Probability – JEE Main MathematicsPractice Questions & PYQs

When dealing with a fixed number of independent trials where each item is returned to the bag after selection, the number of successes follows a binomial distribution. Here, drawing a white ball is defined as a success with probability $p = \frac{3}{4}$, while drawing a red ball represents a failure with probability $q = \frac{1}{4}$ for each of the $n = 16$ trials.

The mean of this distribution is calculated as the product of the number of trials and the probability of success:
$\text{Mean} = np = 16 \times \frac{3}{4} = 12$
The standard deviation is the square root of the variance, $npq$, which is determined as follows:
$\text{Standard Deviation} = \sqrt{npq} = \sqrt{16 \times \frac{3}{4} \times \frac{1}{4}} = \sqrt{3}$
The ratio of the mean to the standard deviation is therefore $\frac{12}{\sqrt{3}}$, which simplifies by rationalizing the denominator to $4\sqrt{3}$.

The probability of a candidate failing to solve a given number of problems follows a binomial distribution with $n=50$ independent trials, where the probability of failure on any single problem is $q = 1 - \frac{4}{5} = \frac{1}{5}$ and the probability of success is $p = \frac{4}{5}$. To determine the probability that the candidate is unable to solve fewer than two problems, the combined probability of having exactly zero failures and exactly one failure must be calculated, expressed as $P(X=0) + P(X=1)$.

The probability of zero failures is given by $\binom{50}{0} (\frac{4}{5})^{50} (\frac{1}{5})^0$, which simplifies to $(\frac{4}{5})^{50}$. The probability of exactly one failure is given by $\binom{50}{1} (\frac{4}{5})^{49} (\frac{1}{5})^1$, which calculates to $50 \times (\frac{4}{5})^{49} \times \frac{1}{5}$, reducing to $10 (\frac{4}{5})^{49}$. Summing these two terms results in $(\frac{4}{5})^{50} + 10 (\frac{4}{5})^{49}$. By rewriting the first term as $\frac{4}{5} (\frac{4}{5})^{49}$ and factoring out the common term $(\frac{4}{5})^{49}$, the expression becomes $(\frac{4}{5} + 10) (\frac{4}{5})^{49}$, which simplifies to $\frac{54}{5} (\frac{4}{5})^{49}$.

The probability that the target is hit at least once is best calculated using the complement rule, where we subtract the probability of total failure from one. Because each person acts independently, we find the probability of everyone missing by multiplying their individual probabilities of failure, which are obtained by subtracting their hit rates from one. This leaves us with miss probabilities of $\frac{1}{2}$, $\frac{2}{3}$, $\frac{3}{4}$, and $\frac{7}{8}$. Multiplying these together reveals the likelihood that no one hits the target:

$P(\text{no hit}) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{7}{8} = \frac{7}{32}$

Subtracting this value from the total probability of one confirms that the target will be hit with a probability of $1 - \frac{7}{32} = \frac{25}{32}$.

Calculating the probability of an event occurring at least once is most effectively approached by finding the complement-that is, the probability of the event never occurring. Given that the probability of hitting the target is $1/3$, the probability of missing in any single attempt is $2/3$. Because the shots are independent, the probability of missing all $n$ shots is $(2/3)^n$, making the probability of hitting at least once equal to $1 - (2/3)^n$.

Setting the requirement that this probability must exceed $5/6$ leads to the inequality $1 - (2/3)^n > 5/6$, which simplifies to $(2/3)^n < 1/6$, or approximately 0.1666. Testing integer values for $n$, we find that for $n=4$, $(2/3)^4$ results in $16/81$, which is approximately 0.197 and still too large. However, when $n=5$, $(2/3)^5$ results in $32/243$, which is approximately 0.131. Since this value is less than 0.1666, 5 is the minimum number of shots required.

Conditional probability effectively reduces the universal set to only those occurrences that meet the specified criteria. For two families with two children each, there are $2^4 = 16$ total gender combinations possible, assuming each child is equally likely to be a boy or a girl. To satisfy the condition that at least two children are girls, we aggregate the outcomes representing two, three, and four girls, which corresponds to $\binom{4}{2} = 6$, $\binom{4}{3} = 4$, and $\binom{4}{4} = 1$ scenarios, respectively. Summing these values reveals that there are $6 + 4 + 1 = 11$ eligible outcomes within the restricted sample space. Since the event of all children being girls happens in only 1 of these cases, the probability is the ratio of that favorable outcome to the total valid outcomes, which is $\frac{1}{11}$.

