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)
(5 pont)
Deadline expired on 10 November 2011.
Solution. Let the probabilities be p_{1},...,p_{n}, where p_{1}+...+p_{n}=4. Let x_{j}=1 if the jth event occurs, otherwise let x_{j}=0. Let N=x_{1}+...+x_{n} be the number of occuring events and let (r=0,1,2,3). We have to show that q_{0}+q_{3}>q_{1}+q_{2}.
Consider the random complex number S=i^{N}. This number is respectively 1, i, 1 or i, if the remainder of N modulo 4 is 0, 1, 2, or 3.
Since
E(S)=q_{0}+q_{1}i+q_{2}(1)+q_{3}(i),
the statement is equivalent to
 (1) 
By the definition of S we have
The factors on the righthand side are independent, so
Let (). To prove (1) it is sufficient to prove
 (2) 
Leet if 0t<1 and let . Then _{j}=f(p_{j}) for every j. We show that
 (3) 
From this (2) immediately follows because
and
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 .
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