Probability – JEE Main MathematicsPractice Questions & PYQs

Bayes' Theorem provides the necessary framework for this problem, as it allows us to update the prior probability of each location-each starting with a probability of $\frac{1}{2}$-based on the specific observed evidence of the sequence "AN" on the envelope. First, consider the word "KANPUR," which contains 6 letters and thus offers 5 consecutive letter pairs; since the sequence "AN" occurs only once among these pairs, the conditional probability of seeing it, given the letter is from Kanpur, is $\frac{1}{5}$. In contrast, "ANANTPUR" has 8 letters and 7 consecutive pairs, containing the sequence "AN" twice, which results in a conditional probability of $\frac{2}{7}$.

Calculating the posterior probability requires dividing the likelihood associated with Anantpur by the sum of probabilities from both cities. This is mathematically expressed as $\frac{\frac{1}{2} \times \frac{2}{7}}{\frac{1}{2} \times \frac{1}{5} + \frac{1}{2} \times \frac{2}{7}}$. Since the factor of $\frac{1}{2}$ is present in every term, it cancels out, simplifying the calculation to $\frac{\frac{2}{7}}{\frac{1}{5} + \frac{2}{7}}$. Combining the fractions in the denominator gives a common denominator of 35, resulting in $\frac{7}{35} + \frac{10}{35} = \frac{17}{35}$. Dividing $\frac{2}{7}$ by $\frac{17}{35}$ effectively means multiplying $\frac{2}{7}$ by $\frac{35}{17}$, which simplifies to $\frac{10}{17}$.

Selecting any three distinct days out of 31 is equivalent to calculating the total number of combinations, given by $\binom{31}{3} = \frac{31 \times 30 \times 29}{3 \times 2 \times 1} = 4495$. For three distinct numbers chosen from this set to form an arithmetic progression $x, y, z$, the condition $x+z=2y$ must hold. This implies that the sum $x+z$ must be even, which forces the first and third terms to share the same parity; they must either both be odd or both be even for the middle term $y$ to be an integer.

Within the set of 31 days, there are 16 odd numbers and 15 even numbers. Picking two odd numbers to define the progression provides $\binom{16}{2} = \frac{16 \times 15}{2} = 120$ ways, and picking two even numbers provides $\binom{15}{2} = \frac{15 \times 14}{2} = 105$ ways. Summing these possibilities gives 225 favorable outcomes. The probability of selecting an arithmetic progression is therefore $\frac{225}{4495}$, which reduces to the fraction $\frac{45}{899}$ after dividing both numerator and denominator by 5. Since $a = 45$ and $b = 899$ are coprime, their sum $a + b$ is 944.

For a set $S$ with $n$ elements, the total number of ordered pairs $(A, B)$ in the Cartesian product $P(S) \times P(S)$ is $(2^n) \times (2^n) = 2^{2n}$. With $n = 5$, the size of the sample space is $2^{10} = 1024$. The condition $A \cap B = \emptyset$ requires that for each of the $5$ elements, there are $3$ possible assignments: the element belongs to $A$ only, $B$ only, or neither set. Consequently, the number of favorable pairs is $3^5 = 243$. The probability of the event $E$ is given by $\frac{3^5}{2^{10}}$, which corresponds to $\frac{3^p}{2^q}$ with $p = 5$ and $q = 10$. The sum $p + q$ is $5 + 10 = 15$.

The probability that the product $a b$ is divisible by 3 is most easily found by calculating the probability of the complement event, where neither $a$ nor $b$ is divisible by 3. In the set of integers from 1 to 50, there are 16 multiples of 3, leaving $50 - 16 = 34$ numbers that are not divisible by 3. The total number of ways to choose 2 distinct numbers from the set of 50 is given by $\binom{50}{2} = \frac{50 \times 49}{2} = 1225$. The number of ways to choose 2 distinct numbers such that neither is a multiple of 3 is $\binom{34}{2} = \frac{34 \times 33}{2} = 561$. Subtracting the probability of the complement event from 1 yields the desired probability, $1 - \frac{561}{1225} = \frac{664}{1225}$.

The total number of ways to distribute 12 distinct balls into 6 unordered pairs is calculated using the formula $\frac{12!}{2^6 \cdot 6!}$, where we account for the 2 balls in each group and divide by the $6!$ ways to arrange the pairs themselves. For the condition where each pair must contain one blue and one green ball, we treat the blue balls as fixed and arrange the green balls among them in $6!$ distinct ways to form the required mixed pairs. By finding the ratio of these favorable arrangements to the total number of partitions, we arrive at the probability $\frac{6! \cdot 2^6 \cdot 6!}{12!}$, which simplifies to the final value of $\frac{16}{231}$.

