A card is selected from a pack of 52 cards. (a) How many points are there in the sample space? (b) Calculate t

Exercise 14.2 · Q4

Q. A card is selected from a pack of 52 cards.

(a) How many points are there in the sample space?

(b) Calculate the probability that the card is an ace of spades.

(i) an ace

(ii) black card.

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Concept understanding — Classical Probability

Classical Probability: The "Fair Game" Definition

Imagine you roll a fair six-sided die. Before it lands, you know there are exactly six possible outcomes — 1, 2, 3, 4, 5, or 6 — and you have no reason to believe any one face is more likely than another. That gut feeling of "all outcomes are equally likely" is the entire foundation of classical probability.

The Intuition

Classical probability was born from games of chance — dice, coins, cards. In these settings, the physical symmetry of the objects (a balanced die, a fair coin) guarantees that no outcome is favoured. So the probability of an event is simply:

Number of ways the event can happen, divided by the total number of possible outcomes.

If you want the chance of rolling an even number on a die, count the evens: 2, 4, 6 — that's 3 ways. Total outcomes: 6. So probability = $3/6 = 1/2$ .

This is the "counting" approach. It works beautifully when the underlying experiment is symmetric and finite.

The Precise Statement

P(E) = \frac{\text{Number of outcomes favourable to event } E}{\text{Total number of equally likely outcomes}}

This is called the classical definition (or a priori definition) of probability. It was formalised by Pierre-Simon Laplace in the 18th century.

Three conditions must hold for this definition to apply:

Finite sample space — there are only a fixed, countable number of possible outcomes.
Equally likely outcomes — each outcome has the same chance of occurring (the "fairness" condition).
Mutually exclusive outcomes — no two outcomes can happen at the same time.

Watch out

The biggest mistake students make is applying classical probability to situations where outcomes are not equally likely. For example: "I can either pass or fail the exam — two outcomes, so probability of passing is $1/2$ ." That's nonsense, because passing and failing are not equally likely. The die works only because the die is fair.

A Simple Example

Problem: A bag contains 3 red marbles and 2 blue marbles. You pick one marble at random. What is the probability it is red?

Step 1 — Identify the sample space: There are 5 marbles total. If the marbles are physically identical except for colour, and you pick without looking, each marble is equally likely to be chosen. So total outcomes = 5.

Step 2 — Count favourable outcomes: 3 marbles are red. So favourable outcomes = 3.

Step 3 — Apply the formula:

P(\text{red}) = \frac{3}{5}

That's it. No deeper theory needed for this case.

When Classical Probability Fails

Classical probability cannot handle: …

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The sample space has 52 equally likely outcomes. The probability of drawing the ace of spades is $\frac{1}{52}$ ; of any ace is $\frac{1}{13}$ ; of a black card is $\frac{1}{2}$ .

Classical probability works here because every card in a well-shuffled pack has the same chance of being selected. The sample space is simply the set of all possible outcomes — in this case, all 52 distinct cards. No card is favoured over another, so we can count favourable outcomes and divide by the total number of equally likely outcomes.

Let’s go through each part.

Sample space size

A standard pack has 52 cards, each unique by suit and rank. So the number of points in the sample space is $52$ .
Probability of the ace of spades

There is exactly one ace of spades in the pack.

P(\text{ace of spades}) = \frac{\text{number of favourable outcomes}}{\text{total outcomes}} = \frac{1}{52}

Probability of an ace There are 4 aces in the pack (one per suit).

P(\text{ace}) = \frac{4}{52} = \frac{1}{13}

Probability of a black card Half the cards are black (spades and clubs). That’s 26 cards.

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 43 on this concept.

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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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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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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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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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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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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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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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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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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CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{2}{3}$
(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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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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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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Q.A card is selected from a pack of 52 cards.

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