Q.One card is drawn from a well-shuffled deck of 52 cards. If its outcome are equally likely, calculate the probability of a diamond card.

Showing the 12 most recent of 43 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.If a pair of dice is thrown, then the probability of getting an even prime number on each die will be

   (a) $1/3$

   (b) $1/12$

   (c) $1/36$

   (d) $0$

   ›Reveal solutionSolution

   The only even prime number is $2$; the probability both dice show $2$ is $1/6\times1/6$.

   Even prime numbers on a die (1–6): only $2$ qualifies (2 is even and prime).

   $P(\text{die shows }2)=\dfrac16$ for each die; the dice are independent.

   …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.A bag contains 3 black and 4 white balls. Probability of drawing one white ball is:

   (a) 4/7

   (b) 3/7

   (c) 1/7

   (d) 7/4

   ›Reveal solutionSolution

   Classical probability = favourable outcomes / total outcomes.

   Working: Total balls $= 3 \text{ black} + 4 \text{ white} = 7$. …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.Three coins are tossed once. Probability of getting 2 heads is

   (a) $\dfrac{3}{8}$

   (b) $\dfrac{1}{8}$

   (c) $\dfrac{1}{4}$

   (d) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   List the sample space (8 equally likely outcomes) and count how many have exactly 2 heads.

   Tossing 3 coins has sample space of $2^3=8$ equally likely outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.

   Outcomes with exactly 2 heads: HHT, HTH, THH — that's 3 outcomes.

   …

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4. CBSE 2026Set ANNUAL1 mark
   Q.A fair coin is tossed 3 times. Find the probability of getting 2 heads and 1 tail.

   ›Reveal solutionSolution

   Use the binomial-style count: choose which 2 of the 3 tosses are heads, over the 8 equally likely outcomes.

   When a fair coin is tossed 3 times, there are $2^3=8$ equally likely outcomes.

   …

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5. CBSE 2026Set ANNUAL1 mark
   Q.Case study: In Bhiwani zoo the lioness Geeta gave birth to three cubs. Their gender was not known at the time of the birth. The probability of male or female cub is equal. Probability of all female cubs.

   ›Reveal solutionSolution

   Each cub is independently male or female with probability 1/2; multiply the probabilities for 3 independent female outcomes.

   Each cub's gender is independent, with $P(\text{female}) = P(\text{male}) = \dfrac12$.

   For all three cubs to be female: …

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6. CBSE 2026Set ANNUAL1 mark
   Q.(Same lioness-cubs case study as 38(i).) Probability of two male and one female cub.

   ›Reveal solutionSolution

   Count the 3 arrangements (MMF, MFM, FMM) out of 8 equally likely gender outcomes for the three cubs.

   Each cub is independently male (M) or female (F) with probability $\dfrac12$ each, giving $2^3=8$ equally likely outcomes in total.

   …

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7. CBSE 2026Set ANNUAL1 mark
   Q.(Same lioness-cubs case study as 38(i).) Probability of one male and two female cubs.

   ›Reveal solutionSolution

   By symmetry with the 2-male case, there are 3 arrangements (MFF, FMF, FFM) out of 8 equally likely outcomes.

   …

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8. CBSE 2026Set ANNUAL1 mark
   Q.(Same lioness-cubs case study as 38(i).) Probability of all male cubs.

   ›Reveal solutionSolution

   By the same reasoning as the all-female case, all three independent male outcomes multiply to 1/8.

   By the same logic as "all female", with $P(\text{male})=\dfrac12$ per cub, independently: …

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9. CBSE 2026Set ANNUAL1 markMCQ
   Q.Ten coins are tossed. The Probability of getting at least 8 heads is:

   (a) $\dfrac{7}{16}$

   (b) $\dfrac{7}{64}$

   (c) $\dfrac{7}{128}$

   (d) $\dfrac{7}{32}$

   ›Reveal solutionSolution

   $P(\text{at least 8 heads})=\dfrac{{}^{10}C_8+{}^{10}C_9+{}^{10}C_{10}}{2^{10}}=\dfrac{56}{1024}=\dfrac{7}{128}$.

   Each coin toss is a fair Bernoulli trial with $p=1/2$. For 10 tosses, $P(\text{exactly }k\text{ heads})=\dfrac{{}^{10}C_k}{2^{10}}$.

   $P(\text{at least 8 heads})=P(8)+P(9)+P(10)=\dfrac{{}^{10}C_8+{}^{10}C_9+{}^{10}C_{10}}{2^{10}}$

   …

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10. CBSE 2026Set ANNUAL1 markMCQ
    Q.Assertion (A): The probability of getting a prime number when a die is thrown once is $\dfrac{2}{3}$. Reason (R): On the faces of a die, prime numbers are 2, 3, 5.

    (a) Both Assertion (A) and Reason (R) are correct and Reason (R) is the correct explanation of Assertion (A).

    (b) Both Assertion (A) and Reason (R) are correct, but Reason (R) is not the correct explanation of Assertion (A).

    (c) Assertion (A) is correct, but Reason (R) is incorrect.

    (d) Assertion (A) is incorrect, but Reason (R) is correct.

    ›Reveal solutionSolution

    Reason (R) correctly lists 2, 3, 5 as the primes on a die, but the resulting probability is $\dfrac{3}{6}=\dfrac12$, not $\dfrac23$ as Assertion (A) claims — so (A) is false and (R) is true.

    Checking Reason (R): A standard die shows faces $1,2,3,4,5,6$. Among these, the prime numbers are $2, 3, 5$ (note $1$ is not prime). So (R) correctly identifies the primes — (R) is TRUE.

    …

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11. CBSE 2025Set E1 markMCQ
    Q.The chance of getting a doublet in a throw of 2 dice is

    (a) $\frac{2}{3}$

    (b) $\frac{1}{6}$

    (c) $\frac{5}{6}$

    (d) $\frac{5}{36}$

    ›Reveal solutionSolution

    $6$ doublets among $36$ equally-likely outcomes gives probability $\frac{6}{36} = \frac{1}{6}$.

    Throwing two dice gives $6\times6 = 36$ equally likely outcomes. The doublets are $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$ — $6$ o …

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12. CBSE 2025Set A1 markMCQ
    Q.The probability of obtaining an odd prime number on each die, when a pair of dice is rolled, is:

    (a) $0$

    (b) $\dfrac{1}{3}$

    (c) $\dfrac{1}{18}$

    (d) $\dfrac{1}{36}$

    ›Reveal solutionSolution

    On a standard die (faces 1–6), the odd prime numbers are $3$ and $5$ — that's 2 favourable faces out of 6. For BOTH dice to show an odd prime, multiply the per-die probabilities.

    Step 1 — identify odd primes on a die. The numbers on a die are $1,2,3,4,5,6$. The primes among these are $2, 3, 5$. Of these, the ODD primes are $3$ and $5$ (2 is prime but even). So 2 out of 6 faces are favourable.

    $P(\text{odd prime on one die}) = \dfrac{2}{6} = \dfrac13$

    Step 2 — both dice. Since the two dice are independent, the probability that EACH die shows an odd prime is

    $P = \dfrac13 \times \dfrac13 = \dfrac19 = \dfrac{4}{36}$

    (Directly: favourable ordered outcomes are $(3,3),(3,5),(5,3),(5,5)$ — 4 outcomes out of the 36 equally likely outcomes of rolling two dice.)

    …

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