Probability – JEE Main MathematicsPractice Questions & PYQs

The probability of the second plane hitting the target depends on an alternating sequence where the first plane attempts to bomb the target, and if it fails, the second plane takes its turn. In the initial round, the second plane scores a hit only if the first plane misses with a probability of 0.7 and the second succeeds with 0.2, resulting in a probability of $(0.7)(0.2) = 0.14$. If both planes miss, they repeat the attempt, establishing a common ratio for subsequent rounds equal to the probability that both planes miss, which is $(0.7)(0.8) = 0.56$. This situation forms an infinite geometric series:

$P = 0.14(1 + 0.56 + 0.56^2 + \dots) = \frac{0.14}{1-0.56} = \frac{7}{22} \approx 0.32$

Summing these probabilities accounts for all cycles in which the second plane ultimately succeeds, yielding a final result of 0.32.

This problem is a classic application of binomial probability, which models the likelihood of achieving a specific number of successes over a set sequence of independent trials. First, we identify the probability of rolling a sum of 9 with two fair dice, where the successful combinations are (3, 6), (4, 5), (5, 4), and (6, 3) out of 36 total possibilities, yielding a probability of success $p = \frac{1}{9}$. Consequently, the probability of failing to roll a 9 is $q = \frac{8}{9}$.

Since we are looking for exactly two successes in three trials, we apply the binomial distribution formula, where $n=3$ and $k=2$:

$P(X=2) = {^3C_2} \times \left(\frac{1}{9}\right)^2 \times \left(\frac{8}{9}\right)$

Calculating this, we have $3 \times \frac{1}{81} \times \frac{8}{9}$, which simplifies to $\frac{24}{729}$ or $\frac{8}{243}$.

The probability of events occurring in a fixed interval follows the Poisson distribution, defined by the mean rate of 5 calls per 10 minutes. To find the chance of receiving at most one call, we must sum the probabilities of receiving exactly zero calls and exactly one call, expressed as $P(X=0) + P(X=1)$. Using the Poisson formula $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$ where $\lambda = 5$, the probability for zero calls becomes $\frac{e^{-5} 5^0}{0!}$, which simplifies to $e^{-5}$, and the probability for one call becomes $\frac{e^{-5} 5^1}{1!}$, which is $5 e^{-5}$. Adding these components results in $e^{-5} + 5 e^{-5}$, which equates to $\frac{6}{e^5}$.

A Poisson distribution models the occurrence of discrete events, meaning the random variable $X$ only takes non-negative integer values. To find $P(X > 1.5)$, we are looking for the total probability associated with $X$ taking any integer value starting from 2, which is $P(X=2) + P(X=3) + \dots$. Since this sum extends to infinity, it is computationally simpler to use the complement rule, recognizing that the sum of all possible probabilities is 1 and subtracting the cases where $X \le 1.5$, namely $P(X=0)$ and $P(X=1)$.

Using the probability mass function $P(X=r) = \frac{e^{-\lambda} \lambda^r}{r!}$ with a mean of $\lambda = 2$, we evaluate the individual terms for the complement. Calculating these values gives $P(X=0) = \frac{e^{-2} 2^0}{0!} = e^{-2}$ and $P(X=1) = \frac{e^{-2} 2^1}{1!} = 2e^{-2}$. Combining these yields $3e^{-2}$, so the final probability $P(X > 1.5)$ is found by subtracting this sum from 1, resulting in $1 - \frac{3}{e^2}$.

Two events are independent if the occurrence of one does not affect the probability of the other, verified by the condition $P(A \cap B) = P(A)P(B)$, while they are considered equally likely only if their individual probabilities are identical. First, the probability of the union is $P(A \cup B) = 1 - 1/6 = 5/6$ and the probability of $A$ is $P(A) = 1 - 1/4 = 3/4$. Substituting these into the addition rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, we get $5/6 = 3/4 + P(B) - 1/4$, which simplifies to $P(B) = 1/3$. Since $3/4 \neq 1/3$, the events are not equally likely, but their product $(3/4) \times (1/3) = 1/4$ matches the given probability $P(A \cap B) = 1/4$, confirming that the events are independent.

