Q.A black and a red dice are rolled.

Showing the 12 most recent of 45 on this concept.

1. COMEDK 2023Set 2023-E1 markMCQ
   Q.A die is thrown twice and the sum of numbers appearing is observed to be 8 . What is the conditional probability that the number 5 has appeared atleast once? (A) $\frac{5}{36}$ (B) $\frac{2}{5}$ (C) $\frac{1}{18}$ (D) $\frac{1}{3}$

   ›Reveal solutionSolution

   Given the sum is $8$, the sample space is the $5$ ordered pairs summing to $8$; two of them include a $5$, giving probability $\frac{2}{5}$.

   The ordered outcomes with sum $8$ are

   $(2,6),(3,5),(4,4),(5,3),(6,2)\quad(5\text{ outcomes}).$ …

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2. KCET 2021Set A-11 markMCQ
   Q.Two dice are thrown. If it is known that the sum of numbers on the dice was less than 6 the probability of getting a sum as 3 is (A) $\dfrac{1}{18}$ (B) $\dfrac{5}{18}$ (C) $\dfrac{1}{5}$ (D) $\dfrac{2}{5}$

   ›Reveal solutionSolution

   We use conditional probability: reduce the sample space to only those outcomes where the sum is less than 6, then find the fraction of those that give a sum of exactly 3. The answer is $\frac{1}{5}$.

   The key idea here is conditional probability — the probability of one event given that another event has already occurred. When we say "if it is known that the sum was less than 6", we are no longer considering all 36 possible outcomes of throwing two dice. Instead, we restrict ourselves to only those outcomes that satisfy the condition, and then ask: within this smaller set, what fraction gives a sum of 3?

   Let’s work it out step by step.

   1. List all possible outcomes when two dice are thrown.

      Each die shows a number from 1 to 6. The total number of ordered pairs $(a,b)$ is $6 \times 6 = 36$.

   2. Identify the condition: sum less than 6.

      The possible sums are 2, 3, 4, and 5 (since sum = 1 is impossible with two dice). Let’s list all ordered pairs that give each sum:

      • Sum = 2: $(1,1)$ → 1 outcome
      • Sum = 3: $(1,2), (2,1)$ → 2 outcomes
      • Sum = 4: $(1,3), (2,2), (3,1)$ → 3 outcomes
      • Sum = 5: $(1,4), (2,3), (3,2), (4,1)$ → 4 outcomes

      Total outcomes with sum < 6: $1 + 2 + 3 + 4 = 10$.

      Watch out

      A common mistake is to forget that $(1,2)$ and $(2,1)$ are different outcomes. Dice are distinct, so order matters. Always count ordered pairs unless the problem explicitly says "identical dice" (which is rare in such problems).

   3. Identify the event of interest: sum = 3. …

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3. COMEDK 2025Set 2025-M1 markMCQ
   Q.Three fair dice are thrown. What is the probability of getting a total of 15 given that they exhibit three different numbers that are in arithmetic progression? (A) $\frac{1}{8}$ (B) $\frac{1}{6}$ (C) $\frac{1}{4}$ (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   The problem asks for the conditional probability that three dice sum to 15, given that the three numbers shown are distinct and form an arithmetic progression. By listing all such triples and counting those summing to 15, the probability is found to be 1/2.

   Concept and intuition

   We need $P(\text{sum}=15 \mid \text{distinct AP})$. The condition restricts the outcomes to only those triples $(a,b,c)$ with $a<b<c$ (since dice are fair, order matters but we can treat ordered rolls; however the condition “three different numbers” and “arithmetic progression” is symmetric). The key is to enumerate all possible ordered triples of distinct numbers from $\{1,2,3,4,5,6\}$ that are in arithmetic progression, then count how many of those sum to 15. Because the dice are fair, each ordered triple is equally likely, so the conditional probability is just the ratio of favorable ordered triples to total ordered triples satisfying the condition.

   Step-by-step reasoning

   1. Characterize arithmetic progressions of three distinct numbers from 1 to 6. Three numbers $x, y, z$ (with $x<y<z$) are in AP iff $2y = x+z$. Since they are distinct and from 1 to 6, the possible triples (unordered) are:

   $(1,2,3),\ (1,3,5),\ (2,3,4),\ (2,4,6),\ (3,4,5),\ (4,5,6).$

   Check: For (1,2,3): $2\cdot2=1+3$; (1,3,5): $2\cdot3=1+5$; etc. That’s all.

