Probability – JEE Main MathematicsPractice Questions & PYQs

Complementary probability simplifies this task by identifying the undesirable outcomes and removing them from the total sample space. Since each coin flip has two independent possibilities, eight tosses result in $2^8 = 256$ total outcomes. The only scenarios that fail the requirement are obtaining all heads or obtaining all tails, and each of these two restricted cases occurs in exactly one way. The sum of the probabilities for these unwanted outcomes is $\frac{1}{256} + \frac{1}{256} = \frac{1}{128}$. Subtracting this sum from the total probability of one leaves $1 - \frac{1}{128} = \frac{127}{128}$.

For independent events $E$ and $F$, the joint probability is $P(E \cap F) = P(E)P(F) = \frac{1}{12}$ and the probability that neither occurs is $P(E' \cap F') = (1 - P(E))(1 - P(F)) = \frac{1}{2}$. Letting $x = P(E)$ and $y = P(F)$, the second equation expands to $1 - (x + y) + xy = \frac{1}{2}$. Substituting $xy = \frac{1}{12}$ into this expression results in $1 - (x + y) + \frac{1}{12} = \frac{1}{2}$, which simplifies to $x + y = \frac{7}{12}$. These values $x$ and $y$ are the roots of the quadratic equation $t^2 - (\frac{7}{12})t + \frac{1}{12} = 0$, which can be written as $12t^2 - 7t + 1 = 0$. Factoring the quadratic gives $(4t - 1)(3t - 1) = 0$, yielding roots $t = \frac{1}{4}$ and $t = \frac{1}{3}$. Consequently, the ratio $\frac{P(E)}{P(F)}$ can be $\frac{1/3}{1/4} = \frac{4}{3}$ or $\frac{1/4}{1/3} = \frac{3}{4}$, with $\frac{4}{3}$ matching the available option.

Determining the probability that at least one event occurs among $A$, $B$, and $C$ relies on the inclusion-exclusion principle, expressed as $P(A \cup B \cup C) = \sum P(A) - \sum P(A \cap B) + P(A \cap B \cap C)$. The condition describing exactly one event occurring between two variables, such as $A$ and $B$, is represented by $P(A) + P(B) - 2P(A \cap B) = \frac{1}{4}$. For the three pairs provided, we have the following relations:

$P(A) + P(B) - 2P(A \cap B) = \frac{1}{4}$
$P(B) + P(C) - 2P(B \cap C) = \frac{1}{4}$
$P(C) + P(A) - 2P(C \cap A) = \frac{1}{4}$

Summing these equations results in $2 \left( \sum P(A) \right) - 2 \left( \sum P(A \cap B) \right) = \frac{3}{4}$. Dividing both sides by 2 reveals that the difference between the sum of individual probabilities and the sum of pairwise intersections is $\sum P(A) - \sum P(A \cap B) = \frac{3}{8}$. By substituting this value and the given probability of the triple intersection, $P(A \cap B \cap C) = \frac{1}{16}$, into the inclusion-exclusion formula, the calculation becomes $\frac{3}{8} + \frac{1}{16}$. Adding these fractions provides the final probability of $\frac{7}{16}$.

The probability that the target is hit by P or Q but not by R relies on the independent nature of the shooters and the requirement that R fails to hit, which occurs with a probability of $1 - \frac{5}{8} = \frac{3}{8}$. There are three mutually exclusive scenarios that satisfy this condition: P hits while Q and R miss, Q hits while P and R miss, or both P and Q hit while R misses. Calculating these scenarios individually leads to $(\frac{3}{4} \times \frac{1}{2} \times \frac{3}{8}) + (\frac{1}{4} \times \frac{1}{2} \times \frac{3}{8}) + (\frac{3}{4} \times \frac{1}{2} \times \frac{3}{8}) = \frac{9}{64} + \frac{3}{64} + \frac{9}{64}$, which sums to $\frac{21}{64}$.

The probability of selecting committees with more women than men given that each committee must contain at least one woman is the ratio of favorable outcomes to the total number of possible committees containing at least one woman. The total number of ways to form a committee of 4 from 15 people is $\binom{15}{4} = 1365$, and the number of committees with no women is $\binom{10}{4} = 210$, resulting in $1365 - 210 = 1155$ valid committees. Favorable outcomes occur when the committee has 3 women and 1 man or 4 women and 0 men, which is calculated as $\binom{5}{3} \binom{10}{1} + \binom{5}{4} \binom{10}{0} = 10 \times 10 + 5 \times 1 = 105$. The probability is thus $\frac{105}{1155} = \frac{1}{11}$.

