Probability – JEE Main MathematicsPractice Questions & PYQs

This problem relies on the binomial distribution, which determines the probability of a specific number of independent events occurring within a fixed set. Since we are evaluating the probability of at most two machines being out of service out of a total of five, we sum the probabilities for exactly zero, one, and two failures, where the probability of failure is $p = \frac{1}{4}$ and the probability of remaining in service is $q = \frac{3}{4}$. The total probability is given by the expression:

$P(X \le 2) = \binom{5}{0} \left(\frac{3}{4}\right)^5 + \binom{5}{1} \left(\frac{1}{4}\right) \left(\frac{3}{4}\right)^4 + \binom{5}{2} \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^3$

By factoring out the common term $\left(\frac{3}{4}\right)^3$, we can simplify the values inside the sum. The expression transforms into $\left(\frac{3}{4}\right)^3 \left[ \left(\frac{3}{4}\right)^2 + 5 \left(\frac{1}{4}\right) \left(\frac{3}{4}\right) + 10 \left(\frac{1}{4}\right)^2 \right]$, which evaluates to $\left(\frac{3}{4}\right)^3 \left( \frac{9}{16} + \frac{15}{16} + \frac{10}{16} \right)$. Summing these fractions yields $\frac{34}{16}$, which simplifies to $\frac{17}{8}$. Comparing this directly to the required form $\left(\frac{3}{4}\right)^3 k$, it is clear that $k = \frac{17}{8}$.

The expected value of a random variable is the weighted average of all its potential outcomes, calculated by summing the products of each value and its corresponding probability. In this scenario, we evaluate the 32 equally likely sequences resulting from tossing a coin five times. The variable $X$ assumes the value $k$ when the longest streak of consecutive heads is exactly $k$ for $k = 3, 4, 5$, while all other outcomes assign $X$ a value of $-1$.

By examining the possible combinations, we find that the longest run of heads is 3 in 5 outcomes, 4 in 2 outcomes, and 5 in 1 outcome. The remaining 24 outcomes, where the longest sequence of heads is less than 3, result in a value of $-1$. Calculating the expectation involves multiplying these outcomes by their frequencies and summing the results:

$E[X] = \left( 3 \cdot \frac{5}{32} \right) + \left( 4 \cdot \frac{2}{32} \right) + \left( 5 \cdot \frac{1}{32} \right) + \left( -1 \cdot \frac{24}{32} \right)$

$E[X] = \frac{15 + 8 + 5 - 24}{32} = \frac{4}{32} = \frac{1}{8}$

Sampling with replacement ensures that each draw is an independent event, where the probability of selecting an A-card or a B-card is defined as $P(A) = P(B) = 1/2$. To determine the probability that the second A-card appears before the third B-card, we must identify all possible sequences ending in the second A-card that contain at most two B-cards.

If the second A-card occurs on the second draw, the only sequence is AA, which has a probability of $(1/2)^2 = 4/16$. If it occurs on the third draw, the sequences ABA and BAA are valid, each with a probability of $(1/2)^3 = 2/16$, contributing a combined $4/16$. When the second A-card appears on the fourth draw, the sequences ABBA, BABA, and BBAA are valid, each with a probability of $(1/2)^4 = 1/16$, contributing $3/16$ in total.

Summing these probabilities across all distinct valid scenarios, we combine $4/16 + 4/16 + 3/16$. This calculation results in a final probability of $11/16$.

Independence of two events $A$ and $B$ implies that the occurrence of $B$ or its complement $B'$ has no influence on the probability of $A$. By the definition of conditional probability for independent events, the expression $P(A|B')$ is calculated as follows:
$P(A|B') = \frac{P(A \cap B')}{P(B')}$
Since independence ensures $P(A \cap B') = P(A)P(B')$, the formula simplifies to:
$P(A|B') = \frac{P(A)P(B')}{P(B')} = P(A)$
Given $P(A) = \frac{1}{3}$, it follows directly that $P(A|B') = \frac{1}{3}$, which confirms the validity of the third option. Conversely, option A is incorrect because $P(A|B) = P(A) = \frac{1}{3}$, option B is incorrect because $P(A|(A \cup B)) = \frac{P(A)}{P(A \cup B)} = \frac{1/3}{1/3 + 1/6 - 1/18} = \frac{3}{4}$, and option D is incorrect because $P(A'|B') = P(A') = 1 - \frac{1}{3} = \frac{2}{3}$.

