Advanced book on Mathematics Olympiad

780 Combinatorics and Probability

Second solution: Using the Poisson scheme

(p 1 x+ 1 −p 1 )(p 2 x+ 1 −p 2 )(p 3 x+ 1 −p 3 )=

2

5

+

13

30

x+

3

20

x^2 +

1

60

x^3 ,

we deduce that 1−p^1 i,i = 1 , 2 ,3, are the roots ofx^3 + 9 x^2 + 26 x+ 24 =0 and

p 1 p 2 p 3 = 601. The three roots are− 2 ,− 3 ,−4, which again givespi’s equal to^13 ,^14 ,
and^15.
(N. Negoescu,Probleme cu...Probleme(Problems with...Problems), Editura Fa-
cla, 1975)

921.Setqi= 1 −pi,i= 1 , 2 ,...,n, and consider the generating function

Q(x)=

∏n

i= 1

(pix+qi)=Q 0 +Q 1 x+···+Qnxn.

The probability for exactlykof the independent eventsA 1 ,A 2 ,...,Anto occur is equal
to the coefficient ofxkinQ(x), hence toQk. The probabilityPfor an odd number of
events to occur is thus equal toQ 1 +Q 3 +···.Let us compute this number in terms of
p 1 ,p 2 ,...,pn.
We have

Q( 1 )=Q 0 +Q 1 +···+Qn and Q(− 1 )=Q 0 −Q 1 +···+(− 1 )nQn.

Therefore,

P=

Q( 1 )−Q(− 1 )

2

=

1

2

(

1 −

∏n

i= 1

( 1 − 2 pi)

)

.

(Romanian Mathematical Olympiad, 1975)

922.It is easier to compute the probability of the contrary event, namely that the batch
passes the quality check. Denote byAithe probability that theith checked product has
the desired quality standard. We then have to computeP(∩^5 i= 1 Ai). The events are not
independent, so we use the formula

P(∩^5 i= 1 Ai)=P(A 1 )P (A 2 /A 1 )(A 3 /A 1 ∩A 2 )P (A 4 /A 1 ∩A 2 ∩A 3 )

×P(A 5 /A 1 ∩A 2 ∩A 3 ∩A 4 ).

We find successivelyP(A 1 )= 10095 ,P(A 2 /A 1 )=^9499 (because ifA 1 occurs then we are left
with 99 products out of which 94 are good),P(A 3 /A 1 ∩A 2 )=^9398 ,P(A 4 /A 1 ∩A 2 ∩A 3 )=
92
97 ,P(A^5 /A^1 ∩A^2 ∩A^3 ∩A^4 )=

91

96. The answer to the problem is