Hi Chelsea, for P(A≤B≤C), lets assume that A,B,C all lie in the range of 0 to 1, and the value of B = p. Now for A≤B≤C to hold true, A must ≤ p and C must be≥ p

P(A≤p) = p and P(C≥p) = (1-p).

So the joint probability for the condition P(A≤B≤C|B=p) = p*(1-p)

Now this should hold for all ranges of p in 0 to 1. Hence,

P(A≤B≤C) = ∫ p*(1-p) for p = 0 to 1, or (p²/2 - p³/3) for p = 0 to 1, which is 1/6;

By symmetry, P(C≤B≤A) would also be equal to the same value. So

P(A≤B≤C or C≤B≤A) = 1/6 + 1/6 = 1/3

The other problem too is very similar to this.