   2. Count ordered triples satisfying the condition.

      For each unordered triple, the three numbers are distinct, so they can appear in any order on the three dice. That gives $3! = 6$ permutations per triple.

      Total ordered triples satisfying the condition = $6 \times 6 = 36$.

   3. Identify which of these triples sum to 15.

      Compute sums of the unordered triples:

      $\begin{aligned} (1,2,3) &\to 6,\\ (1,3,5) &\to 9,\\ (2,3,4) &\to 9,\\ (2,4,6) &\to 12,\\ (3,4,5) &\to 12,\\ (4,5,6) &\to 15. \end{aligned}$ …

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4. KCET 2018Set A-11 markMCQ
   Q.In a simultaneous throw of a pair of dice, the probability of getting a total more than $7$ is (A) $\dfrac{7}{12}$ (B) $\dfrac{5}{36}$ (C) $\dfrac{5}{12}$ (D) $\dfrac{7}{36}$

   ›Reveal solutionSolution

   Count the ordered outcomes whose sum exceeds 7 out of the 36 equally likely outcomes of a pair of dice.

   Step 1 — The sample space.

   Each die shows one of $6$ faces, and the two dice are independent, so

   $n(S) = 6 \times 6 = 36$

   All $36$ ordered pairs $(a,b)$ are equally likely, so classical probability applies:

   $P(E) = \frac{n(E)}{n(S)}$

   Step 2 — Count the favourable outcomes (sum $> 7$, i.e. sum $= 8,9,10,11,12$).

   Sum	Ordered pairs	Count
   $8$	$(2,6),(3,5),(4,4),(5,3),(6,2)$	$5$
   $9$	$(3,6),(4,5),(5,4),(6,3)$	$4$
   $10$	$(4,6),(5,5),(6,4)$	$3$
   $11$	$(5,6),(6,5)$	$2$
   $12$	$(6,6)$	$1$

   $n(E) = 5+4+3+2+1 = 15$

   Step 3 — Probability.

   $P(\text{sum} > 7) = \frac{15}{36} = \frac{5}{12}$ …

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5. COMEDK 2024Set 2024-M1 markMCQ
   Q.A and B each have a calculator which can generate a single digit random number from the set $\{1,2,3,4,5,6,7,8\}$. They can generate a random number on their calculator. Given that the sum of the two numbers is 12 , then the probability that the two numbers are equal is (A) $\frac{5}{64}$ (B) $\frac{1}{5}$ (C) $\frac{1}{16}$ (D) $\frac{1}{8}$

   ›Reveal solutionSolution

   We are asked for the conditional probability that two numbers are equal given their sum is 12.

   The only equal pair summing to 12 is (6,6), and there are 5 total pairs summing to 12.

   So the probability is $\frac{1}{5}$, which corresponds to option (B).

   ---

   Concept and intuition

   This is a classic conditional probability problem: we are not interested in all possible outcomes, only those where the sum is exactly 12. The phrase “given that the sum is 12” means we restrict our universe to those pairs. Then we count how many of those restricted outcomes have the two numbers equal. The trap is to forget to restrict the denominator — many students mistakenly use the total number of all possible pairs (64) instead of only the favorable-sum pairs.

   ---

   Step-by-step solution

   1. Identify the sample space

      Each of A and B picks a digit from $\{1,2,\dots,8\}$.

      Total possible ordered pairs $(a,b)$: $8 \times 8 = 64$.

   2. List all pairs with sum 12

      We need $a + b = 12$, with $1 \le a,b \le 8$.

      Possible values for $a$:

      • If $a = 4$, then $b = 8$
      • If $a = 5$, then $b = 7$
      • If $a = 6$, then $b = 6$
      • If $a = 7$, then $b = 5$
      • If $a = 8$, then $b = 4$

      So the pairs are:

   $(4,8),\ (5,7),\ (6,6),\ (7,5),\ (8,4)$

   That’s 5 ordered pairs.