The Principle of Inclusion-Exclusion is the fundamental tool for relating the probability of the union of multiple events to their individual and intersection probabilities. To determine the intersection of events A and B, we apply the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, which yields $P(A \cap B) = 0.6 + 0.4 - 0.8 = 0.2$. We then extend this logic to the union of three events using $P(A \cup B \cup C) = P(A) + P(B) + P(C) - (P(A \cap B) + P(A \cap C) + P(B \cap C)) + P(A \cap B \cap C)$.

Substituting the provided values into this expression results in $\alpha = 0.6 + 0.4 + 0.5 - (0.2 + 0.3 + \beta) + 0.2$. Simplifying this equation gives $\alpha = 1.2 - \beta$. Because the problem constrains $\alpha$ within the interval $[0.85, 0.95]$, we set up the inequality $0.85 \leq 1.2 - \beta \leq 0.95$. Subtracting $1.2$ from each part produces $-0.35 \leq -\beta \leq -0.25$, and multiplying throughout by negative one reverses the inequality signs, demonstrating that $\beta$ must fall within the range $[0.25, 0.35]$.

Conditional probability requires narrowing the sample space to satisfy a specific condition, which in this case is that the sum of the two dice scores must be a multiple of 4. Listing all possible outcomes that result in a sum of 4, 8, or 12 provides the restricted set of (1, 3), (2, 2), (3, 1), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), and (6, 6), which consists of 9 distinct possibilities. Within this specific subset, we identify which outcomes contain the score 4 at least once. Checking these values reveals that only (4, 4) satisfies the requirement. The final probability is determined by the ratio of these favorable outcomes to the total number of outcomes in the restricted sample space:

$P = \frac{1}{9}$

Bayes' Theorem provides the standard framework for calculating the probability of a specific source event given an observed outcome. Because the box selection is random, the probability of selecting Box I is $P(B_1) = 1/2$ and the probability of selecting Box II is $P(B_2) = 1/2$.

Evaluating the contents of the boxes identifies the likelihood of drawing a non-prime card from each. Box I contains 30 cards, and subtracting the 10 prime numbers (2, 3, 5, 7, 11, 13, 17, 19, 23, 29) leaves 20 non-prime cards. Therefore, the probability of selecting a non-prime card from Box I is $P(E|B_1) = 20/30 = 2/3$. Box II contains 20 cards, and subtracting the 5 prime numbers (31, 37, 41, 43, 47) leaves 15 non-prime cards. Consequently, the probability of selecting a non-prime card from Box II is $P(E|B_2) = 15/20 = 3/4$.

The probability that the card originated from Box I is the ratio of the probability of picking a non-prime card from Box I to the total probability of picking a non-prime card from either box. This is expressed as:
$P(B_1|E) = \frac{P(E|B_1) \times P(B_1)}{P(E|B_1) \times P(B_1) + P(E|B_2) \times P(B_2)}$
Substituting the values calculated above, the equation becomes:
$P(B_1|E) = \frac{1/2 \times 2/3}{(1/2 \times 2/3) + (1/2 \times 3/4)} = \frac{2/3}{2/3 + 3/4} = \frac{8}{17}$

The probability that at least one of the two events occurs, denoted as $P(A \cup B)$, follows the addition rule $P(A) + P(B) - P(A \cap B) = \frac{1}{2}$. In contrast, the probability that exactly one of the two events occurs is calculated by removing the intersection twice from the sum of the individual probabilities, expressed as $P(A) + P(B) - 2P(A \cap B) = \frac{2}{5}$. By subtracting the equation representing the probability of exactly one event occurring from the equation for the union of the events, the individual terms for $P(A)$ and $P(B)$ cancel out, leaving $P(A \cap B) = \frac{1}{2} - \frac{2}{5}$. Converting these fractions to a common denominator produces $\frac{5}{10} - \frac{4}{10}$, which yields $0.10$.

The probability of the event $E_2^C \cap E_3^C$ occurring given $E_1$ is defined by the ratio of the joint probability of all three components to the probability of the condition $E_1$, which is written as $P\left(\frac{E_{2}^{C} \cap E_{3}^{C}}{E_{1}}\right) = \frac{P(E_1 \cap E_2^C \cap E_3^C)}{P(E_1)}$. The numerator represents the portion of $E_1$ that does not overlap with $E_2$ or $E_3$, which can be calculated by subtracting the union $E_2 \cup E_3$ from the total probability of $E_1$, expressed as $P(E_1) - P(E_1 \cap (E_2 \cup E_3))$. By applying the inclusion-exclusion principle, this intersection expands to $P(E_1 \cap E_2) + P(E_1 \cap E_3) - P(E_1 \cap E_2 \cap E_3)$.

