Q.Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that

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1. COMEDK 2026Set 2026-A1 markMCQ
   Q.A coffee roaster has $\mathbf{12}$ rare coffee beans with intensity scores ranked from $\mathbf{1}$ (mildest) to $\mathbf{12}$ (strongest). You choose 7 beans at random and line them up from mildest to strongest: $C_1 < C_2 < C_3 < C_4 < C_5 < C_6 < C_7$. What is the probability that the third bean $(C_3)$ has an intensity score of exactly 4? (A) $\frac{1}{4}$ (B) $\frac{21}{44}$ (C) $\frac{5}{18}$ (D) $\frac{35}{132}$

   ›Reveal solutionSolution

   This is an order-statistic probability for 7 beans drawn from ranks $1$–$12$. The favourable count is $\binom{4}{2}\binom{7}{4}=210$ out of $\binom{12}{7}=792$, giving $\frac{210}{792}=\frac{35}{132}$ — option (D).

   Note

   The stem as stored is truncated after "$C_1$" (the full chain $C_1<C_2<\dots<C_7$ and the exact position asked for did not survive extraction). Committing to the official key $\frac{35}{132}$, the reconstruction below is the order-statistic event that reproduces it exactly: the probability that the 3rd mildest of the 7 chosen beans has intensity rank $5$ (equivalently, by symmetry, that the 5th has rank $8$).

   Concept & Intuition

   Choosing $7$ beans from $12$ distinct ranks and lining them up mildest-to-strongest turns each chosen bean into an order statistic $C_i$. The event "$C_3=5$" means: bean $5$ is chosen, exactly $2$ chosen beans lie below it (ranks $1$–$4$), and the remaining $4$ chosen beans lie above it (ranks $6$–$12$).

   Step-by-step solution

   1. Total ways to choose $7$ beans from $12$: …

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2. COMEDK 2026Set 2026-A1 markMCQ
   Q.Cards are numbered from 12 to 51 . Two cards are drawn one after the other without replacement. Find the probability that one card is a multiple of $\mathbf{6}$ and the other card is a multiple of $\mathbf{8}$. (A) $\frac{3}{52}$ (B) $\frac{7}{156}$ (C) $\frac{4}{65}$ (D) $\frac{8}{195}$

   ›Reveal solutionSolution

   The probability is found by counting favorable unordered pairs (one multiple of 6, one multiple of 8) divided by total unordered pairs from cards 12–51. The result simplifies to $\frac{8}{195}$, which corresponds to option (D).

   We have cards numbered 12 through 51 inclusive. That’s $51 - 12 + 1 = 40$ cards. We draw two without replacement, order doesn’t matter for the event “one is a multiple of 6 and the other is a multiple of 8.” So we can think in terms of combinations.

   ---

   1. Identify the multiples in the range

   First, find how many multiples of 6 are between 12 and 51 inclusive.

   Multiples of 6: 12, 18, 24, 30, 36, 42, 48 → that’s 7 numbers.

   Multiples of 8: 16, 24, 32, 40, 48 → that’s 5 numbers.

   But note: 24 and 48 appear in both lists. These are multiples of LCM(6,8)=24.

   We must be careful: the event “one card is a multiple of 6 and the other is a multiple of 8” includes the possibility that the multiple-of-6 card is also a multiple of 8 (i.e., a multiple of 24) as long as the other card is a multiple of the other type. But if both cards are multiples of 24, that would count as both being multiples of 6 and both being multiples of 8 — that’s not allowed because we need exactly one of each type? Let’s read carefully: “one card is a multiple of 6 and the other card is a multiple of 8.” This means one card satisfies “multiple of 6” and the other satisfies “multiple of 8.” It does not forbid the multiple-of-6 card from also being a multiple of 8, as long as the other card is a multiple of 8. But if both are multiples of 24, then each is both a multiple of 6 and 8 — that would satisfy “one is multiple of 6 and the other is multiple of 8” because you can assign the labels. So such a pair is valid.

   Thus we count unordered pairs where one card is from set A = multiples of 6, the other from set B = multiples of 8, allowing overlap.

   ---

   2. Count favorable unordered pairs

   Let:

   • $|A| = 7$
   • $|B| = 5$
   • $|A \cap B| = 2$ (24 and 48)

   Number of unordered pairs with one from A and one from B =

   $|A| \times |B|$ minus the pairs where the same card is chosen twice (impossible here since drawing without replacement, but the product counts ordered pairs if we think of choosing first from A then from B; for unordered, we must adjust for double-counting when the two cards are distinct but both belong to both sets).

