Q.Probability of getting even numbers in a single throw of die.
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🔒 Start your 14-day free trial to unlock the full solution →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)=Total number of equally likely outcomesNumber of outcomes favourable to event E
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.
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(red)=53
That's it. No deeper theory needed for this case.
When Classical Probability Fails
Classical probability cannot handle: …
A die shows an even number in 3 of its 6 equally likely faces, so the probability is 63=21. …
P(even)=21; none of the printed options matches, so the options appear corrupted.
A fair die has outcomes {1,2,3,4,5,6}, all equally likely. The even outcomes are {2,4,6} — three of them.
P(even)=63=21.
…
Showing the 12 most recent of 43 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.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 1/6×1/6.
Even prime numbers on a die (1–6): only 2 qualifies (2 is even and prime).
P(die shows 2)=61 for each die; the dice are independent.
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- CBSE 2026Set ANNUAL1 markMCQQ.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 black+4 white=7. …
- CBSE 2026Set ANNUAL1 markMCQQ.Three coins are tossed once. Probability of getting 2 heads is(a) 83(b) 81(c) 41(d) 21
›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 23=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 markQ.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 23=8 equally likely outcomes.
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- CBSE 2026Set ANNUAL1 markQ.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(female)=P(male)=21.
For all three cubs to be female: …
- CBSE 2026Set ANNUAL1 markQ.(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 21 each, giving 23=8 equally likely outcomes in total.
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- CBSE 2026Set ANNUAL1 markQ.(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 markQ.(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(male)=21 per cub, independently: …
- CBSE 2026Set ANNUAL1 markMCQQ.Ten coins are tossed. The Probability of getting at least 8 heads is:(a) 167(b) 647(c) 1287(d) 327
›Reveal solutionSolution
P(at least 8 heads)=21010C8+10C9+10C10=102456=1287.
Each coin toss is a fair Bernoulli trial with p=1/2. For 10 tosses, P(exactly k heads)=21010Ck.
P(at least 8 heads)=P(8)+P(9)+P(10)=21010C8+10C9+10C10
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- CBSE 2026Set ANNUAL1 markMCQQ.Assertion (A): The probability of getting a prime number when a die is thrown once is 32. 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 63=21, not 32 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 markMCQQ.The chance of getting a doublet in a throw of 2 dice is(a) 32(b) 61(c) 65(d) 365
›Reveal solutionSolution
6 doublets among 36 equally-likely outcomes gives probability 366=61.
Throwing two dice gives 6×6=36 equally likely outcomes. The doublets are (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) — 6 o …
- CBSE 2025Set A1 markMCQQ.The probability of obtaining an odd prime number on each die, when a pair of dice is rolled, is:(a) 0(b) 31(c) 181(d) 361
›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(odd prime on one die)=62=31
Step 2 — both dice. Since the two dice are independent, the probability that EACH die shows an odd prime is
P=31×31=91=364
(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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