For any events $X$ and $Y$, the conditional probability $P(X|Y)$ is defined as the quotient $\frac{P(X \cap Y)}{P(Y)}$. Applying this definition to the expression $P(A | (A' \cup B'))$, we identify the ratio as $\frac{P(A \cap (A' \cup B'))}{P(A' \cup B')}$. Through the distributive law of sets, the numerator $A \cap (A' \cup B')$ expands to $(A \cap A') \cup (A \cap B')$, and because the intersection of $A$ and its complement is empty, this simplifies directly to $P(A \cap B')$. The value of $P(A \cap B')$ is determined by subtracting the intersection from the primary event, expressed as $P(A) - P(A \cap B)$, which results in $\frac{2}{5} - \frac{3}{20}$, or $\frac{5}{20}$.

Regarding the denominator, De Morgan's Law establishes that $A' \cup B'$ is equivalent to $(A \cap B)'$, meaning $P(A' \cup B') = 1 - P(A \cap B)$. Using the given probability of $\frac{3}{20}$, the denominator becomes $1 - \frac{3}{20}$, which equals $\frac{17}{20}$. Dividing the simplified numerator by this denominator yields $\frac{5/20}{17/20}$, resulting in the final value of $\frac{5}{17}$.

Since the probability of success is twice that of failure, we assign the probability of success as $p = \frac{2}{3}$ and failure as $q = \frac{1}{3}$, ensuring their sum equals 1. We apply the binomial probability formula, $P(X=k) = \binom{n}{k} p^k q^{n-k}$, across 6 trials to find the cumulative probability of at least 5 successes, which includes the cases for 5 and 6 successes. For exactly 5 successes, the calculation is $\binom{6}{5} (\frac{2}{3})^5 (\frac{1}{3})^1 = 6 \cdot \frac{32}{243} \cdot \frac{1}{3} = \frac{192}{729}$, while for 6 successes it is $\binom{6}{6} (\frac{2}{3})^6 = 1 \cdot \frac{64}{729} = \frac{64}{729}$. Adding these two outcomes yields:

$\frac{192}{729} + \frac{64}{729} = \frac{256}{729}$

Two events are statistically independent if the occurrence of one does not affect the probability of the other, defined by the condition $P(E_i \cap E_j) = P(E_i) \cdot P(E_j)$. When dealing with three events, they are mutually independent only if they are pairwise independent and also satisfy the condition $P(E_1 \cap E_2 \cap E_3) = P(E_1) \cdot P(E_2) \cdot P(E_3)$.

For the given dice rolls, the individual probabilities are $P(E_1) = \frac{1}{6}$, $P(E_2) = \frac{1}{6}$, and $P(E_3) = \frac{1}{2}$. Calculating the pairwise intersections yields $P(E_1 \cap E_2) = \frac{1}{36}$, $P(E_1 \cap E_3) = \frac{1}{12}$, and $P(E_2 \cap E_3) = \frac{1}{12}$. Since each of these results matches the product of the probabilities of the individual events within each pair, all pairs of events are independent.

Finally, mutual independence requires the intersection of all three events to match the product of their individual probabilities. Because obtaining 4 on die A and 2 on die B results in a total sum of 6, which is an even number, it is impossible for $E_1$, $E_2$, and $E_3$ to occur together. Therefore, the triple intersection probability is zero:

$P(E_1 \cap E_2 \cap E_3) = 0 \neq \frac{1}{6} \times \frac{1}{6} \times \frac{1}{2} = \frac{1}{72}$

This confirms that the three events are not mutually independent.

The total number of ways to distribute 12 balls into 3 boxes, assuming each ball has 3 independent options, is given by $3^{12}$. To find the probability that at least one box contains exactly 3 balls, we identify the favorable outcomes by selecting one box to contain exactly 3 balls and distributing the remaining balls, while subtracting the overcounted cases where multiple boxes contain 3 balls to prevent double-counting. This counting process leads to the probability expression $\frac{ \binom{3}{1} \binom{12}{3} 2^{9} - \binom{3}{2} \binom{12}{2} \binom{9}{3} }{ 3^{12} }$. Comparing this result to the provided choices reveals that none of the listed options represents the correct probability.

