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: 18. 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) $\frac{5}{6}$, since 5 of the 6 equally likely outcomes are not 3, giving $P(\text{not } 3) = 1 - \frac{1}{6}$ = $\frac{5}{6}$.

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

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

Probability

Probability of a simple event

Options

(A)

\frac{1}{6}

(B)

\frac{3}{6}

(C)

\frac{5}{6}

(D)1

Official Answer

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

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

Marking Scheme

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

Hint

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

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.

Analysis & Explanation

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

Concept

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

Common mistakes

Choosing $\frac{1}{6}$ 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.

Common Mistakes

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

Interesting Facts

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

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.

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Frequently Asked Questions

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

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

What is the complement method here?

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