Question 181 markMCQSection A

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 (Die Throw)

Options

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

Official answer

(c) 5/6 — favourable outcomes for 'not 3' = 5 out of 6 total; equivalently 1−P(3)=1−16=561 - P(3) = 1 - \frac{1}{6} = \frac{5}{6}.

Answer keywords:probability of a diecomplement ruleequally likely outcomesclassical probabilitysample space of a dieP(not A) = 1 − P(A)

Marking scheme

  1. 1.1 mark: correctly selecting option (c) 56\frac{5}{6} using either direct counting (5 favourable outcomes out of 6) or the complement rule 1−161 - \frac{1}{6}.
  2. 2.No partial marks for MCQs; working may be shown for verification only.

Hint

Use the complement rule: P(not getting 3)=1−P(getting 3)=1−16P(\text{not getting } 3) = 1 - P(\text{getting } 3) = 1 - \frac{1}{6}.

Quick oral answer

There are 6 equally likely outcomes on a die; 5 of them are not 3, so P(number other than 3)=56P(\text{number other than } 3) = \frac{5}{6}, which also equals 1 minus P(getting 3)=1−16P(\text{getting } 3) = 1 - \frac{1}{6}.

Common mistakes

  1. 1.Selecting 16\frac{1}{6} (the probability of getting exactly 3) instead of the probability of NOT getting 3, due to careless reading of the question.
  2. 2.Incorrectly counting the favourable outcomes for 'other than 3' as 4 or 6 instead of the correct 5.
  3. 3.Forgetting that all six faces of a fair die are equally likely, leading to an incorrect total in the denominator.

Analysis and explanation

Tests the complement rule of probability using a single die throw.


Concept

  • Total equally likely outcomes on one die throw = 6; the event 'not 3' is the complement of 'getting 3'.

Key points

  • Favourable outcomes for 'other than 3' = 6−1=56 - 1 = 5 (1, 2, 4, 5, 6).
  • P(not 3)=56P(\text{not } 3) = \frac{5}{6}, equivalently 1−P(3)=1−16=561 - P(3) = 1 - \frac{1}{6} = \frac{5}{6}.

Common mistakes

  • Misreading the question and directly picking P(getting 3) = 1/6 instead of its complement.

Real-world

  • The complement rule (P(not A) = 1 − P(A)) is widely used in reliability engineering, risk assessment, and quality control, where it's easier to compute failure probability and subtract from 1 to get success probability.

Interesting facts

A standard six-sided die has been used as a randomizing tool since at least 3000 BCE, with dice found in ancient Mesopotamian archaeological sites.

Since each of the six faces is equally likely, the probability of any single specific outcome (like rolling a 3) is exactly 16\frac{1}{6}, making the probability of 'not that outcome' exactly 56\frac{5}{6}.

Dice problems like this one form the historical foundation of probability theory, which began in the 17th century with correspondence between Pascal and Fermat about gambling with dice.

Question details

Q18 of 56 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Standard)
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Probability
Topic
Probability of a Simple Event (Die Throw)
Bloom's level
apply
Syllabus topic
Unit VII: Statistics and Probability — Probability: Classical (theoretical) probability using a single die
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

How many total outcomes are there when a die is thrown once?

There are 6 equally likely outcomes: 1, 2, 3, 4, 5, and 6.

What is the complement rule in probability?

P(not A)=1−P(A)P(\text{not } A) = 1 - P(A). Here, P(not getting 3)=1−P(getting 3)=1−16=56P(\text{not getting } 3) = 1 - P(\text{getting } 3) = 1 - \frac{1}{6} = \frac{5}{6}.