Probability is calculated as the ratio of favorable outcomes to the total number of possible outcomes. Since each of the three individuals independently chooses one of three houses, the total number of possible combinations is $3 \times 3 \times 3 = 27$.

Favorable outcomes occur only when all three people apply for the same house, which happens in exactly 3 ways: they all apply for house 1, all for house 2, or all for house 3. The probability is therefore $\frac{3}{27}$, which simplifies to $\frac{1}{9}$.

A binomial distribution is characterized by its mean $np$ and variance $npq$, which provides a straightforward path to identifying the underlying parameters $n$, $p$, and $q$. By dividing the variance by the mean, specifically $\frac{npq}{np} = \frac{2}{4}$, we determine that $q = \frac{1}{2}$, and consequently $p = \frac{1}{2}$. Using the mean formula $np=4$, we find that $n = 8$ since $8 \times \frac{1}{2} = 4$.

To determine the probability of exactly 2 successes, apply the binomial formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$ with $n=8$ and $k=2$. This leads to the expression:
$P(X=2) = \binom{8}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^6$
The binomial coefficient $\binom{8}{2}$ evaluates to 28, and multiplying this by the exponentiated probabilities yields $28 \times (\frac{1}{2})^8$, which results in $\frac{28}{256}$.

Two individuals contradict each other when their accounts differ, which occurs if one person tells the truth and the other provides a false statement. Since the probability of $A$ telling the truth is $\frac{4}{5}$, the chance of $A$ lying is $1 - \frac{4}{5} = \frac{1}{5}$, and with $B$ telling the truth with probability $\frac{3}{4}$, the likelihood of $B$ lying is $1 - \frac{3}{4} = \frac{1}{4}$.

A contradiction manifests in two distinct, mutually exclusive scenarios: $A$ tells the truth while $B$ lies, or $A$ lies while $B$ tells the truth. The probability for the first scenario is $\frac{4}{5} \times \frac{1}{4} = \frac{4}{20}$, and the probability for the second is $\frac{1}{5} \times \frac{3}{4} = \frac{3}{20}$. By summing these separate possibilities, the total probability that they contradict each other is $\frac{4}{20} + \frac{3}{20} = \frac{7}{20}$.

In a binomial distribution, the mean is defined by the product $np$ and the variance is defined by $npq$, where $n$ is the number of trials, $p$ is the probability of success, and $q$ is the probability of failure. Dividing the variance by the mean allows us to isolate $q$ as $q = \frac{npq}{np} = \frac{2}{4} = \frac{1}{2}$, which establishes that the probability of success $p = 1 - q = \frac{1}{2}$. By substituting this value into the mean formula $np = 4$, we find that $n \times \frac{1}{2} = 4$, meaning $n = 8$. We then apply the binomial probability formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$ for $k=1$ to calculate $P(X=1) = \binom{8}{1} (\frac{1}{2})^1 (\frac{1}{2})^7$, which simplifies to $8 \times (\frac{1}{2})^8$ or $\frac{8}{256}$, resulting in $\frac{1}{32}$.

Calculating the probability of picking the winning horse is most efficient by finding the complement-the probability of failing to select the winner at all-and subtracting it from the total probability of one. With 4 non-winning horses out of 5 total, the chance of the first selection missing the winner is $\frac{4}{5}$, and if that fails, there are 3 non-winning horses remaining out of 4, leaving a $\frac{3}{4}$ probability of missing again. The combined probability of excluding the winner is $\frac{4}{5} \times \frac{3}{4} = \frac{3}{5}$. Subtracting this from the whole results in a probability of $1 - \frac{3}{5} = \frac{2}{5}$ for selecting the winning horse.

Probability axioms dictate that for any event, the likelihood must fall between 0 and 1. Additionally, when events are mutually exclusive, the probability of their union is defined as the sum of their individual probabilities, which must also not exceed 1.

Evaluating the restrictions on the variable $x$ requires each individual probability to remain within the valid range. For event A, $0 \leq \frac{3x+1}{3} \leq 1$ simplifies to $-\frac{1}{3} \leq x \leq \frac{2}{3}$. For event B, $0 \leq \frac{1-x}{4} \leq 1$ leads to $-3 \leq x \leq 1$, and for event C, $0 \leq \frac{1-2x}{2} \leq 1$ results in $-\frac{1}{2} \leq x \leq \frac{1}{2}$.