This problem requires applying Bayes' Theorem to determine the likelihood of a specific bag composition given an observed result. Since no prior information is provided regarding the initial distribution of the balls, we assume that every possible state that makes the observed event feasible is equally likely to have occurred. Given that drawing three black balls is only possible if there are at least three black balls in the bag, the number of red balls $k$ must be in the range $0 \leq k \leq 7$.

The conditional probability of the bag having 1 red ball and 9 black balls given the observed event is proportional to the ratio of ways to select three black balls in that specific configuration compared to the total number of ways to select three black balls across all valid configurations. For the specific case where $k=1$, the number of ways to draw three black balls is $\binom{9}{3}$, which evaluates to 84. To find the denominator, we sum the combinations $\binom{10-k}{3}$ for all valid values of $k$ from 0 to 7. By letting $j = 10-k$, this sum becomes $\sum_{j=3}^{10} \binom{j}{3}$. Applying the hockey-stick identity, this summation is equivalent to $\binom{11}{4}$, which equals 330. Dividing the favorable outcomes for $k=1$ by this total results in the fraction $\frac{84}{330}$, which simplifies to $\frac{14}{55}$.

The outcome of this experiment relies on the law of total probability, which dictates that the chance of observing a specific result is the weighted sum of the probabilities for each mutually exclusive scenario. Here, we must account for the two ways to draw a coin: selecting one of the fair coins or selecting the unique double-headed coin. There are $N$ fair coins and one double-headed coin, making the total number of coins $(N+1)$. Selecting a fair coin occurs with probability $\frac{N}{N+1}$, and if chosen, it yields a head with probability $\frac{1}{2}$. Selecting the double-headed coin occurs with probability $\frac{1}{N+1}$, and this coin guarantees a head with probability 1. Aggregating these outcomes provides the overall probability of observing a head:

$P(\text{Head}) = \left(\frac{N}{N+1}\right)\left(\frac{1}{2}\right) + \left(\frac{1}{N+1}\right)(1) = \frac{9}{16}$

Simplifying the expression on the left side leads to the equation $\frac{N+2}{2(N+1)} = \frac{9}{16}$. By cross-multiplying, we obtain $16(N+2) = 18(N+1)$, which expands to $16N + 32 = 18N + 18$. Subtracting $16N$ from both sides and $18$ from both sides results in the equation $2N = 14$, which yields $N = 7$.

The sum of all probabilities in a valid probability distribution must equal 1, leading to the equation $2k + k + 3k + 2k^2 + 2k + (k^2 + k) + 7k^2 = 1$, which simplifies to $10k^2 + 9k - 1 = 0$. Factoring this as $(10k - 1)(k + 1) = 0$ reveals the valid solution $k = 0.1$, as probabilities cannot be negative. The required probability $P(3 < x \le 6)$ is the sum of $P(4) + P(5) + P(6)$, which is $2k^2 + 2k + (k^2 + k) = 3k^2 + 3k$. Substituting $k = 0.1$ into this expression yields 0.33.

To solve this problem, we analyze the sequence of eight coin tosses by dividing them into three distinct, non-overlapping segments: the first three tosses, the three middle tosses, and the last two tosses. Let $k$ represent the number of heads occurring in the middle segment of three tosses, which are shared by the two conditions given. For the first condition to be satisfied, there must be a total of four heads in the first six tosses, implying that there must be $4-k$ heads in the first three tosses. Similarly, the second condition requires a total of three heads in the last five tosses, meaning there must be $3-k$ heads in the final two tosses.

Because $k$ is the number of heads in three tosses, it must be an integer between 0 and 3, but the requirement for non-negative heads in the other segments limits $k$ to the values 1, 2, and 3. For $k=1$, we choose 1 head from the middle three, 3 heads from the first three, and 2 heads from the final two, giving $\binom{3}{1} \times \binom{3}{3} \times \binom{2}{2} = 3 \times 1 \times 1 = 3$ ways. For $k=2$, we choose 2 heads from the middle three, 2 heads from the first three, and 1 head from the final two, resulting in $\binom{3}{2} \times \binom{3}{2} \times \binom{2}{1} = 3 \times 3 \times 2 = 18$ ways. Finally, for $k=3$, we choose 3 heads from the middle three, 1 head from the first three, and 0 heads from the final two, resulting in $\binom{3}{3} \times \binom{3}{1} \times \binom{2}{0} = 1 \times 3 \times 1 = 3$ ways. Adding these gives 24 favorable outcomes out of a total sample space of $2^8 = 256$, so $p = \frac{24}{256}$. Calculating the final value, $96p = 96 \times \frac{24}{256} = 9$.

