Advanced book on Mathematics Olympiad

Combinatorics and Probability 779

P(min(x◦,y◦)= 70 ◦)=P(x◦= 70 ◦)+P(y◦= 70 ◦)−P(max(x◦,y◦)= 70 ◦)

=a+b−c.

(29th W.L. Putnam Mathematical Competition, 1968)

919.In order fornblack marbles to show up inn+xdraws, two independent events
should occur. First, in the initialn+x−1 draws exactlyn−1 black marbles should be
drawn. Second, in the(n+x)th draw a black marble should be drawn. The probability
of the second event is simplyq. The probability of the first event is computed using the
Bernoulli scheme; it is equal to
(
n+x− 1
x

)

pxqn−^1.

The answer to the problem is therefore
(
n+m− 1
m

)

pmqn−^1 q=

(

n+m− 1

m

)

pmqn.

(Romanian Mathematical Olympiad, 1971)

920.First solution: Denote byp 1 ,p 2 ,p 3 the three probabilities. By hypothesis,

P(X= 0 )=

∏

i

( 1 −pi)= 1 −

∑

i

pi+

∑

i   =j

pipj−p 1 p 2 p 3 =

2

5

,

P(X= 1 )=

∑

{i,j,k}={ 1 , 2 , 3 }

pi( 1 −pj)( 1 −pk)=

∑

i

pi− 2

∑

i   =j

pipj+ 3 p 1 p 2 p 3 =

13

30

,

P(X= 2 )=

∑

{i,j,k}={ 1 , 2 , 3 }

pipj( 1 −pk)=

∑

i   =j

pipj− 3 p 1 p 2 p 3 =

3

20

,

P(X= 3 )=p 1 p 2 p 3 =

1

60

.

This is a linear system in the unknowns

∑

ipi,

∑

i   =jpipj, andp^1 p^2 p^3 with the solution

∑

i

pi=

47

60

,

∑

i   =j

pipj=

1

5

,p 1 p 2 p 3 =

1

60

.

It follows thatp 1 ,p 2 ,p 3 are the three solutions to the equation

x^3 −

47

60

x^2 +

1

5

x−

1

60

= 0.

Searching for solutions of the form^1 qwithqdividing 60, we find the three probabilities

to be equal to^13 ,^14 , and^15.