For one particular process it has been found in the past that 2 per cent of the packets are underweight. An inspector takes a random sample of ten packets. Calculate (a) the expected number of packets in the sample which are underweight, (b) the probability that none of the packets in the sample are underweight. (c) the probability that more than one of the packets in the sample is underweight.

Step 1 ：This problem can be solved using the binomial distribution model, which is appropriate when the trials are independent, the number of trials is fixed, each trial outcome can be classified as a success or failure, and the probability of success is the same for each trial. In this case, the probability of success (a packet being underweight) is 0.02, and the number of trials is 10.

Step 2 ：For part (a), the expected number of successes in a binomial distribution is given by \(np\), where \(n\) is the number of trials and \(p\) is the probability of success. So, the expected number of packets in the sample which are underweight is \(10 \times 0.02 = \boxed{0.2}\).

Step 3 ：For part (b), the probability that none of the packets in the sample are underweight is given by the binomial probability formula \(P(X = k) = C(n, k) \times (p^k) \times ((1-p)^{(n-k)})\), where \(n\) is the number of trials, \(p\) is the probability of success, and \(k\) is the number of successes. In this case, \(k = 0\). So, the probability that none of the packets in the sample are underweight is approximately \(\boxed{0.817}\).

Step 4 ：For part (c), the probability that more than one of the packets in the sample is underweight is 1 minus the probability that none or only one of the packets in the sample is underweight. This can be calculated using the binomial probability formula for \(k = 0\) and \(k = 1\), and then subtracting these probabilities from 1. So, the probability that more than one of the packets in the sample is underweight is approximately \(\boxed{0.016}\).