Mathematics Challenge Questions
(I) ABC plays two games, Game X and Game Y.
Playing Game X and game Y are independent
events.
1. The
probability of ABC winning both games is 9/25.
2. The
probability of ABC winning Game Y is four times greater than the probability of
losing Game X.
Find the probability of ABC winning only one of
the two games he plays.
Solution -
(II) A Bag contains some black and white balls, all of the same size and shape.
The chance
of getting the two same color balls is exactly 0.5; if I put my hands in the
bag and simultaneously and randomly pull out two balls.
Prove the above statement and show the full steps.
Solution
Suppose that there are b black balls
and w white balls in the
bag and that there are at least two of each (we can think about the other
cases at the end).
Then
P(bb)P(wb)P(bw)P(ww)====bb+w⋅b−1b+w−1bb+w⋅wb+w−1wb+w⋅bb+w−1wb+w⋅w−1b+w−1
Although these look complicated, the denominators in each case are the same.
Since the probabilities are mutually exclusive we know that
P(bb)+P(ww)=0.5
Thus,
2b(b−1)+2w(w−1)= (b+w)(b+w−1)
Expanding and rearranging gives us
b2−b+w2−w−2wb=0
So, where are we now? We've applied the probability and are left with an
expression involving w and b. This can be factorized as
(b−w)2=b+w
Some thought should convince you that this expression is satisfied if b and w are consecutive
triangle numbers. But we need to ask: are there any other solutions? We need to
ask this because for any choice of w there will be up to 2 solutions for b upon solving the equation. The quadratic formula tells us that for
any given w
b=1+2w±1+8w−−−−−−√2
Now, we need to impose the condition that b and w are natural numbers. This would require that 1+8w is a square number, say N2. Then
w=N2−18=(N+1)(N−1)8
For w to be a natural
number, N must be an odd
number, which means that (N+1) and (N−1) are consecutive
even numbers, which means that N+12 and N−12 are consecutive natural numbers. Thus, w is half the
product of two natural numbers which means that w, by definition must be triangular. Thus there is a solution in this case
if and only if w and b are consecutive triangle numbers bigger than 1.
Finally, we consider the special cases where there are less than two of each
ball.
Clearly, the balls cannot all be the same color, as then the chance of drawing
two balls of the same color would be 1. Clearly, there cannot be exactly one ball of each color, as then the
chance of drawing two balls of the same color would be 0. If there were exactly
one black ball and w>1 white ball then
the chance of drawing two balls of the same color would be
ww+1⋅w−1w=0.5
This has exactly the two solutions w=0,3. Since 3 is also
triangular, our rule still holds.
Putting this all together proves that a magic bag can, and only can,
contain consecutive triangle numbers of white and black balls.
(III) There are 03 circles with centers A, B, and C.
Each Circle has a radius 4.
Find out the area of a blue-colored sector.
Solution -
(IV) Digits 1, 2,3,4,5,6,7,8, and 9 are represented by
a different letter in the figure below:
A
B C
D
E
F G
H
I
Each of
A+B+C, C+D+E, E+F+G, and G+H+I is equal to 13.
Which digit does E represent?
