Advanced book on Mathematics Olympiad

Combinatorics and Probability 777

E(X^2 i)=E(Xi)=

1

n

and E(XiXj)= 1 · 1 ·P(Xi= 1 ,Xj= 1 )=

1

n(n− 1 )

.

HenceE(X^2 )= 1 + 1 =2.
On the other hand,

E(X^2 )=

∑n

k= 1

k^2

1

k!

n∑−k

i= 0

(− 1 )i

i!

,

which proves the second identity.
(proposed for the USA Mathematical Olympiad by T. Andreescu)

915.Denote byAithe event “the student solves correctly exactlyiof the three proposed
problems,’’i= 0 , 1 , 2 ,3. The eventAwhose probability we are computing is

A=A 2 ∪A 3 ,

and its probability is

P (A)=P(A 2 )+P(A 3 ),

sinceA 2 andA 3 exclude each other.
Because the student knows how to solve half of all the problems,

P(A 0 )=P(A 3 ) and P(A 1 )=P(A 2 ).

The equality

P(A 0 )+P(A 1 )+P(A 2 )+P(A 3 )= 1

becomes

2 [P(A 2 )+P(A 3 )]= 1.

It follows that the probability we are computing is

P (A)=P(A 2 )+P(A 3 )=

1

2

.

(N. Negoescu,Probleme cu... Probleme(Problems with... Problems), Editura Facla,
1975)

916.For the solution we will use Bayes’ theorem for conditional probabilities. We denote
byP (A)the probability that the eventAholds, and byP(BA)the probability that the event
Bholds given thatAin known to hold. Bayes’ theorem states that