ESAT Maths 1 · free worked solution

ESAT Maths 1 conditional probability: restrict the denominator

Conditional-probability solution: condition on green draws before counting the 12 favourable outcomes.

Conditional probabilitySpecification M7.7Reviewed 23 August 2026

Answer first

Option B

The correct answer is

$12/49$

Once the question gives a condition, count only outcomes that satisfy that condition in the denominator.

This is original UniGenius practice material from the reviewed Maths 1 bank. It is not an official UAT-UK question, past paper, or recalled question.

The question

Read the setup, then choose the shortest valid route

A bag contains $4$ red, $3$ blue and $2$ green counters. Three counters are drawn without replacement. Given that at least one is green, what is the probability that exactly two are red?

A $6/49$
B $12/49$ Answer
C $14/49$
D $5/14$
E $1/7$

Worked solution

The route in 4 steps

The number of three-counter draws containing green is $\binom{9}{3}-\binom{7}{3}=84-35=49$ .
Exactly two reds with at least one green means two red and one green, giving $\binom{4}{2}\binom{2}{1}=6\times2=12$ favourable draws.
The conditional probability is therefore $12/49$ .
The restricted denominator is essential because the question conditions on observing at least one green.

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