Q18

1 markMCQSection A

A die is thrown once. Probability of getting a number other than 3 is :

(a) $\frac{1}{6}$ (b) $\frac{3}{6}$ (c) $\frac{5}{6}$ (d) 1

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

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

Q: What is the complement rule in probability?

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

Probability

Probability of a Simple Event (Die Throw)

Options

(A)

\frac{1}{6}

(B)

\frac{3}{6}

(C)

\frac{5}{6}

(D)1

Official Answer

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

Marking Scheme

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

Hint

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

Quick Oral Answer

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

Analysis & 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' = $6 - 1 = 5$ (1, 2, 4, 5, 6).
$P(\text{not } 3) = \frac{5}{6}$ , equivalently $1 - P(3) = 1 - \frac{1}{6} = \frac{5}{6}$ .

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.

Common Mistakes

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

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 $\frac{1}{6}$ , making the probability of 'not that outcome' exactly $\frac{5}{6}$ .

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.

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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?

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