Because the events are pairwise independent, we can replace the pairwise intersections with the products $P(E_1)P(E_2)$ and $P(E_1)P(E_3)$, and substitute the given value $P(E_1 \cap E_2 \cap E_3) = 0$. This leads to the following expression:

$\frac{P(E_1) - [P(E_1)P(E_2) + P(E_1)P(E_3) - 0]}{P(E_1)}$

Dividing the entire expression by $P(E_1)$ simplifies the result to $1 - P(E_2) - P(E_3)$. Rearranging the terms into $(1 - P(E_3)) - P(E_2)$ allows for the substitution of the complement, identifying the final value as $P(E_3^C) - P(E_2)$.

This scenario follows a binomial distribution where each bomb drop acts as an independent trial with a $0.5$ probability of hitting the target. Since destruction requires at least two hits, the probability of the target remaining intact involves the sum of probabilities for failing scenarios, specifically having zero hits or exactly one hit. The target is destroyed with at least $99\%$ certainty if the complement of these failing scenarios meets or exceeds $0.99$.

Expressing this probability as $1 - [P(0 \text{ hits}) + P(1 \text{ hit})] \ge 0.99$, we substitute the binomial probabilities to obtain $1 - \left( \binom{n}{0} (0.5)^n + \binom{n}{1} (0.5)^n \right) \ge 0.99$, which simplifies to $1 - \frac{1 + n}{2^n} \ge 0.99$. Rearranging the terms leads to $\frac{n+1}{2^n} \le \frac{1}{100}$, or equivalently, $2^n \ge 100(n+1)$. Testing integer values for $n$, the inequality $2^{10} \ge 100(11)$ results in $1024 \ge 1100$, which is false. However, for $n = 11$, the inequality $2^{11} \ge 100(12)$ produces $2048 \ge 1200$, which confirms the minimum number of bombs required is 11.

The probability depends on identifying all possible 5-digit numbers formed by exactly two distinct digits and dividing that count by the total number of 5-digit numbers, which is $9 \times 10^4$. Because the leading digit cannot be zero, it is necessary to separate the calculation into two distinct scenarios based on whether zero is used as one of the two digits. First, consider cases where neither digit is zero. We select two digits from the set of nine non-zero possibilities in ${}^{9}C_{2} = 36$ ways, and for each pair, we form $2^5 - 2 = 30$ valid 5-digit numbers that contain both chosen digits, resulting in $36 \times 30 = 1080$ total possibilities.

Second, consider cases where one of the two digits is zero. We choose one non-zero digit from the nine available options in ${}^{9}C_{1} = 9$ ways. Since the first position must be the non-zero digit and the remaining four positions can be filled by either zero or that non-zero digit, there are $2^4 = 16$ possible strings. Excluding the single case where the non-zero digit is used exclusively, we obtain $16 - 1 = 15$ valid numbers for each selection, giving $9 \times 15 = 135$ additional possibilities. Summing these favorable outcomes yields $1080 + 135 = 1215$, and dividing by the total count of $9 \times 10^4$ provides the final probability:

$\frac{1215}{9 \times 10^4} = \frac{135}{10^4}$

Calculating the probability of hitting a target at least once is best approached by considering the complement: missing the target entirely in every attempt. Given the probability of a single miss is $9/10$, the probability of missing $n$ times in a row is $(\frac{9}{10})^n$, making the probability of at least one hit $1 - (\frac{9}{10})^n$. Setting this probability to be greater than $1/4$ leads to the inequality $1 - (\frac{9}{10})^n > \frac{1}{4}$, which simplifies to $(\frac{9}{10})^n < 0.75$. Testing small integers shows that for $n=1$ the result is $0.9$ and for $n=2$ it is $0.81$, both exceeding the threshold, while for $n=3$ the result is $0.729$, which successfully satisfies the condition.

Every probability distribution must satisfy the fundamental rule that the sum of all individual probabilities is exactly equal to one. Applying this to the given distribution, we add the probabilities $K^2 + 2K + K + 2K + 5K^2$ and set the total to $1$, which simplifies to the quadratic equation $6K^2 + 5K - 1 = 0$. Factoring this leads to $(6K - 1)(K + 1) = 0$, giving potential solutions of $K = 1/6$ and $K = -1$; since probability cannot be negative, we reject the negative value and use $K = 1/6$.

The probability $P(X > 2)$ covers the cases where $X$ takes the values $3$, $4$, and $5$, requiring us to sum the corresponding expressions $K + 2K + 5K^2$. This total simplifies to $3K + 5K^2$, and substituting $K = 1/6$ yields $3(1/6) + 5(1/6)^2$. Performing the arithmetic results in $1/2 + 5/36$, which combines to $18/36 + 5/36$, totaling $23/36$.

