Question 181 markMCQSection A

  1. A die is thrown once. Probability of getting a number other than 3 is : (a) 16\frac{1}{6} (b) 36\frac{3}{6} (c) 56\frac{5}{6} (d) 1

Probability
Probability of a simple event

Options

(A)16\frac{1}{6}
(B)36\frac{3}{6}
(C)56\frac{5}{6}
(D)1

Official answer

(c) 56\frac{5}{6}, since 5 of the 6 equally likely outcomes are not 3, giving P(not 3)=1−16P(\text{not } 3) = 1 - \frac{1}{6} = 56\frac{5}{6}.

Answer keywords:probabilitydie thrown oncefavourable outcomessample spacecomplementary event5/61 − P(3)equally likely

Marking scheme

  1. 1.1 mark for correct option (c) 56\frac{5}{6}.
  2. 2.Expected reasoning: 5 favourable outcomes out of 6, or 1−16=561 - \frac{1}{6} = \frac{5}{6}.
  3. 3.Accept either the direct count or the complement method.

Hint

Favourable outcomes are {1,2,4,5,6}\{1,2,4,5,6\} out of {1,2,3,4,5,6}\{1,2,3,4,5,6\}; or use 1−P(3)1 - P(3).

Quick oral answer

A die has six equally likely outcomes; five of them are not a 3, so the probability of getting a number other than 3 is five-sixths, or equivalently one minus one-sixth.

Common mistakes

  1. 1.Selecting 16\frac{1}{6}, which is the probability of getting a 3 rather than 'other than 3'.
  2. 2.Choosing 1, wrongly assuming a non-3 outcome is certain even though rolling a 3 is possible.
  3. 3.Miscounting favourable outcomes as 4 (giving 46\frac{4}{6}) by forgetting one of {1,2,4,5,6}\{1,2,4,5,6\}.

Analysis and explanation

Tests basic probability using the complement/favourable-outcomes approach on a single die throw.


Concept

  • Sample space for one die throw: {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}, so total outcomes = 6.
  • 'Number other than 3' excludes only the outcome 3, leaving 5 favourable outcomes: {1,2,4,5,6}\{1, 2, 4, 5, 6\}.
  • P(not 3)=56P(\text{not } 3) = \frac{5}{6}, also obtainable as 1−P(3)=1−16=561 - P(3) = 1 - \frac{1}{6} = \frac{5}{6}.

Common mistakes

  • Choosing 16\frac{1}{6} by mistake, which is actually the probability of getting a 3 (the opposite event).
  • Confusing 'other than 3' with 'greater than 3', which would give a different, smaller count.

Interesting facts

The probability of an event and its complement always add up to 1 — here P(3)+P(not 3)=16+56=1P(3) + P(\text{not } 3) = \frac{1}{6} + \frac{5}{6} = 1.

A standard cubic die has opposite faces that always sum to 7 (1–6, 2–5, 3–4), a design tradition dating back thousands of years.

Question details

Q18 of 53 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Basic)
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Probability
Topic
Probability of a simple event
Bloom's level
understand
Syllabus topic
Unit 6: Statistics and Probability — Probability: Probability of a simple event (complementary events)
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

How many favourable outcomes are there for 'a number other than 3'?

Five: the outcomes {1,2,4,5,6}\{1, 2, 4, 5, 6\}. Only the outcome 3 is excluded from the six faces of the die.

What is the complement method here?

P(not 3)=1−P(3)=1−16=56P(\text{not } 3) = 1 - P(3) = 1 - \frac{1}{6} = \frac{5}{6}. The complement of getting a 3 is getting any other number.

Why can't the answer be 1?

A probability of 1 means certainty, but it is possible to roll a 3, so 'not getting a 3' is not certain. Its probability is 56\frac{5}{6}.