Problem A. 543. (October 2011)
A. 543. The sum of the probabilities of some (finitely many) independent events is 4. Prove that with probability higher than 1/2, the number of occurred events is congruent to 0 or 3 modulo 4.
(Suggested by Endre Csóka, Budapest)
Deadline expired on 10 November 2011.
Solution. Let the probabilities be p1,...,pn, where p1+...+pn=4. Let xj=1 if the jth event occurs, otherwise let xj=0. Let N=x1+...+xn be the number of occuring events and let (r=0,1,2,3). We have to show that q0+q3>q1+q2.
Consider the random complex number S=iN. This number is respectively 1, i, -1 or -i, if the remainder of N modulo 4 is 0, 1, 2, or 3.
the statement is equivalent to
By the definition of S we have
The factors on the right-hand side are independent, so
Let (). To prove (1) it is sufficient to prove
Leet if 0t<1 and let . Then j=f(pj) for every j. We show that
From this (2) immediately follows because
Let g(t)=f(t)-t. For 0<t<1 we have , the function g increases. Therefore, g(t)g(0)=0 and f(t)t.
Now consider the function . Since , we have h'<0 in the intervals (0,2/7) and (5/7,1), and we have h'>0 in the interval (2/7,5/7). Therefore, h decreases in [0,2/7] and [5/7,1], and increases in [2/7,5/7]. Since h(0)=0 and , it follows that h is never positive, so h(t)0 and hence .