# 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 *p*_{1},...,*p*_{n}, where *p*_{1}+...+*p*_{n}=4. Let *x*_{j}=1 if the *j*th 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 right-hand side are independent, so

Let (). To prove (1) it is sufficient to prove

(2) |

Leet if 0*t*<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 .

### 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