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E(Xa)· Furthermore, for each i, E(X1) = (~) 2 • Therefore, < (t2) r4log2 tl (~) 2(1og2t)(4log2t-l) (14log 2 t l ) ! (t2) r4log2 tl c~) (41og2t-l) (14log 2 tl)! t4 < (4log t)! 2 ' which is less than ~ for t sufficiently large. : 4log 2 t and the number of edges in the subgraph is at least ~I Ul (4log 2 t - 1). : 4log 2 t has a probability of at most ( ~ )2r log2 t- 5 of being counted towards Y 1 • 36 3. The First Moment Method Therefore, 1 E(Y) :S 2 L t r r= 4log2 < (t2) ( r tl t 2r r r= 4log2 tr - r!

In this case, the events Ae are mutually independent, and so the probability that none of them hold is exactly (1- 2-(k- 1 ) )m which is positive no matter how large m is. Therefore, the hypergraph is 2-colourable. 1 Of course for a general hypergraph, H, the events {Aele E E(1i)} are not independent as many pairs of hyperedges intersect. The Lovasz Local Lemma is a remarkably powerful tool which says that in such situations, as long as there is a sufficiently limited amount of dependency, we can still claim a positive probability of success.

5: We choose a random B-set by choosing for each w E 4 B one of the (88 ) possible lists, with each list equally likely to be chosen. Consider a particular vertex v E A. We will bound the probability that v is surrounded. Let X denote the number of subsets of {1, ... , s 4 } of sizes which do not 4 appear as a list on a neighbour of v. There are (88 ) possibilities for such a subset, and the probability that one particular subset does not appear 4 on any neighbour of vis at most (1- 1/(88 ))d. e.