Probability problems of this type are often simplified by considering the complement of the subset, as the total sum of all elements in the set $\{1, 2, \dots, 20\}$ is fixed at 210. A subset $B$ is nice if its elements sum to 203, which implies its complement must contain elements that sum to $210 - 203 = 7$. The distinct subsets of $\{1, 2, \dots, 20\}$ that sum to 7 are $\{7\}$, $\{1, 6\}$, $\{2, 5\}$, $\{3, 4\}$, and $\{1, 2, 4\}$. With 5 such favorable subsets out of $2^{20}$ total possible outcomes, the probability is $\frac{5}{2^{20}}$.

The probability of drawing a red ball second depends on the result of the first draw, which we account for using the law of total probability. Initially, with 5 red and 2 green balls, the first draw results in a red ball with probability $\frac{5}{7}$ and a green ball with probability $\frac{2}{7}$. If a red ball is drawn first, it is replaced by a green one, leaving 4 red and 3 green balls, making the probability of drawing a red ball next $\frac{4}{7}$. If a green ball is drawn first, it is replaced by a red one, leaving 6 red and 1 green ball, so the probability of drawing a red ball next is $\frac{6}{7}$. Totaling these probabilities gives the final result:

$\left(\frac{5}{7} \times \frac{4}{7}\right) + \left(\frac{2}{7} \times \frac{6}{7}\right) = \frac{20}{49} + \frac{12}{49} = \frac{32}{49}$

The concept of expected value in probability is determined by calculating the sum of every possible outcome's net value multiplied by the specific probability of that outcome occurring. In this game, rolling a five or a six constitutes a success with a probability of $p = \frac{2}{6} = \frac{1}{3}$, while rolling any other number results in a loss with a probability of $q = \frac{4}{6} = \frac{2}{3}$. The game concludes as soon as a success is achieved or after a maximum of three attempts have been made.

Four distinct scenarios exist based on when the game terminates. A player might succeed on the first throw, winning $100$ with a probability of $\frac{1}{3}$. If the first throw is a loss but the second is a win, the net gain is $-50 + 100 = 50$, which occurs with a probability of $q \cdot p = \frac{2}{3} \cdot \frac{1}{3} = \frac{2}{9}$. If the first two throws are failures and the third is a success, the net gain is $-50 - 50 + 100 = 0$, occurring with a probability of $q^2 \cdot p = (\frac{2}{3})^2 \cdot \frac{1}{3} = \frac{4}{27}$. Finally, if the player fails on all three throws, the net result is $-50 - 50 - 50 = -150$, with a probability of $q^3 = (\frac{2}{3})^3 = \frac{8}{27}$.

Summing the product of each net gain and its corresponding probability gives the total expected value:

$E = \left(\frac{1}{3} \times 100\right) + \left(\frac{2}{9} \times 50\right) + \left(\frac{4}{27} \times 0\right) + \left(\frac{8}{27} \times -150\right)$

This expression resolves to $\frac{100}{3} + \frac{100}{9} + 0 - \frac{1200}{27}$, which further simplifies to $\frac{300}{9} + \frac{100}{9} - \frac{400}{9}$, resulting in an overall expected gain or loss of $0$.

Drawing cards with replacement makes each selection an independent event where the probability of picking an ace is $4/52$, or $1/13$, and the probability of picking a non-ace is $48/52$, or $12/13$. To calculate the probability of obtaining exactly one ace, we sum the two mutually exclusive sequences of drawing an ace followed by a non-ace and drawing a non-ace followed by an ace, resulting in $(\frac{1}{13} \times \frac{12}{13}) + (\frac{12}{13} \times \frac{1}{13}) = \frac{24}{169}$. Since the probability of drawing two aces is simply $\frac{1}{13} \times \frac{1}{13} = \frac{1}{169}$, the combined probability is calculated as:

$\frac{24}{169} + \frac{1}{169} = \frac{25}{169}$

To find the number of tosses needed to guarantee at least one head with $90\%$ confidence, we focus on the complementary scenario where only tails appear. Because a fair coin lands on tails half the time, the chance of failing to get a head in $n$ attempts is $\frac{1}{2^n}$.

Setting our target probability as $1 - \frac{1}{2^n} \ge 0.9$ allows us to isolate the failure rate as $\frac{1}{2^n} \le 0.1$. This simplified form implies that $2^n$ must be greater than or equal to $10$. Calculating the powers of two, we find that $2^3$ is $8$, which is insufficient, whereas $2^4$ is $16$, confirming that four tosses are required to cross the threshold.

A random experiment ending exactly on the fifth throw requires the fourth and fifth outcomes to both be four, while the first three throws must avoid creating a consecutive pair. This configuration dictates that the third throw cannot be a four, as any four appearing there would force the experiment to conclude on the fourth throw. Consequently, we determine the number of valid sequences for the first three throws that do not contain the forbidden pair and end with a non-four digit.

