Four items are chosen uniformly without replacement from ten, of which three are defective. Find the probability of at least two defectives.

01 / Standard#Your turn

Official paper · jm01-2024 · II.1(a) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Count samples containing two or three defectives.

Hint 2

Divide by the total number of four-item samples.

Worked solution

The two cases are disjoint.
$N=\binom32\binom72+\binom33\binom71=63+7=70$
Use the equally likely sample count.
$P=\frac{70}{\binom{10}4}=\frac{70}{210}=\frac13$

Probability 1/3.

Checks and common pitfalls: Sampling is without replacement, so a binomial model is inappropriate.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
The two cases are disjoint.
$N=\binom32\binom72+\binom33\binom71=63+7=70$
Use the equally likely sample count.
$P=\frac{70}{\binom{10}4}=\frac{70}{210}=\frac13$

Think first. Reveal a hint when the class is ready.

2024 JM01