Mathematical and Physical Journal
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# 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 November 10, 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.

Since

E(S)=q0+q1i+q2(-1)+q3(-i),

the statement is equivalent to

 (1)

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

 (2)

Leet if 0t<1 and let . Then j=f(pj) 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 .

### Statistics:

 6 students sent a solution. 5 points: Ágoston Tamás, Gyarmati Máté, Omer Cerrahoglu, Strenner Péter. 3 points: 1 student. 2 points: 1 student.

Problems in Mathematics of KöMaL, October 2011