The valid sequences for the first three throws include cases where no fours appear, which gives $5 \times 5 \times 5 = 125$ possibilities. If a four appears in the first position, there are $1 \times 5 \times 5 = 25$ ways to fill the remaining spots with non-four values. Similarly, if a four appears in the second position, there are $5 \times 1 \times 5 = 25$ ways, provided the third spot remains a non-four. Adding these cases results in $125 + 25 + 25 = 175$ favorable sequences.

Each valid three-throw sequence occurs with a probability of $\frac{1}{6^3}$. Since the experiment specifically ends on the fifth throw, the fourth and fifth throws must both be four, adding a probability factor of $\frac{1}{6} \times \frac{1}{6} = \frac{1}{6^2}$. Multiplying the count of valid three-throw sequences by the probability of the final two rolls yields the total probability as $\frac{175}{6^5}$.

For any binomial distribution, the mean $np$ and variance $npq$ provide sufficient information to determine the parameters $n$, $p$, and $q$. With the mean equal to $8$ and the variance equal to $4$, dividing the variance by the mean gives the probability of failure $q = \frac{4}{8} = \frac{1}{2}$. Consequently, the probability of success is $p = 1 - q = \frac{1}{2}$, and substituting these into the mean equation reveals the total number of trials $n = 16$.

The probability of achieving at most two successes is the sum of probabilities for $x=0$, $x=1$, and $x=2$, determined by the binomial formula $\binom{n}{x} p^x q^{n-x}$. Since $p$ and $q$ are both $\frac{1}{2}$, each term simplifies to $\binom{16}{x} (\frac{1}{2})^{16}$. Evaluating the combinations $\binom{16}{0}$, $\binom{16}{1}$, and $\binom{16}{2}$ results in $1$, $16$, and $120$, respectively. Summing these values yields $1 + 16 + 120 = 137$, establishing that $k = 137$ is the required numerator for the total probability over $2^{16}$.

To find the probability that a student opted for neither NCC nor NSS, we first determine how many students participated in at least one program using the principle of inclusion-exclusion. Adding the number of students in NCC and NSS and subtracting those counted in both provides the total number of students involved in at least one activity, which is $40 + 30 - 20 = 50$.

Because there are 60 students in total, we find the number of students who chose neither program by performing the calculation $60 - 50 = 10$. The required probability is the ratio of these students to the total class, yielding $\frac{10}{60}$, which simplifies to $\frac{1}{6}$.

The law of total probability governs this situation because the outcome of the second draw is directly dependent on the result of the first. We evaluate the two mutually exclusive scenarios-drawing a red ball first or a black ball first-and sum their respective probabilities of leading to a red ball on the second draw.

If a red ball is picked initially, which happens with a probability of $\frac{4}{10}$, we replace it and add two more red balls, changing the bag's composition to 6 red and 6 black balls out of 12. The likelihood of pulling a red ball next is $\frac{6}{12}$, contributing $\frac{4}{10} \times \frac{6}{12} = \frac{24}{120}$ to the total probability. If a black ball is chosen first with a probability of $\frac{6}{10}$, we add two black balls, resulting in 4 red and 8 black balls in the bag. The chance of drawing a red ball then is $\frac{4}{12}$, contributing $\frac{6}{10} \times \frac{4}{12} = \frac{24}{120}$. Combining these two paths yields the final result:

$\frac{24}{120} + \frac{24}{120} = \frac{48}{120} = \frac{2}{5}$

The probability of selecting 3 children out of a total of $2n$ children, all belonging to family B, is determined by the ratio of the number of ways to choose 3 children from the $n$ children in family B to the total number of ways to choose 3 children from $2n$, defined by the equation $\frac{\binom{n}{3}}{\binom{2n}{3}} = \frac{1}{12}$. Expanding the combination formula gives the expression $\frac{n(n-1)(n-2)}{2n(2n-1)(2n-2)} = \frac{1}{12}$, which simplifies to $\frac{n-2}{4(2n-1)} = \frac{1}{12}$ after canceling the common factors $n(n-1)$. Cross-multiplying results in $12(n-2) = 4(2n-1)$, which simplifies to $3n - 6 = 2n - 1$ and results in $n = 5$.