Two events are classified as independent if the probability of their intersection equals the product of their individual probabilities, whereas they are equally likely only if their individual probabilities are identical. Determining the relationship between these events requires finding the probability of event B first. Since the probability of the complement of A is $1/4$, the probability of event A is $P(A) = 3/4$. Similarly, knowing that the probability of the complement of the union of A and B is $1/6$ establishes that the union probability is $P(A \cup B) = 5/6$.

Using the addition rule, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ we substitute the known values into the equation to get $5/6 = 3/4 + P(B) - 1/4$. This simplifies to $5/6 = 1/2 + P(B)$, which leads to $P(B) = 5/6 - 1/2$, resulting \in $P(B) = 1/3$.

Comparing the product of the individual probabilities, $P(A) \cdot P(B) = 3/4 \cdot 1/3 = 1/4$, with the given intersection $P(A \cap B) = 1/4$, confirms the events are independent. Because the probability of event A, $3/4$, is distinct from the probability of event B, $1/3$, the two events are not equally likely.

Since guessing on each question is an independent event with a success probability of $\frac{1}{3}$ and a failure probability of $\frac{2}{3}$, we use the binomial distribution to model the outcomes. The probability of getting four or more correct answers is calculated by summing the likelihood of getting exactly four correct answers and the likelihood of getting exactly five correct answers. For the first case, we use the expression $\binom{5}{4}(\frac{1}{3})^4(\frac{2}{3})^1$, which simplifies to $\frac{10}{3^5}$, and for the second case, we use $\binom{5}{5}(\frac{1}{3})^5$, which equals $\frac{1}{3^5}$. Adding these two values together yields a total probability of $\frac{11}{3^5}$.

Conditional probability allows us to narrow our focus solely on the cases where the maximum number is 6, effectively treating this condition as the new sample space. To pick three numbers such that 6 is the maximum, we must select two additional numbers from the available set $\{1, 2, 3, 4, 5\}$, which results in ${}^5C_2$ distinct combinations. This gives a total of 10 possible outcomes where the maximum is 6.

Within these outcomes, we require the minimum to be 3. Since we have already fixed 6 as the maximum and 3 as the minimum, we only need to choose one more number from the values strictly between 3 and 6, which are $\{4, 5\}$. Because there are 2 such numbers to pick from, there are ${}^2C_1$ ways to satisfy the requirement. The probability is therefore the ratio of these favorable outcomes to the total outcomes satisfying the condition, calculated as $\frac{2}{10}$ or $\frac{1}{5}$.

When an event $C$ is entirely contained within an event $D$, the conditional probability $P \left( \frac{C}{D} \right)$ is calculated as the ratio $\frac{P(C \cap D)}{P(D)}$. Because $C \subset D$, the intersection of the two events is simply $C$ itself, allowing us to rewrite the expression as $\frac{P(C)}{P(D)}$. Since the probability of any event $P(D)$ is always less than or equal to 1, dividing $P(C)$ by $P(D)$ yields a value that must be greater than or equal to $P(C)$, which confirms that $P \left( \frac{C}{D} \right) \geq P(C)$.

Calculating the probability of at least one failure is easiest by subtracting the probability of experiencing no failures from the total probability of 1. With 5 independent trials, each having a success probability $p$, the probability of all trials resulting in success is $p^5$, making the probability of at least one failure $1 - p^5$. Given that this probability is at least $\frac{31}{32}$, we set up the inequality $1 - p^5 \geq \frac{31}{32}$. Subtracting this fraction from 1 reveals $p^5 \leq \frac{1}{32}$, which simplifies to $p \leq \frac{1}{2}$ upon taking the fifth root. Considering that any probability must remain within the range of 0 to 1, the value of $p$ is constrained to the interval $[0, \frac{1}{2}]$.

The total number of ways to choose four distinct integers from the set $\{1, 2, \dots, 20\}$ is calculated using the combination formula ${}^{20}C_4$, which equals 4845. An arithmetic progression of four terms is defined by the sequence $a, a+d, a+2d, a+3d$, and for these numbers to exist within the set, the inequality $a+3d \le 20$ must be satisfied.

By systematically counting the possible sets for each common difference $d$, we find that for $d=1$, the first term $a$ can range from 1 to 17, yielding 17 sets. Applying the same logic, $d=2$ provides 14 sets, $d=3$ provides 11 sets, $d=4$ provides 8 sets, $d=5$ provides 5 sets, and $d=6$ provides 2 sets. Summing these counts, there are $17 + 14 + 11 + 8 + 5 + 2 = 57$ total sets of four numbers that form an arithmetic progression. Consequently, the probability of selecting such a set at random is $\frac{57}{4845}$, which simplifies to $\frac{1}{85}$, validating the first statement.

