A carton of 12 bulbs contains 3 defective ones. An electrician picks 2 bulbs at random, without replacement. What is the probability that at least one of the two is defective?

\frac{1}{22}

Correct answerYour answer

\frac{5}{11}

Correct answerYour answer

\frac{7}{16}

Correct answerYour answer

\frac{1}{4}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

Ways to pick 2 of 12: ${}^{12}C_{2} = 66$ .
Ways to pick 2 good bulbs (none defective): ${}^{9}C_{2} = 36$ .
P(at least one defective) $= 1 - \frac{36}{66} = \frac{30}{66} = \frac{5}{11}$ .

Worth noting

$\frac{7}{16} = 1 - \left(\frac{3}{4}\right)^2$ treats the draws as WITH replacement. $\frac{1}{4}$ is the chance for a single bulb, and $\frac{1}{22}$ is the chance that both are defective.

Concept: 'At least one' without replacement via combinations
Typical timeTypical: 120 s
Your timeYours: —

The rule this testsBasic probability

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A carton of 12 bulbs contains 3 defective ones. An electrician picks 2 bulbs at random, without replacement. What is the probability that at least one of the two is defective?

Explanation