   3. Count the favorable outcomes “The two numbers are equal” means $a = b$. …

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6. COMEDK 2025Set 2025-E1 markMCQ
   Q.Two numbers are selected at random from integers 1 to 9 . If their sum is even, what is the probability that both the numbers are odd? (A) $\frac{4}{9}$ (B) $\frac{5}{8}$ (C) $\frac{1}{6}$ (D) $\frac{2}{3}$

   ›Reveal solutionSolution

   This is a conditional probability problem: given that the sum of two numbers from 1–9 is even, we want the probability both are odd. The answer is 5/8, option (B).

   We are selecting two numbers from 1 to 9 without replacement (since "selected at random" from distinct integers usually implies no repetition). The sum is even only if both numbers are odd or both are even. So the condition restricts us to those pairs. The question asks: among those pairs with an even sum, what fraction consists of two odd numbers?

   ---

   1. Count total possible pairs (without replacement)

   From 1 to 9, there are 9 numbers. The number of ways to choose any two distinct numbers is

   $\binom{9}{2} = 36.$

   2. Count pairs with an even sum

   A sum is even when both numbers have the same parity.

   • Odd numbers from 1 to 9: 1, 3, 5, 7, 9 → 5 odds.

     Number of odd–odd pairs: $\binom{5}{2} = 10$.

   • Even numbers from 1 to 9: 2, 4, 6, 8 → 4 evens.

     Number of even–even pairs: $\binom{4}{2} = 6$.

   So total pairs with an even sum:

   $10 + 6 = 16.$

   3. Apply conditional probability

   We want

   $$$$

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7. COMEDK 2024Set 2024-E1 markMCQ
   Q.Suppose we have three cards identical in form except that both sides of the first card are coloured red, both sides of the second are coloured black, and one side of the third card is coloured red and the other side is coloured black. The three cards are mixed and a card is picked randomly. If the upper side of the chosen card is coloured red, what is the probability that the other side is coloured black. (A) $\frac{1}{6}$ (B) $\frac{1}{2}$ (C) 0 (D) $\frac{1}{3}$

   ›Reveal solutionSolution

   This is a classic conditional probability problem (often called the "three cards" or "Bertrand's box" variant). The key is that seeing a red side updates the probability space to only the red sides, and among those, only one belongs to the mixed card. The answer is 1/3.

   We are asked: given that the visible side is red, what is the probability that the other side is black? This is not simply "one of the two remaining cards has a black other side" because the cards are not equally likely once we condition on the observation.

   ---

   Concept and Intuition

   The pitfall is to think: "We see red, so the card is either the all-red or the mixed card. That's two possibilities, so the chance is 1/2." But this ignores that the all-red card has two red sides, while the mixed card has only one red side. When we pick a card at random and then look at a random side, each of the six sides is equally likely to be the one we see. Seeing red eliminates the three black sides, leaving only the three red sides. Among those three red sides, two belong to the all-red card and only one belongs to the mixed card. So the probability that the other side is black is 1/3.

   ---

   Step-by-step reasoning

   1. Label the sides.

      Let the cards be:

      • Card A: both sides red (sides $R_1, R_2$)
      • Card B: both sides black (sides $B_1, B_2$)
      • Card C: one red, one black (sides $R_3, B_3$)

   2. Count equally likely outcomes.

      When we pick a card uniformly at random and then look at a random side, there are $3 \times 2 = 6$ equally likely side-views. Each of the six sides has probability $1/6$ of being the one we see.

   3. Condition on seeing a red side.

      The red sides are: $R_1, R_2$ (from card A) and $R_3$ (from card C). That's 3 red sides. So the conditional space has 3 equally likely possibilities. …

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8. COMEDK 2026Set 2026-M1 markMCQ
   Q.Advika chooses one of three scarves every morning: Red, Blue, or Green. The probability she chooses Red is $20 \%$. The probability she chooses Blue is twice the probability of choosing Red. On the remaining days she wears a Green scarf. Once a scarf is chosen, she decides whether to wear a Hat (H) and Sunglasses (S). These choices are independent of each other but depend on the scarf colour: $\begin{array}{|l|l|l|} \hline \text { Scarf colour } & \mathbf{P ( H )} & \mathbf{P ( S )} \\ \hline \text { Red } & 0.5 & 0.8 \\ \hline \text { Blue } & 0.4 & 0.5 \\ \hline \text { Green } & 0.1 & 0.5 \\ \hline \end{array}$ Advika is spotted outdoors wearing both a Hat and Sunglasses. What is the probability that she is wearing the Red scarf? (A) $\frac{13}{313}$ (B) $\frac{8}{21}$ (C) $\frac{4}{9}$ (D) $\frac{8}{13}$