Distributing 10 different balls into 4 distinct boxes results in $4^{10}$ possible arrangements since each ball has 4 independent placement choices. The probability that two boxes contain exactly 2 and 3 balls depends on selecting the specific boxes and assigning the balls using combinations like $\binom{10}{2}$ and $\binom{8}{3}$ for the designated containers. The remaining balls are then allocated across the other boxes, and the total number of favorable outcomes is compared against the entire sample space. This approach yields a final probability of $\frac{945}{2^{10}}$.

The total number of ways to select three vertices from the six available on a regular hexagon is determined by the combination formula $^{6}C_{3}$, which yields 20 unique groups. Within these combinations, an equilateral triangle is formed only when the vertices are chosen such that they are separated by an equal distance along the perimeter, specifically the sets $(A_1, A_3, A_5)$ and $(A_2, A_4, A_6)$. Since there are only 2 such favorable triangles out of the 20 total possibilities, the probability is calculated as $\frac{2}{20}$, which simplifies to $\frac{1}{10}$.

Expected value represents the long-term average outcome of a random event, determined by calculating the sum of every potential result multiplied by its respective probability of occurring. When rolling two dice, there are 36 equally likely outcomes in total. A doublet consists of 6 specific pairs-(1,1), (2,2), (3,3), (4,4), (5,5), and (6,6)-which corresponds to a probability of $6/36$ for gaining Rs. 15. Getting a sum of 9 happens with the pairs (3,6), (4,5), (5,4), and (6,3), providing 4 favorable outcomes or a probability of $4/36$ for winning Rs. 12. Every other combination, totaling $36 - 6 - 4 = 26$ outcomes, results in a loss of Rs. 6.

Calculating the expectation requires summing these individual weighted values using the formula $E = \frac{(15 \times 6) + (12 \times 4) + (-6 \times 26)}{36}$. This simplifies to $\frac{90 + 48 - 156}{36}$, which results in $\frac{-18}{36}$ or $-\frac{1}{2}$. The negative sign indicates that the person faces an expected loss of $\frac{1}{2}$ rupee.

The total probability is determined by partitioning the experiment into two equally likely branches based on the coin flip, each weighted by a probability of $\frac{1}{2}$. For the first branch where the coin lands on heads, rolling a pair of dice results in a sum of 7 or 8 with a probability of $\frac{11}{36}$. In the second branch where the coin lands on tails, choosing a card numbered 7 or 8 from a pack of nine yields a probability of $\frac{2}{9}$. Combining these branches gives the final probability as $\frac{1}{2} \times \frac{11}{36} + \frac{1}{2} \times \frac{2}{9}$, which simplifies to $\frac{11}{72} + \frac{8}{72}$ and results in $\frac{19}{72}$.

The most straightforward approach to determining the probability of getting at least one head is to subtract the probability of the complementary event-getting only tails-from the total probability of one. For $n$ independent tosses of a fair coin, the probability of not getting a single head is $(\frac{1}{2})^n$, which leads to the inequality $1 - (\frac{1}{2})^n > 0.99$. This expression simplifies to $(\frac{1}{2})^n < 0.01$, or equivalently, $2^n > 100$. Calculating the powers of two, we find that $2^6 = 64$, which is not enough to exceed 100, whereas $2^7 = 128$ satisfies the condition, confirming that seven is the minimum number of tosses required.

The fundamental relationship $A \subset B$ ensures that the intersection $A \cap B$ is identical to $A$ itself, as all outcomes in $A$ are necessarily contained within $B$. By substituting this identity into the definition of conditional probability, we find that $P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)}$. Since the probability $P(B)$ represents a value no greater than 1, dividing $P(A)$ by this value results in a quantity that is at least as large as $P(A)$ itself, establishing the inequality $P(A \mid B) \geq P(A)$.

For the sum of two integers to be even, both selected numbers must share the same parity, which means the pair must consist of either two even numbers or two odd numbers. Within the given set $\{1, 2, \dots, 11\}$, there are 5 even numbers and 6 odd numbers available. The total number of favorable outcomes for an even sum is the sum of picking two even numbers, $\binom{5}{2}$, and picking two odd numbers, $\binom{6}{2}$. Therefore, the conditional probability that both numbers are even is given by the ratio $\frac{\binom{5}{2}}{\binom{5}{2}+\binom{6}{2}}$, which simplifies to $\frac{10}{10+15}$ or $\frac{2}{5}$.