Bayes' Theorem establishes the posterior probability of a hypothesis given observed evidence, expressed as $P(B|E) = \frac{P(E|B)P(B)}{P(E|A)P(A) + P(E|B)P(B)}$, where $E$ is the event of drawing one white and one red ball. The probability of selecting either box A or box B is $P(A) = P(B) = \frac{1}{2}$. For box A, which contains 7 balls, the probability of selecting one white and one red ball is $P(E|A) = \frac{\binom{2}{1} \binom{3}{1}}{\binom{7}{2}} = \frac{6}{21} = \frac{2}{7}$. For box B, containing 9 balls, the probability is $P(E|B) = \frac{\binom{4}{1} \binom{2}{1}}{\binom{9}{2}} = \frac{8}{36} = \frac{2}{9}$. Substituting these values into the theorem yields:

$P(B|E) = \frac{\frac{2}{9} \cdot \frac{1}{2}}{\left(\frac{2}{7} \cdot \frac{1}{2}\right) + \left(\frac{2}{9} \cdot \frac{1}{2}\right)} = \frac{\frac{1}{9}}{\frac{1}{7} + \frac{1}{9}} = \frac{\frac{1}{9}}{\frac{16}{63}} = \frac{63}{9 \cdot 16} = \frac{7}{16}$

The probability of winning the game for player X is represented by an infinite geometric series, as the game can continue for an indefinite number of rounds until someone throws a head. The total probability $P(X)$ is calculated by summing the probabilities of X winning on the first, third, fifth, and subsequent turns, which is given by $P(X) = p + (1-p)(\frac{1}{2})p + (1-p)(\frac{1}{2})(1-p)(\frac{1}{2})p + \dots$. This expression forms a geometric series with the first term $a = p$ and the common ratio $r = (\frac{1-p}{2})$, so the sum is $S = \frac{p}{1 - \frac{1-p}{2}} = \frac{2p}{1+p}$. Since the probability of winning for both players is equal, $P(X) = \frac{1}{2}$, which gives the equation $\frac{2p}{1+p} = \frac{1}{2}$. Solving this, $4p = 1 + p$, which simplifies to $3p = 1$, yielding $p = \frac{1}{3}$.

The conditional probability $P\left(\overset{-}{A}\cap\overset{-}{B}|C\right)$ is calculated using the definition of conditional probability as $\frac{P\left(\overset{-}{A}\cap\overset{-}{B}\cap C\right)}{P(C)}$. According to De Morgan's laws, $\overset{-}{A}\cap\overset{-}{B} = \overline{A\cup B}$, which allows the numerator to be expressed as $P(C) - P(C\cap(A\cup B))$. Using the principle of inclusion-exclusion for the intersection, this expands to $P(C) - [P(A\cap C) + P(B\cap C) - P(A\cap B\cap C)]$.

Since the events are pair-wise independent, $P(A\cap C) = P(A)P(C)$ and $P(B\cap C) = P(B)P(C)$, and given $P(A\cap B\cap C) = 0$, the expression for the numerator becomes $P(C) - [P(A)P(C) + P(B)P(C) - 0]$. Dividing this entire expression by $P(C)$ results in $1 - P(A) - P(B)$. Because $P\left(\overset{-}{A}\right) = 1 - P(A)$, the final result simplifies to $P\left(\overset{-}{A}\right) - P(B)$.

Since the balls are drawn with replacement, the probability of selecting a green ball remains constant for every draw, which confirms that the scenario follows a binomial distribution. With a total of 25 balls available and 15 being green, the probability of success is $p = \frac{3}{5}$ and the probability of failure is $q = \frac{2}{5}$ across $n = 10$ trials.

The variance of a binomial distribution is calculated using the formula $npq$. Applying the known values yields the following:
$10 \times \frac{3}{5} \times \frac{2}{5} = \frac{12}{5}$

For two distinct numbers chosen from the set 0, 1, 2, ..., 10 to have both their sum and their absolute difference as multiples of 4, they must satisfy specific divisibility requirements. If we represent the two numbers as $a$ and $b$, the conditions $a + b = 4k$ and $a - b = 4m$ imply that their sum $2a = 4(k + m)$ and their difference $2b = 4(k - m)$ are both multiples of 4, which fundamentally requires both $a$ and $b$ to be even numbers.

The total number of ways to select two distinct numbers from the 11 available values is given by the combination formula $^{11}C_2 = 55$. By focusing on the even integers within the given set, which are 0, 2, 4, 6, 8, and 10, we can systematically identify the pairs that satisfy the criteria. Testing these combinations reveals that the valid pairs are (0, 4), (0, 8), (2, 6), (2, 10), (4, 8), and (6, 10).

There are 6 such favorable pairs that fulfill both conditions. The probability of selecting one of these pairs is the ratio of these favorable cases to the total number of possible outcomes in the sample space, resulting in $\frac{6}{55}$.