   ›Reveal solutionSolution

   Bayes' theorem on scarf colour given that both a hat and sunglasses are worn. Priors $P(R)=0.2,\ P(B)=0.4,\ P(G)=0.4$; likelihoods $P(H\cap S\mid\text{colour})=P(H)\,P(S)$. The posterior $P(R\mid H\cap S)=\frac{4}{9}$ — option (C).

   Concept. Hat and sunglasses are independent given the scarf, so $P(H\cap S\mid\text{colour})=P(H\mid\text{colour})\cdot P(S\mid\text{colour})$. Bayes' theorem then reverses the conditioning to give the probability of the scarf colour from the observed accessories.

   Step 1 — Priors.

   $P(R)=20\%=0.2,\qquad P(B)=2P(R)=0.4,\qquad P(G)=1-0.2-0.4=0.4.$

   Step 2 — Likelihood of wearing both accessories for each colour.

   $P(H\cap S\mid R)=0.5\times0.8=0.40,$

   $P(H\cap S\mid B)=0.4\times0.5=0.20,$

   $P(H\cap S\mid G)=0.1\times0.5=0.05.$ …

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9. COMEDK 2021Set 2021-B1 markMCQ
   Q.An urn contains 5 red and 5 black coloured balls. A ball is picked at random, its colour noted and then is put back into the urn. Now, additional two balls of the same colour are put into the urn and then a ball is drawn at random. What is the probability that the ball now chosen is red. (A) 7/24 (B) 1/2 (C) 5/8 (D) 2/3

   ›Reveal solutionSolution

   By symmetry the second-draw red probability averages to $\tfrac12$.

   Start: 5 red, 5 black (10 total). First ball is drawn, replaced, then 2 more of the same colour are added, so the urn always has 12 balls before the second draw.

   • First ball red ($P=\tfrac{5}{10}=\tfrac12$): urn now 7R, 5B ⇒ $P(\text{red})=\tfrac{7}{12}$. …

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10. COMEDK 2026Set 2026-A1 markMCQ
    Q.Vishnu has two jars of marbles, Jar A and Jar B. Jar A contains 3 yellow marbles and 2 green marbles. Jar B contains 4 yellow marbles and 3 green marbles. Vishnu flips a fair coin. If it lands heads, he picks two marbles at random without replacement from Jar A. If it lands tails, he picks two marbles at random with replacement from Jar B. Given that Vishnu picked one yellow and one green marble, what is the probability that they came from Jar B? (A) $\frac{21}{41}$ (B) $\frac{49}{89}$ (C) $\frac{40}{89}$ (D) $\frac{20}{41}$

    ›Reveal solutionSolution

    $P(E\mid A)=\tfrac35$ (without replacement), $P(E\mid B)=\tfrac{24}{49}$ (with replacement); Bayes gives $P(B\mid E)=\dfrac{40}{89}$ — option (C).

    Likelihoods of drawing one yellow and one green ($E$).

    Jar A (3 yellow, 2 green; two draws without replacement):

    $P(E\mid A)=\frac{\binom{3}{1}\binom{2}{1}}{\binom{5}{2}}=\frac{6}{10}=\frac{3}{5}.$

    Jar B (4 yellow, 3 green; two draws with replacement):

    $P(E\mid B)=2\cdot\frac{4}{7}\cdot\frac{3}{7}=\frac{24}{49}.$

    Bayes' theorem with $P(A)=P(B)=\tfrac12$ (the $\tfrac12$ cancels): …

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11. COMEDK 2025Set 2025-E1 markMCQ
    Q.A pot contains 5 red and 2 green balls. A ball is drawn at random from this pot. If a drawn ball is green, then a red ball is added to the pot. If a drawn ball is red, then a green ball is added to the pot, while the original ball drawn is not replaced in the pot. Now a second ball is drawn at random from the pot, what is the probability that the second ball drawn is a red ball? (A) $\frac{12}{49}$ (B) $\frac{32}{49}$ (C) $\frac{3}{7}$ (D) $\frac{27}{49}$

    ›Reveal solutionSolution

    Condition on the first draw: $P(\text{2nd red}) = \tfrac{2}{7}\cdot\tfrac{6}{7} + \tfrac{5}{7}\cdot\tfrac{4}{7} = \tfrac{32}{49}$.