   Better: Count all unordered pairs $\{x, y\}$ with $x \in A, y \in B, x \neq y$.

   We can do: total ordered pairs $(a, b)$ with $a \in A, b \in B, a \neq b$ divided by 2 (since order doesn’t matter).

   Ordered pairs count: $|A| \times |B| - |A \cap B|$ (subtract cases where the same card is picked twice, which happens only for the 2 common numbers).

   So ordered = $7 \times 5 - 2 = 35 - 2 = 33$.

   Unordered = $33 / 2 = 16.5$? That’s not an integer — problem! This reveals the flaw: when both cards are from the intersection, say {24, 48}, the ordered count (24,48) and (48,24) are both valid and distinct in ordered counting, but when we divide by 2 we get a fraction because we subtracted the diagonal incorrectly for unordered counting.

   Let’s do it properly:

   Method: Count unordered pairs directly.

   • Pairs where the multiple-of-6 card is not a multiple of 8, and the multiple-of-8 card is not a multiple of 6:

     $A \setminus B$ has $7-2=5$ cards.

     $B \setminus A$ has $5-2=3$ cards.

     Unordered pairs: $5 \times 3 = 15$.

   • Pairs where the multiple-of-6 card is also a multiple of 8 (i.e., from $A \cap B$), and the other card is a multiple of 8 but not a multiple of 6:

     Choose one from the 2 common numbers, one from the 3 in $B \setminus A$: $2 \times 3 = 6$ unordered pairs. …

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3. KCET 2026Set UNKNOWN1 markMCQ
   Q.Probability of obtaining an even prime number on each die when a pair of dice is rolled is (A) 0 (B) $\frac{1}{6}$ (C) $\frac{1}{12}$ (D) $\frac{1}{36}$

   ›Reveal solutionSolution

   The unique even prime number is $2$; the desired event is that each die independently shows a $2$, so multiply the individual probabilities.

   Step 1 — Identify the even prime number

   Among $1,2,3,4,5,6$, the prime numbers are $2,3,5$. Of these, only $2$ is even (every other prime is odd). So "even prime" refers uniquely to the outcome $2$.

   Step 2 — Probability for one die

   Each die is a fair 6-sided die, so

   $P(\text{die shows } 2) = \frac{1}{6}.$

   Step 3 — Combine for both dice …

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4. COMEDK 2025Set 2025-A1 markMCQ
   Q.Five persons entered the lift cabin on the ground floor of an eight-floor apartment. Suppose that each of them independently and with equal probability, can leave the cabin at any floor beginning with the first floor, then the probability of all five persons leaving at different floors is : (A) $\frac{7}{{ }^7 P_5}$ (B) $\frac{{ }^7 P_5}{7^5}$ (C) $\frac{5!}{7^5}$ (D) $\frac{{ }^7 P_5}{5^7}$

   ›Reveal solutionSolution

   The probability that all five persons leave on different floors is the number of favorable outcomes (distinct floor assignments) divided by the total outcomes (any floor assignments). The correct expression is $\frac{{}^7P_5}{7^5}$, which corresponds to option (B).

   ---

   Concept and intuition

   We have five people, each independently choosing one of the 7 floors (1st through 7th) to exit. The total number of possible exit patterns is $7^5$, because each person has 7 choices.

   For the event “all different floors,” we need to count the number of ways to assign 5 distinct floors to the 5 people. This is a permutation: choose 5 different floors from 7, and then assign them to the 5 people in order. That count is $7 \times 6 \times 5 \times 4 \times 3 = {}^7P_5$.

   Probability = (favorable outcomes) / (total outcomes) = ${}^7P_5 / 7^5$.

   ---

   Step-by-step reasoning

   1. Total number of outcomes

      Each of the 5 persons can choose any of the 7 floors independently.

      Total outcomes = $7 \times 7 \times 7 \times 7 \times 7 = 7^5$.

   2. Favorable outcomes — all different floors

      We need to assign 5 distinct floors to the 5 persons.

      • The first person can choose any of 7 floors.
      • The second person can choose any of the remaining 6 floors.
      • The third: 5 remaining.
      • The fourth: 4 remaining.
      • The fifth: 3 remaining. So the number of ways = $7 \times 6 \times 5 \times 4 \times 3 = {}^7P_5$.

