Find the probability that at least 2 defective bulbs are drawn, if 4 bulbs are drawn from a box containing 10 bulbs of which 3 are defective.

Question

Solution: 1 - (probability of 1 + probability of 0 defective).

• What is wrong in

(3C2 * 8C2) / 10C4

answer: 1/3

Answer 1

$$\frac{\binom{3}{2}\times \binom{7}{2}+ \binom{3}{3}\times \binom{7}{1}}{\binom{10}{4}}=\frac{1}{3}$$

Number of ways choosing $2$ defective out of $3$ and $2$ nondefective out of $7$ is $\binom{3}{2}\times \binom{7}{2}$

Number of ways choosing $3$ defective out of $3$ and $1$ nondefective out of $7$ is $\binom{3}{3}\times \binom{7}{1}$