    Start: $5$ red, $2$ green, total $7$. The first ball is not replaced, and a ball of the other colour is added.

    Case 1 — first ball is green. $P = \dfrac{2}{7}$.

    Remove that green ($5R,1G$), then add a red $\Rightarrow 6R,1G$ (total $7$).

    $P(\text{2nd red}\mid \text{green first}) = \frac{6}{7}$

    Case 2 — first ball is red. $P = \dfrac{5}{7}$.

    Remove that red ($4R,2G$), then add a green $\Rightarrow 4R,3G$ (total $7$). …

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12. COMEDK 2024Set 2024-A1 markMCQ
    Q.P and Q are considering to apply for a job. The probability that P applies for the job is $\frac{1}{4}$. The probability that $\mathrm{P}$ applies for the job given that $\mathrm{Q}$ applies for the job is $\frac{1}{2}$, and the probability that Q applies for the job given that P applies for the job is $\frac{1}{3}$. Then the probability that $\mathrm{P}$ does not apply for the job given that $\mathrm{Q}$ does not apply for the job is (A) $\frac{4}{5}$ (B) $\frac{7}{8}$ (C) $\frac{5}{6}$ (D) $\frac{11}{12}$

    ›Reveal solutionSolution

    We are given conditional probabilities and need to find $P(\overline{P} \mid \overline{Q})$. Using the definitions of conditional probability and the law of total probability, we compute $P(Q)$ and $P(\overline{Q})$, then apply Bayes' theorem to get $\frac{4}{5}$, which corresponds to option (A).

    Concept and intuition:

    This is a classic problem of working backwards from conditional probabilities to find a joint probability table. We know $P(P)$, $P(P \mid Q)$, and $P(Q \mid P)$. From these, we can find $P(P \cap Q)$ in two ways, which lets us solve for $P(Q)$. Then we can compute the desired conditional probability $P(\overline{P} \mid \overline{Q})$ using the complement rule and the definition of conditional probability.

    Step-by-step solution:

    1. Write down what is given. Let $P$ = event that P applies, $Q$ = event that Q applies. We have:

    $P(P) = \frac{1}{4}, \quad P(P \mid Q) = \frac{1}{2}, \quad P(Q \mid P) = \frac{1}{3}.$

    2. Use the definition of conditional probability to express $P(P \cap Q)$ in two ways. From $P(P \mid Q) = \frac{P(P \cap Q)}{P(Q)}$, we get

    $P(P \cap Q) = P(P \mid Q) \cdot P(Q) = \frac{1}{2} P(Q).$

    From $P(Q \mid P) = \frac{P(P \cap Q)}{P(P)}$, we get

    $P(P \cap Q) = P(Q \mid P) \cdot P(P) = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}.$

    3. Equate the two expressions for $P(P \cap Q)$ to find $P(Q)$.

    $\frac{1}{2} P(Q) = \frac{1}{12} \quad \Rightarrow \quad P(Q) = \frac{1}{6}.$

    4. Find $P(\overline{Q})$ and $P(P \cap \overline{Q})$.

    $P(\overline{Q}) = 1 - P(Q) = 1 - \frac{1}{6} = \frac{5}{6}.$

    Also, $P(P \cap \overline{Q}) = P(P) - P(P \cap Q) = \frac{1}{4} - \frac{1}{12} = \frac{3}{12} - \frac{1}{12} = \frac{2}{12} = \frac{1}{6}.$

    5. Find $P(\overline{P} \cap \overline{Q})$. Since $\overline{P} \cap \overline{Q}$ is the complement of $P \cup Q$, we can use:

    $P(\overline{P} \cap \overline{Q}) = 1 - P(P \cup Q).$

    First, $P(P \cup Q) = P(P) + P(Q) - P(P \cap Q) = \frac{1}{4} + \frac{1}{6} - \frac{1}{12}.$ …

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