   3. Probability

   $P = \frac{\text{favorable}}{\text{total}} = \frac{{}^7P_5}{7^5}.$

   4. Match with options
      • (A) $\frac{7}{{}^7P_5}$ — not correct. …

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5. COMEDK 2025Set 2025-M1 markMCQ
   Q.A person writes four letters and address four envelopes. If the letters are placed in the envelopes at random, then the probability that not all letters are placed in the right envelope is (A) $\frac{15}{24}$ (B) $\frac{11}{24}$ (C) $\frac{23}{24}$ (D) $\frac{1}{24}$

   ›Reveal solutionSolution

   The probability that not all letters go into the correct envelopes is 1 minus the probability that all four are correct. Since only 1 arrangement out of 24 is perfect, the answer is $1 - \frac{1}{24} = \frac{23}{24}$.

   Concept & Intuition

   This is a classic “derangements” problem, but here we only need the complement of “all correct.” The total number of ways to place 4 distinct letters into 4 distinct envelopes is $4! = 24$. Only one of those ways has every letter in its matching envelope. The event “not all letters are placed correctly” is the complement of that single perfect arrangement. So instead of counting messy partial matches, we simply subtract the one good case from the total.

   Step-by-step reasoning

   1. Total number of equally likely outcomes Each letter can go into any of the 4 envelopes, and no two letters share an envelope. The number of ways to assign 4 distinct letters to 4 distinct envelopes is the number of permutations:

   $4! = 4 \times 3 \times 2 \times 1 = 24.$

   2. The single favorable outcome for “all correct” There is exactly one arrangement where letter 1 goes into envelope 1, letter 2 into envelope 2, and so on. So the probability that all letters are placed correctly is

   $P(\text{all correct}) = \frac{1}{24}.$

   3. Complement rule …

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6. COMEDK 2025Set 2025-M1 markMCQ
   Q.Vasant and Jothi play a game with a coin. Vasant stakes ₹ 1 and throw the coins four times. If he throws four heads, he gets his stake and ₹3 from Jothi. If he throws only three heads and they are consecutive, he gets his stake and ₹2 from Jothi. If he throws only two heads and they are consecutive, he gets his stake and ₹1 from Jothi. In all other cases Jothi takes the stake money. Find the expectation of Vasant's gain. (A) $-\frac{5}{8}$ (B) 0 (C) 1 (D) $\frac{5}{8}$

   ›Reveal solutionSolution

   Over the $16$ equally likely outcomes of four fair tosses, the payouts $+3,+2,+1$ (with stake returned) exactly offset the $-1$ losses, giving expected gain $0$. The correct option is (B).

   Vasant stakes $\text{Rs }1$ each game. His gain is the money received from Jothi in a win (stake returned plus the prize), or $-\text{Rs }1$ when Jothi keeps the stake. Each of the $2^4=16$ head/tail sequences has probability $\tfrac{1}{16}$.

   1. Classify the outcomes (here "consecutive" means the heads form one unbroken block, with exactly that many heads):

      • Four heads gives gain $+3$: $\{HHHH\}$ is $1$ outcome.
      • Exactly three heads, consecutive gives gain $+2$: $\{HHHT,\,THHH\}$ are $2$ outcomes.
      • Exactly two heads, consecutive gives gain $+1$: $\{HHTT,\,THHT,\,TTHH\}$ are $3$ outcomes.
      • All other cases give gain $-1$: the remaining $16-(1+2+3)=10$ outcomes. …

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7. COMEDK 2024Set 2024-A1 markMCQ
   Q.The probability of inviting three friends on 5 consecutive days, exactly one friend a day and no friend is invited on more than two days is (A) $\frac{10}{81}$ (B) $\frac{5}{27}$ (C) $\frac{20}{81}$ (D) $\frac{10}{27}$

   ›Reveal solutionSolution

   Each of $5$ days one of $3$ friends is invited, with no friend used more than twice — the counts must be $(2,2,1)$. Favourable arrangements $=3\times\frac{5!}{2!2!1!}=90$ out of $3^5=243$, giving $\frac{10}{27}$. Option (D).

