2 bulbs are drawn together from 10, of which 3 are defective. $X$ is the number of defective bulbs drawn. What is $P(X = 0)$ ?

\frac{7}{30}

Correct answerYour answer

\frac{3}{10}

Correct answerYour answer

\frac{7}{15}

Correct answerYour answer

\frac{49}{100}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

Both bulbs must come from the 7 good ones: ${}^{7}C_2 = 21$ of ${}^{10}C_2 = 45$ ways.
$P(X = 0) = \frac{21}{45} = \frac{7}{15}$ .

Worth noting

0.49 is $0.7 \times 0.7$ , which assumes the first bulb is put back.

Ten questions from Random Variables, Expectation & Variance, at your own pace.

Ten questions from Random Variables, Expectation & Variance, at your own pace.

2 bulbs are drawn together from 10, of which 3 are defective. XXX is the number of defective bulbs drawn. What is P(X=0)P(X = 0)P(X=0)?