Furthermore, because these events are mutually exclusive, their combined probability must satisfy $P(A) + P(B) + P(C) \leq 1$. Calculating the sum yields:

$\frac{3x+1}{3} + \frac{1-x}{4} + \frac{1-2x}{2} \leq 1$

Multiplying by the common denominator 12, the inequality becomes $\frac{13-3x}{12} \leq 1$, which implies $13-3x \leq 12$, or $x \geq \frac{1}{3}$. Also, since the total probability must be non-negative, we have $13-3x \geq 0$, or $x \leq \frac{13}{3}$.

Identifying the intersection of all these specific conditions reveals the definitive range for $x$, where the lower limit is determined by the maximum of all lower bounds and the upper limit is determined by the minimum of all upper bounds. This calculation yields a range where the lower bound is $\frac{1}{3}$ and the upper bound is $\frac{1}{2}$, placing $x$ within the interval $[\frac{1}{3}, \frac{1}{2}]$.

This scenario describes a binomial distribution, where each die toss serves as an independent trial with a success probability of $p = \frac{3}{6} = \frac{1}{2}$ and a failure rate of $q = 1 - \frac{1}{2} = \frac{1}{2}$. With a total of $n = 5$ trials, the variance of this distribution is determined by the formula $\sigma^2 = npq$. Substituting the relevant values into this expression results in a calculation of $5 \times \frac{1}{2} \times \frac{1}{2}$, giving a total variance of $\frac{5}{4}$.

When calculating the probability that at least one event occurs among several independent trials, it is mathematically simpler to subtract the probability of all trials failing from the total certainty of one. The individual failure probabilities for students $A$, $B$, and $C$ are calculated as $1 - 1/2 = 1/2$, $1 - 1/3 = 2/3$, and $1 - 1/4 = 3/4$, respectively. Because the students work independently, the probability that none of them solve the problem is the product of these individual failure rates, computed as $1/2 \times 2/3 \times 3/4 = 1/4$. Subtracting this combined failure probability from the total certainty results in the probability that the problem is solved, which is $1 - 1/4 = 3/4$.

The probability of $B$ occurring outside of $A$, written as $P(\overline{A} \cap B)$, is found by taking the entire probability of $B$ and removing the portion it shares with $A$. First, find the probability of event $A$ by considering that $P(A) = 1 - P(\overline{A})$, which results in $1 - \frac{2}{3} = \frac{1}{3}$. Using the addition rule for probabilities, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, substitute the known values to get $\frac{3}{4} = \frac{1}{3} + P(B) - \frac{1}{4}$. Rearranging this equation reveals that $P(B) = \frac{3}{4} + \frac{1}{4} - \frac{1}{3} = 1 - \frac{1}{3} = \frac{2}{3}$. Subtracting the intersection from this total probability yields the final result, $P(\overline{A} \cap B) = \frac{2}{3} - \frac{1}{4} = \frac{5}{12}$.

For three numbers to form a geometric progression, the square of the middle term must equal the product of the first and third terms. Given the sequence $N-2$, $\sqrt{3N}$, $N+2$, the condition for the progression is that the square of $\sqrt{3N}$ equals the product of $N-2$ and $N+2$. This relationship is defined by:
$(\sqrt{3N})^2 = (N-2)(N+2)$
Simplifying this expression leads to $3N = N^2 - 4$, which is equivalent to the quadratic equation $N^2 - 3N - 4 = 0$. Factoring the quadratic yields $(N-4)(N+1) = 0$, resulting in potential values of $N = 4$ or $N = -1$. Because $N$ represents the sum of two fair dice, it must be a positive integer, so we exclude the negative value and proceed with $N = 4$.

To determine the probability, we count the number of outcomes when two dice are rolled that result in a sum of 4. The possible pairs are (1, 3), (2, 2), and (3, 1), which provide 3 favorable outcomes. With 36 total possible outcomes, the probability is 3/36, which simplifies to 1/12. Since the probability is expressed as k/48, we multiply the numerator and denominator by 4 to obtain 4/48, confirming that $k = 4$.