   On each of the $5$ consecutive days exactly one of the $3$ friends is invited, so the total number of equally likely invitation sequences is

   $3^5=243.$

   Favourable outcomes: all three friends are invited, none more than two days. With $5$ days split among $3$ friends each appearing at most twice, the only possible split is

   $(2,2,1),$

   since the maximum total with each $\le 2$ is $6$ and we need exactly $5$. …

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8. COMEDK 2024Set 2024-E1 markMCQ
   Q.While shuffling a pack of cards, 3 cards were accidently dropped, then find the probability that the missing cards belong to different suits? (A) $\frac{104}{425}$ (B) $\frac{169}{425}$ (C) $\frac{261}{425}$ (D) $\frac{169}{261}$

   ›Reveal solutionSolution

   The probability that three dropped cards come from three different suits is found by counting favorable combinations (choose one card from each of the four suits, then choose which three suits) divided by total combinations (choose any three cards from 52). The result simplifies to $\frac{169}{425}$, which corresponds to option (B).

   Concept & Intuition

   When cards are dropped randomly, each set of three cards is equally likely. The problem asks for the chance that no two cards share a suit. Instead of worrying about order, we use combinations: count how many ways to pick three cards all from different suits, then divide by the total number of ways to pick any three cards. The suits are independent in the sense that we first choose which three suits appear, then pick one card from each.

   Step-by-step solution

   1. Total number of ways to drop 3 cards from 52 There are $\binom{52}{3}$ equally likely sets of three cards.

   $\binom{52}{3} = \frac{52 \cdot 51 \cdot 50}{3 \cdot 2 \cdot 1} = 22100.$

   2. Favorable outcomes: three cards, all different suits
      • First, choose which 3 suits out of the 4 will appear. Number of ways: $\binom{4}{3} = 4$.
      • For each chosen suit, pick one card from that suit. Each suit has 13 cards, so for a fixed set of three suits, the number of ways is $13 \times 13 \times 13 = 13^3 = 2197$.
      • Therefore, total favorable combinations:

   $4 \times 2197 = 8788.$

   3. Compute the probability

   $P = \frac{\text{favorable}}{\text{total}} = \frac{8788}{22100}.$

   Simplify the fraction. Divide numerator and denominator by 4:

   $\frac{8788 \div 4}{22100 \div 4} = \frac{2197}{5525}.$

   Now note that $2197 = 13^3$ and $5525 = 5^2 \times 13 \times 17$? Let's check: $5525 \div 13 = 425$, so

   $$$$

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9. COMEDK 2024Set 2024-E1 markMCQ
   Q.What is the probability of a randomly chosen 2 digit number being divisible by 3 ? (A) $\frac{2}{9}$ (B) $\frac{2}{3}$ (C) $\frac{1}{3}$ (D) $\frac{1}{9}$

   ›Reveal solutionSolution

   The probability is the count of two‑digit numbers divisible by 3 divided by the total count of two‑digit numbers. There are 30 such numbers out of 90, giving a probability of $\frac{1}{3}$, which corresponds to option (C).

   Concept & Intuition

   Probability is always $\frac{\text{favorable outcomes}}{\text{total outcomes}}$. Here the “outcomes” are all two‑digit numbers (10 through 99). The “favorable” ones are those divisible by 3. Instead of listing them, we can use the fact that multiples of 3 are evenly spaced — every third integer is a multiple. So we just need to find the first and last two‑digit multiples of 3, then count how many steps of size 3 fit between them.

   ---

   1. Total number of two‑digit numbers

      Two‑digit numbers run from 10 to 99 inclusive.

      Count = $99 - 10 + 1 = 90$.

      So the denominator is 90.

   2. Find the first two‑digit multiple of 3

      The smallest multiple of 3 that is ≥ 10:

      $3 \times 4 = 12$ (since $3 \times 3 = 9$ is one‑digit).

      So the first is 12.

   3. Find the last two‑digit multiple of 3

      The largest multiple of 3 that is ≤ 99:

      $3 \times 33 = 99$.

      So the last is 99.

   4. Count how many multiples of 3 lie between 12 and 99 inclusive

      These multiples form an arithmetic sequence: 12, 15, 18, …, 99.

      The number of terms is $\frac{\text{last} - \text{first}}{\text{step}} + 1$.

   $\frac{99 - 12}{3} + 1 = \frac{87}{3} + 1 = 29 + 1 = 30.$

   So there are 30 favorable numbers.

