Binomial distribution, In the mass production of bolts, it's found that $5$% are defective. Bolts are selected at random and put into packets of $10$. There's three parts of questions, I've solved the first and second, they are just normal binomial question. But that's the last part I couldn't. $2$ packets are selected at random. Find the probability that there are no defective bolts in both the packets.

Selecting two packets is the same as selecting 20 bolts. That makes this into a regular binomial problem, with $$ p = \binom{20}{20}\cdot 0.05^0\cdot0.95^{20} $$ or even just the product rule directly, with no binomial coefficients: $p = 0.95^{20}$.

Alternatively, if one of the earlier questions asked about the probability that a randomly chosen packet had no defective bolts, say the answer to that was $p_1$, then you can use that as your binomial sample: $$ p = \binom22p_1^2\cdot (1-p_1)^0 = p_1^2 $$