Q.A box contains $6$ good and $4$ defective bulbs. Two bulbs are drawn at random without replacement. Find the probability that

Draws are without replacement, so use the multiplication theorem (or combinations); $10$ bulbs in all.

There are $6+4=10$ bulbs; two are drawn without replacement, so $n(S)=\binom{10}{2}=45$ equally likely pairs.

1. Both defective. Choose $2$ defective from $4$:

   $P(\text{both defective})=\frac{\binom{4}{2}}{\binom{10}{2}}=\frac{6}{45}=\frac{2}{15}.$

2. Exactly one defective. Choose $1$ defective from $4$ and $1$ good from $6$:

   $P(\text{exactly one defective})=\frac{\binom{4}{1}\binom{6}{1}}{\binom{10}{2}}=\frac{4\times 6}{45}=\frac{24}{45}=\frac{8}{15}.$

   Independent check by the multiplication theorem (ordered draws).

(i) $P=\dfrac{4}{10}\times\dfrac{3}{9}=\dfrac{12}{90}=\dfrac{2}{15}$ — matches. …

Combinations ($\binom{}{}$) ignore order and are quickest; the ordered multiplication method is a good cross-chec …

In (ii), counting only "defective then good" and forgetting "good then defective", which halves the answer to $\tfrac{4}{15}$. Also, using replacement fractions (e.g. …