $np$ and $npq$ respectively, where $n$ represents the number of trials, $p$ is the probability of success, and $q = 1 - p$ is the probability of failure. With the mean $M = np$ and variance $V = npq$, the problem states that $M + V = 24$ and $MV = 128$, which implies $M$ and $V$ are roots of the quadratic equation $x^2 - 24x + 128 = 0$. Solving this equation yields the roots 16 and 8; because the mean $np$ must be greater than the variance $npq$ when $q < 1$, it follows that $np = 16$ and $npq = 8$. From these values, we determine that $q = 8/16 = 1/2$, which means $p = 1/2$ and $n = 32$. The probability of achieving exactly one or two successes is the sum $P(X=1) + P(X=2)$, calculated using the binomial probability formula ${}^{n}C_{k} p^k q^{n-k}$ as ${}^{32}C_{1} (\frac{1}{2})^{32} + {}^{32}C_{2} (\frac{1}{2})^{32}$. Simplifying this expression gives $(32 + 496) \cdot (\frac{1}{2})^{32}$, which equals $528 \cdot \frac{1}{2^{32}}$. Given that $528 = 33 \cdot 16 = 33 \cdot 2^4$, the expression simplifies further to $\frac{33 \cdot 2^4}{2^{32}}$, resulting in $\frac{33}{2^{28}}$.

For a quadratic expression $x^2 + \alpha x + \beta$ to remain strictly positive for every real input, the corresponding parabola must lie entirely above the horizontal axis, which occurs only when the discriminant is negative. This condition simplifies to $\alpha^2 - 4\beta < 0$, or $\alpha^2 < 4\beta$. By testing each integer value of $\alpha$ from 1 to 6, we can identify which outcomes for $\beta$ satisfy this inequality.

When $\alpha = 1$, we require $1 < 4\beta$, a condition met by all 6 possible values of $\beta$. For $\alpha = 2$, the inequality becomes $4 < 4\beta$, which holds true for values of $\beta$ from 2 to 6, totaling 5 favorable outcomes. If $\alpha = 3$, we need $9 < 4\beta$, satisfied by $\beta$ values from 3 to 6, adding 4 more cases. With $\alpha = 4$, the condition $16 < 4\beta$ requires $\beta$ to be greater than 4, which provides 2 choices. For $\alpha = 5$ and $\alpha = 6$, the required values for $\beta$ exceed the maximum die face of 6, leaving no possible outcomes. By summing these instances, we obtain 17 successful outcomes out of a total sample space of 36 possible rolls, leading to a final probability of $\frac{17}{36}$.

Geometric probability is determined by the ratio of the area of the favorable region to the total area of the bounded region. The total area is defined by a triangle with vertices at $(0, 3)$, $(0, -5)$, and $(6, 0)$, which yields an area of $\frac{1}{2} \times 8 \times 6 = 24$.

To find the probability for $y<1$, it is efficient to calculate the area of the region where $y \ge 1$ and subtract it from the total. This smaller triangular region is bounded by the $y$-axis and the line $2y+x=6$, extending from $y=1$ to $y=3$, with a base of length $2$ and a height of $4$ at $y=1$, resulting in an area of $\frac{1}{2} \times 2 \times 4 = 4$. Consequently, the favorable area for $y<1$ is $24 - 4 = 20$, and the probability is $\frac{20}{24} = \frac{5}{6}$.

A binomial distribution with parameters $n=5$ and probability $p$ follows the probability mass function $P(X=k) = \binom{n}{k} p^k q^{n-k}$, where $q = 1-p$. The condition $P(X=0) = P(X=1)$ establishes the following relationship:
$\binom{5}{0} p^0 q^5 = \binom{5}{1} p^1 q^4$
Because $\binom{5}{0}$ equals 1 and $\binom{5}{1}$ equals 5, this equation simplifies to $q^5 = 5 p q^4$. Dividing both sides by $q^4$, assuming $q$ is non-zero, reveals that $q = 5p$.

Substituting $q = 1-p$ into this result leads to $1-p = 5p$, which identifies the value of $p$ as $1/6$ and $q$ as $5/6$. The required ratio $\frac{P(X=2)}{P(X=3)}$ is defined by the formula:
$\frac{P(X=2)}{P(X=3)} = \frac{\binom{5}{2} p^2 q^3}{\binom{5}{3} p^3 q^2}$
Since both $\binom{5}{2}$ and $\binom{5}{3}$ equal 10, they cancel out, reducing the expression to $\frac{q}{p}$. Calculating this with the established values of $p$ and $q$ gives $\frac{5/6}{1/6}$, resulting in a final value of 5.

For a binomial distribution, the mean is defined as $np$ and the variance as $npq$, where $q = 1 - p$. The provided conditions $np + npq = 24$ and $(np)(npq) = 128$ imply that $np$ and $npq$ are roots of the quadratic equation $t^2 - 24t + 128 = 0$. Factoring this quadratic reveals the roots to be 16 and 8. Because the mean must be larger than the variance for a binomial distribution, we assign $np = 16$ and $npq = 8$, which results in $q = 0.5$, $p = 0.5$, and $n = 32$.

Calculating $P(X > n-3)$ for $n=32$ involves finding the sum of the probabilities for $X = 30$, $X = 31$, and $X = 32$. Using the binomial probability formula $P(X = r) = \binom{n}{r} p^r q^{n-r}$, the total probability is:
$P(X > 29) = \left( \binom{32}{30} + \binom{32}{31} + \binom{32}{32} \right) \frac{1}{2^{32}}$
Simplifying the binomial coefficients, we get $\binom{32}{2} = 496$, $\binom{32}{1} = 32$, and $\binom{32}{0} = 1$. Summing these results gives $496 + 32 + 1 = 529$. Comparing this to the form $\frac{k}{2^n}$, the value of $k$ is 529.

Five-digit numbers form a sequence starting at 10000 and ending at 99999, which encompasses a total sample space of $90000$ integers. To identify numbers divisible by 7 but not by 5, we calculate the total count of multiples of 7 and subtract the multiples of their least common multiple, which is 35, because numbers divisible by both 7 and 5 are also divisible by 35.

The first five-digit multiple of 7 is 10003 and the last is 99995. Applying the formula for the number of terms in an arithmetic progression, the count of these integers is $\frac{99995 - 10003}{7} + 1 = 12857$. Similarly, the first five-digit multiple of 35 is 10010, while the last remains 99995. The count of these numbers is $\frac{99995 - 10010}{35} + 1 = 2572$.

Subtracting the count of multiples of 35 from the count of multiples of 7 yields the total frequency of numbers satisfying the condition, which is $12857 - 2572 = 10285$. The probability $p$ is defined as $\frac{10285}{90000}$. Multiplying this by 9 simplifies the expression to $\frac{10285}{10000}$, which equals $1.0285$.