Circuit switched: each user needs 1/10 of link, so can reserve only 10 channels on the link, whether they are using it 10% or 100%.
Packet switched: Each user is using the channel 10% of the time, so probability of a given user being active is p = 0.1, and inactive q = 0.9. It's a binomial distribution X ~ B(35, 0.1), so probability Pr(X=k) = C(35, k) pk q(35-k). You need Pr(X>10) which is 1 - Pr(X<=10) which is 1-(Pr(X=0) + Pr(X=1) + ... Pr(X=10)).
I actually get 0.000424, not "less than 0.0004".
Calculate it here: https://stattrek.com/online-calculator/binomial.aspx
[EDIT to add] It needs to be said both of those include a great deal of assumption and modelling simplifications. In the case of circuit switching, you can do overbooking (in the way of airlines), especially if you know that the channels are only used a certain percentage and you have enough of them to average well. In the case of packet switching, the model assumes they are independent and evenly distributed; again these are reasonable if the number of hosts is large.