The probability follows a binomial distribution defined by $P(X=k) = \binom{n}{k} p^k q^{n-k}$, where $n=8$ is the number of trials, $p=0.1$ is the probability of selecting a defective bulb, and $q=0.9$ is the probability of selecting a non-defective bulb. The total probability of obtaining at least $7$ defective bulbs is the sum $P(X \ge 7) = P(X=7) + P(X=8)$.

$P(X=7) = \binom{8}{7} (0.1)^7 (0.9)^1 = 8 \times 10^{-7} \times 0.9 = 72 \times 10^{-8}$
$P(X=8) = \binom{8}{8} (0.1)^8 (0.9)^0 = 1 \times 10^{-8}$
$P(X \ge 7) = 72 \times 10^{-8} + 1 \times 10^{-8} = \frac{73}{10^8}$

This problem relies on the Law of Total Probability, which enables us to determine the total likelihood of an event by aggregating the probabilities of its occurrence across all mutually exclusive scenarios. Here, the team wins the tournament either when player A is captain or when player B is captain, and these possibilities represent the only two ways the team can achieve a win.

To find the overall probability, we calculate the joint probability for each captaincy scenario and then sum them. We determine the chance of winning with player A by multiplying the probability that A is chosen, 0.6, by the probability of winning with A, 0.8, which results in the equation $0.6 \times 0.8 = 0.48$. Similarly, we determine the chance of winning with player B by multiplying the probability that B is chosen, 0.4, by the probability of winning with B, 0.7, yielding $0.4 \times 0.7 = 0.28$. The total probability of winning the tournament is the sum of these contributions, calculated as $0.48 + 0.28 = 0.76$.

Bayes' Theorem provides the method for determining the probability of a specific cause given an observed outcome, which in this case is the candidate arriving late. The total probability of arriving late is the sum of the joint probabilities of being late by each individual transport mode, where each joint probability is the product of the chance of selecting that mode and the chance of being late while using it. Calculating these, the bus contributes $\frac{2}{5} \times \frac{1}{5} = \frac{2}{25}$, the scooter contributes $\frac{1}{5} \times \frac{1}{3} = \frac{1}{15}$, and the car contributes $\frac{2}{5} \times \frac{1}{4} = \frac{1}{10}$.

Summing these individual probabilities using a common denominator of $150$ results in $\frac{12}{150} + \frac{10}{150} + \frac{15}{150} = \frac{37}{150}$. The probability that the candidate travelled by bus given they arrived late is the ratio of the bus-specific joint probability of $\frac{12}{150}$ to the total probability of $\frac{37}{150}$. This division simplifies directly to $\frac{12}{37}$.

The Total Probability Theorem states that the overall probability of an event is the sum of the conditional probabilities weighted by the probability of each mutually exclusive condition. There are two scenarios for the transfer from bag B: drawing a white ball with probability $\frac{6}{10}$ (resulting in 10 white and 8 black balls in bag A) or drawing a black ball with probability $\frac{4}{10}$ (resulting in 9 white and 9 black balls in bag A).

The probability of drawing a white ball from bag A is calculated as $P(W) = \left(\frac{6}{10} \times \frac{10}{18}\right) + \left(\frac{4}{10} \times \frac{9}{18}\right) = \frac{60}{180} + \frac{36}{180} = \frac{96}{180}$. Simplifying $\frac{96}{180}$ by dividing both the numerator and the denominator by 12 yields $\frac{8}{15}$. Given $\gcd(8, 15) = 1$, we identify $p=8$ and $q=15$, leading to $p+q = 23$.

The number of favorable outcomes for selecting two distinct numbers $a$ and $b$ such that $a \geq b + 10$ is calculated by summing the possible values of $a$ for each valid $b$. For a fixed $b \in \{1, 2, \dots, 90\}$, $a$ can range from $b + 10$ to $100$, providing $91 - b$ choices for each $b$. The total number of favorable outcomes is $\sum_{b=1}^{90} (91 - b) = \frac{90 \times 91}{2} = 4095$.