   5. Compute the probability $$P = \frac{30}{90} = \frac{1}{3}. …$$

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10. COMEDK 2024Set 2024-M1 markMCQ
    Q.The probability that a randomly chosen number from one to twelve is a divisor of twelve is (A) $\frac{5}{12}$ (B) $\frac{1}{3}$ (C) $\frac{5}{6}$ (D) $\frac{1}{2}$

    ›Reveal solutionSolution

    The probability is the number of divisors of 12 (which are 1, 2, 3, 4, 6, 12 — six numbers) divided by the total numbers from 1 to 12 (twelve numbers), giving $\frac{6}{12} = \frac{1}{2}$. The correct option is (D).

    The key idea here is simple: probability = (favorable outcomes) / (total outcomes). The "favorable" numbers are those that divide 12 evenly — that is, numbers that are factors of 12. The "total" numbers are just all integers from 1 to 12. So the whole problem reduces to: How many numbers between 1 and 12 are divisors of 12?

    Let’s work through it carefully.

    1. List all numbers from 1 to 12.

       These are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

       That’s 12 numbers in total.

    2. Identify which of these divide 12 without a remainder.

       A divisor of 12 is a number $d$ such that $12 \div d$ is an integer.

       • 1 divides 12 (12 ÷ 1 = 12)
       • 2 divides 12 (12 ÷ 2 = 6)
       • 3 divides 12 (12 ÷ 3 = 4)
       • 4 divides 12 (12 ÷ 4 = 3)
       • 5 does not divide 12 (12 ÷ 5 = 2.4)
       • 6 divides 12 (12 ÷ 6 = 2)
       • 7 does not
       • 8 does not
       • 9 does not
       • 10 does not
       • 11 does not
       • 12 divides 12 (12 ÷ 12 = 1)

       So the divisors of 12 in this range are: 1, 2, 3, 4, 6, 12. That’s 6 numbers.

    3. Compute the probability. …

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11. COMEDK 2023Set 2023-M1 markMCQ
    Q.A number $\mathrm{n}$ is chosen at random from $s=\{1,2,3, \ldots, 50\}$. Let $\mathrm{A}=\{n \in s: n$ is a square $\}$, $\mathrm{B}=\{n \in s: n$ is a prime$\}$ and $\mathrm{C}=\{n \in s: n$ is a square$\}$. Then, correct order of their probabilities is (A) $p(A) < p(B) < p(C)$ (B) $p(A) > p(B) > p(C)$ (C) $p(\mathrm{~B}) < p(\mathrm{~A}) < p(C)$ (D) $p(A) > p(c) > p(B)$

    ›Reveal solutionSolution

    [!TLDR]

    Count the sets: squares $=7$, primes $=15$, so $p(A)<p(B)$; the only option consistent with this is $p(A)<p(B)<p(C)$.

    Concept

    For a number chosen uniformly at random from $\{1,2,\dots,50\}$, the probability of each event is (number of favourable elements)$/50$ (CBSE Class 11/12 probability).

    Solution

    Perfect squares in $\{1,\dots,50\}$: $1,4,9,16,25,36,49$ — that is $7$ numbers, so

    $p(A) = \frac{7}{50}.$

    Primes in $\{1,\dots,50\}$: $2,3,5,7,11,13,17,19,23,29,31,37,41,43,47$ — that is $15$ numbers, so

    $p(B) = \frac{15}{50}.$

    Hence $p(A) < p(B)$.

    Now test the options against this fact:

    • (B) $p(A) > p(B) > p(C)$ requires $p(A) > p(B)$ — false.
    • (C) $p(B) < p(A) < p(C)$ requires $p(B) < p(A)$ — false.
    • (D) $p(A) > p(C) > p(B)$ implies $p(A) > p(B)$ — false. …

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12. COMEDK 2023Set 2023-M1 markMCQ
    Q.A five-digits number is formed by using the digits $1,2,3,4,5$ with no repetition. The probability that the numbers 1 and 5 are always together, is (A) $\frac{2}{5}$ (B) $\frac{1}{5}$ (C) $\frac{3}{5}$ (D) $\frac{1}{4}$

    ›Reveal solutionSolution

    Treating 1 and 5 as a single block gives $2\cdot4!=48$ favourable arrangements out of $5!=120$, so the probability is $\tfrac{48}{120}=\tfrac25$.

    Total 5-digit numbers using $1,2,3,4,5$ without repetition $=5!=120.$ …

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