Postby **amateur** » Thu Oct 09, 2008 11:34 am

Thanks for the compliment.

The following solution is a bit informal:

Regarding Q. 40, since x, y, and z are being selected randomly from [0, 1], we may assume that their values are equal to their probabilities.

Now consider selecting x randomly. The probability that x >= v is 1 - v.

For example, p(x >= 0) = 1 - 0 = 1, p(x >= 1) = 1 - 1 = 0...

Since y and z are selected randomly and independently, their values would be (on average), equal to 1/2 and 1/2 each, respectively. So yz = (1/2)(1/2) = 1/4 (on average).

Therefore p(x >= 1/4) = 1 - 1/4 = 3/4 = 75%

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For Q.66, Observe the following,

In the interval [1, 2n], there are n intervals each containing a pair of even/odd integers.

For example when n = 5, 2n = 10, the intervals are [1,2], [3,4], [5,6], [7,8], [9,10].

Since there are n+1 distinct "x" values to be selected, according to the Pigeonhole Principle, at least two of them must belong to the same pair. So we can deduce II quite easily.

Regarding I, the selection 1, x1=2, x2=3, x3=5, x4=6, x5=8, x6=10, 10 does not satisfy I.

Regarding III, since the number of primes diminishes with increasing n, so I am inclined to rule out III also. As a matter of fact, for n = 10, 2n = 20, the selection of the 11 integers 1, x1=4, x2=6, x3=8, x4=9, x5=10, x6=12, x7=14, x8=15, x9=16, x10=18, x11=20, 20 rules out III also.