Dividing these favorable outcomes by the total sample space of $100 \times 99 = 9900$ gives a probability of $\frac{4095}{9900} = \frac{91}{220}$. Since $\gcd(91, 220) = 1$, we identify $m = 91$ and $n = 220$, resulting in $m + n = 311$.

The expectation of a random variable is defined as $E(X) = \sum x_i P(x_i)$. Substituting the given probability distribution into this formula yields:
$\frac{263}{15} = k \left( 4 \cdot \frac{2}{15} + \frac{30}{7} \cdot \frac{1}{15} + \frac{32}{7} \cdot \frac{2}{15} + \frac{34}{7} \cdot \frac{3}{15} + \frac{36}{7} \cdot \frac{1}{15} + \frac{38}{7} \cdot \frac{2}{15} + \frac{40}{7} \cdot \frac{3}{15} + 6 \cdot \frac{1}{15} \right)$
To solve for $k$, combine the terms using a common denominator of $105$:
$\frac{263}{15} = \frac{k}{105} \left( 8 \cdot 7 + 30 + 64 + 102 + 36 + 76 + 120 + 6 \cdot 7 \right)$
$\frac{263}{15} = \frac{k}{105} (56 + 30 + 64 + 102 + 36 + 76 + 120 + 42) = \frac{526k}{105}$
Equating the terms results in $k = \frac{263 \cdot 105}{15 \cdot 526} = \frac{7}{2} = 3.5$. With $k=3.5$, the values of the random variable $X$ are determined as follows:
$4(3.5)=14, \quad \frac{30}{7}(3.5)=15, \quad \frac{32}{7}(3.5)=16, \quad \frac{34}{7}(3.5)=17, \quad \frac{36}{7}(3.5)=18, \quad \frac{38}{7}(3.5)=19, \quad \frac{40}{7}(3.5)=20, \quad 6(3.5)=21$
The probability $P(X < 20)$ is the sum of the probabilities for the values $14, 15, 16, 17, 18,$ and $19$, which corresponds to the first six terms of the distribution:
$P(X < 20) = \frac{2}{15} + \frac{1}{15} + \frac{2}{15} + \frac{1}{5} + \frac{1}{15} + \frac{2}{15}$
Substituting $\frac{1}{5} = \frac{3}{15}$ into the sum gives:
$P(X < 20) = \frac{2 + 1 + 2 + 3 + 1 + 2}{15} = \frac{11}{15}$

The mean of the observations is $\frac{2+4+10+x+12+14+y}{7} = 8$, which simplifies to $x+y=14$. Using the variance formula $\sigma^2 = \frac{\sum X_i^2}{n} - \bar{X}^2 = 16$, we calculate $\frac{2^{2+}4^{2+}10^{2+}x^{2+}12^{2+}14^{2+}y^2}{7} - 8^2 = 16$, which leads to $\frac{460+x^{2+}y^2}{7} = 80$, and thus $x^{2+}y^2=100$. Solving the system $x+y=14$ and $x^{2+}y^2=100$ yields $x=8$ and $y=6$ because $x > y$.

The set is $S = \{1, 2, 3, x-4, y, 5\} = \{1, 2, 3, 4, 6, 5\}$. When choosing two numbers without replacement from these six values, the total number of possible combinations is $^6C_2 = 15$. The probability that the smaller number is less than 4 is the complement of the event where both selected numbers are greater than or equal to 4. Since the numbers in $S$ that are $\ge 4$ are $\{4, 5, 6\}$, there are $^3C_2 = 3$ combinations where the smaller number is $\ge 4$. Therefore, the probability of this complement is $\frac{3}{15} = \frac{1}{5}$, and the required probability is $1 - \frac{1}{5} = \frac{4}{5}$.

The problem requires identifying all combinations of heads and tails that yield a total of 30 points based on the score values provided. If $h$ represents the number of heads and $t$ represents the number of tails, the equation $10h + 5t = 30$ simplifies to $2h + t = 6$. Testing non-negative integer values reveals four distinct scenarios: 0 heads and 6 tails, 1 head and 4 tails, 2 heads and 2 tails, and 3 heads and 0 tails.