Regarding the second statement, the set of possible common differences is not restricted to values between $\pm 1$ and $\pm 5$. Since an arithmetic progression can also have a common difference of 6, such as the set $\{1, 7, 13, 19\}$, the claim that the values are limited to the range $\pm 1$ to $\pm 5$ is false.

Calculating the probability of an event involves dividing the number of favorable outcomes by the total number of equally likely outcomes in the sample space. To determine the total number of ways to draw three balls from the nine available, we calculate the combination $\binom{9}{3}$, which yields 84 total possible groups. For the specific case where all three balls are of different colors, we must select one red ball from three, one blue ball from four, and one green ball from two. This corresponds to the calculation $\binom{3}{1} \times \binom{4}{1} \times \binom{2}{1}$, resulting in 24 favorable outcomes. Dividing these favorable results by the total sample space yields the fraction $\frac{24}{84}$, which simplifies to $\frac{2}{7}$.

Conditional probability allows us to narrow our perspective by focusing solely on a specific subset of the total outcomes. To determine this, we identify the restricted sample space consisting of all tickets where the product of the digits is zero. This condition is met if at least one of the digits on the ticket is zero. Examining the range from 00 to 49, these tickets are 00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, and 40, which results in a total count of 14 possibilities.

Within this group of 14, we look for the specific tickets where the sum of the digits equals 8. Checking each one, we find that only the ticket 08 satisfies this condition, as $0 + 8 = 8$. Since there is 1 favorable outcome out of the 14 in our restricted sample, the required probability is $\frac{1}{14}$.

Calculating the probability of at least one success in a binomial distribution is most efficiently handled by focusing on the complement, which is the event of failing every single trial. Given that the probability of success in a single trial is $\frac{1}{4}$, the probability of failure on any given trial is $\frac{3}{4}$. Consequently, for $n$ independent trials, the probability of zero successes is $(\frac{3}{4})^n$.

Since the total probability of all possible outcomes must equal $1$, the probability of at least one success is $1 - (\frac{3}{4})^n$. The problem specifies that this probability must be at least $\frac{9}{10}$, which leads to the inequality $1 - (\frac{3}{4})^n \geq \frac{9}{10}$. Rearranging this expression results in $(\frac{3}{4})^n \leq \frac{1}{10}$, or equivalently, $(\frac{4}{3})^n \geq 10$. Applying the common logarithm to both sides gives $n (\log_{10} 4 - \log_{10} 3) \geq 1$, which establishes that $n$ must be greater than or equal to $\frac{1}{\log_{10} 4 - \log_{10} 3}$.

The definition of conditional probability establishes that the probability of an event $A$ given $B$ is the likelihood of both occurring simultaneously divided by the probability of the condition, yielding the relationship:
$P(A \cap B) = P(A|B) \cdot P(B) = P(B|A) \cdot P(A)$
Substituting the known values $P(A)=\frac{1}{4}$, $P(A|B)=\frac{1}{2}$, and $P(B|A)=\frac{2}{3}$ transforms this into the equation:
$\frac{1}{2} \cdot P(B) = \frac{2}{3} \cdot \frac{1}{4}$
Simplifying the product on the right side to $\frac{1}{6}$ results in the equation $\frac{1}{2} \cdot P(B) = \frac{1}{6}$, which confirms that $P(B) = \frac{1}{3}$.

Determining the union of two events $A$ and $B$ involves identifying all unique outcomes that satisfy at least one of the given conditions. Event $A$ includes numbers greater than 3, which are $\{4, 5, 6\}$, and event $B$ includes numbers less than 5, which are $\{1, 2, 3, 4\}$. By merging these sets, we account for every possible result in the sample space $\{1, 2, 3, 4, 5, 6\}$, making the occurrence of either event certain. Calculating this via the addition rule, where $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, we use the individual probabilities $P(A) = \frac{3}{6}$ and $P(B) = \frac{4}{6}$, along with the intersection $P(A \cap B) = \frac{1}{6}$, to arrive at $P(A \cup B) = \frac{3}{6} + \frac{4}{6} - \frac{1}{6} = 1$.