Consider this question:
Screws are sold in packets of 15. Faulty screws occur randomly. A large number of packets are tested for faulty screws and the mean number of faulty screws per packet is found to be 1.2. Damien buys 8 packets of screws at random. Find the probability that there are exactly 7 packets in which there is at least 1 faulty screw.
With this information given, I found out the Variance = 1.104
The probability = p = 0.08
The probability that a packet contains at most 2 faulty screws = 0.887. (I found this because it was a subpart of the question...)
How do I go about solving this question? Any help is appreciated
Let $p$ be the probability of a screw being faulty. Then (as you have already calculated) \begin{align} p &= \frac{1.2}{15} = 0.08\\ 1 - p &= 0.92 \end{align}
Probability of finding no faulty screws in a packet is $15 \choose 0$$(0.92)^{15}$. Hence the probability of finding at least faulty screw in a packet is $1-$$ 15 \choose 0$$(0.92)^{15}$.
Coming to your question, I am a bit confused, but this is what I think -
From 8 packets, we can choose 7 packets in $8\choose7$ ways. The probability of these packets having at least one faulty screw is $1-$$ 15 \choose 0$$(0.92)^{15}$ (as before). The remaining packet must have no faulty screws, with probability $15 \choose 0$$(0.92)^{15}$. Hence the total probability is - $8\choose7$$(1-$$ 15 \choose 0$$(0.92)^{15})$ + $15 \choose 0$$(0.92)^{15}$.