Each combination involves a fixed number of coin tosses, where the probability of a specific outcome is $(\frac{1}{2})^N$ for $N$ total tosses, multiplied by the number of possible arrangements. For 6 tails, the probability is $(\frac{1}{2})^6 = \frac{1}{64}$. For 1 head and 4 tails, there are 5 total tosses, with $\binom{5}{1} = 5$ arrangements, providing a probability of $5 \times (\frac{1}{2})^5 = \frac{10}{64}$. With 2 heads and 2 tails, 4 total tosses exist with $\binom{4}{2} = 6$ arrangements, yielding $6 \times (\frac{1}{2})^4 = \frac{24}{64}$. For 3 heads and 0 tails, there are 3 total tosses and only 1 arrangement, resulting in $1 \times (\frac{1}{2})^3 = \frac{8}{64}$.

Summing these mutually exclusive probabilities yields $\frac{1 + 10 + 24 + 8}{64} = \frac{43}{64}$. Since $m = 43$ and $n = 64$ are coprime, the requested sum $m + n$ equals 107.

For a quadratic expression $ax^{2}+2\sqrt{2} bx + c$ to remain strictly positive for all real values of $x$, the parabola must open upward and have no real roots, which implies $a > 0$ and a discriminant $D < 0$. Since the coefficients $a, b, c$ are selected from $\{1, 2, 3, 4\}$, the condition $a > 0$ is always satisfied. The discriminant condition $(2\sqrt{2}b)^{2} - 4ac < 0$ simplifies to $8b^{2} - 4ac < 0$, which reduces further to the inequality $2b^{2} < ac$.

With three variables each taking four possible values, there are $4 \times 4 \times 4 = 64$ total possible outcomes. Evaluating the condition $2b^{2} < ac$ for each valid integer $b$, we find that when $b=1$, the inequality $ac > 2$ is satisfied by 13 different pairs of $(a, c)$. When $b=2$, the inequality $ac > 8$ is satisfied by 4 specific pairs where the product of $a$ and $c$ is greater than 8. For $b=3$ and $b=4$, the condition requires $ac > 18$ and $ac > 32$ respectively, which is impossible given that the maximum product of $a$ and $c$ is 16. Summing these cases yields 17 favorable outcomes out of 64, resulting in a probability of $\frac{17}{64}$. Since $\gcd(17, 64)=1$, identifying $m=17$ and $n=64$ gives the final sum $m + n = 81$.

This problem is best solved using Bayes' Theorem, which allows us to determine the likelihood of a specific cause-selecting Bag $B_2$-given that we have already observed a specific effect, the drawing of a white ball. First, we identify that there is a $\frac{1}{3}$ probability of choosing any of the three bags. We then calculate the overall probability of drawing a white ball by summing the weighted probabilities from each bag: $\frac{1}{3} \times \frac{6}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}$, which simplifies to $\frac{1}{3} \times \frac{15}{10}$, resulting in an overall probability of $\frac{1}{2}$. Finally, to isolate the probability that the white ball originated from Bag $B_2$, we take the specific contribution of that bag, $\frac{1}{3} \times \frac{4}{10}$, and divide it by the total probability of $\frac{1}{2}$, yielding $\frac{4}{30} \times 2 = \frac{4}{15}$.

Bayes' Theorem provides the standard framework for inverse probability problems where an observed outcome, such as picking a ball of a specific color, requires us to determine the likelihood it originated from a particular source. Because each bag is chosen with an equal probability of $\frac{1}{3}$, we can determine the conditional probabilities by focusing on the relative proportions of each color across the three bags.

For a red ball, the total probability of selection is the weighted sum across all bags, calculated as $P(\text{Red}) = \frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10} = \frac{12}{30}$. The probability that this red ball originated from Bag I is the ratio of the chance of picking a red ball from Bag I to the total probability, resulting in $p = \frac{\frac{1}{3} \times \frac{3}{10}}{\frac{12}{30}} = \frac{3}{12} = \frac{1}{4}$.

Similarly, the total probability of picking a green ball is $P(\text{Green}) = \frac{1}{3} \times \frac{5}{10} + \frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10} = \frac{12}{30}$. Finding the probability that a green ball came from Bag III involves taking the ratio of the chance of drawing a green ball from Bag III against this total, which gives $q = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{12}{30}} = \frac{4}{12} = \frac{1}{3}$. Combining these findings yields the final value of $\frac{1}{p} + \frac{1}{q} = 4 + 3 = 7$.