## Thursday, February 26, 2015

### Unizor - Probability - Easy Problems 3

Problem A

Two rooks of different color are randomly put on a standard chess board (8x8).
What is the probability they are on the same horizontal or vertical line (that is, one can capture another)?

Solution

The sample space contains all the different positions of two rooks on the board. There are 64 positions for one rook and 63 remaining positions for another. That makes 64·63 elementary events in our sample space, all having the same probability of 1/(64·63).
Now, for every position of one rook, there are 7 positions of another to have both rooks on the same horizontal line and another 7 positions of the second rook to have both on the same vertical line. That make 14 positions of the second rook for each (out of 64) position of the first. The total number of positions of our pair of rooks when one can capture another is, therefore, 64·14.
We can determine the probability of two rooks standing in a position of capturing by multiplying the number of elementary events that satisfy this condition by a probability of each getting
P = (64·14)/(64·63) = 14/63 = 2/9

Problem B

26 letters of English alphabet are randomly written in one line, one after another.
What is the probability of letter Z to be to the right of letter A?

Solution 1 (straight forward)

Our random space contains 26! different permutations of letters. All permutations are equally probable with the probability 1/26!.
Let's count the number of positions of two letters, A and Z, when Z is to the right of A.
For A we have 25 different positions. Counting left to right, they are: 1, 2, 3,...,25 (position #26 cannot be taken since there will be no room for Z).
If A is at position #1, there are 25 different positions of Z that satisfy the condition of it being to the right of A.
If A is at position #2, there are 24 different positions of Z that satisfy the condition of it being to the right of A.
etc...
If A is at position #25, there is only 1 position of Z that satisfy the condition of it being to the right of A.
Therefore, the number of positions of a pair (A,Z) when Z is to the right of A equals to
25+24+...+1=26·25/2=325
With each of these there are 24! of different positions of the other letters. That makes the number of permutations of all letter with Z to the right of A equal to 325·24!.
Since the probability of each equals to 1/26!, the probability of Z to be to the right of A equals to
P = 325·24!/26! =
= 325/(26·25) = 13/26 = 1/2

Solution 2 (smart)

Since there are as many permutations of our letters with Z to the right of A as with Z to the left of A (since we can just exchange their positions), the number of permutations when Z is to the right of A equals to the number of permutations when Z is to the left of A. Therefore, the probabilities of both are equal and there can be no other cases.
Therefore, the probability of Z standing to the right of A is 1/2.

Problem C

Take 16 cards out of a standard deck - four Jacks, four Queens, four Kings and four Aces.
Let's disregard their suits and pay attention only to their rank: any Jack is lower then any Queen, any Queen is lower than any King, any King is lower than any Ace.
Now randomly put them in a row.
What is the probability of them being ordered by rank, that is four Jacks, followed by four Queens, then four Kings and, finally, four Aces?

Solution

This is a combinatorial problem. We have to calculate the number of permutations of 16 cards, knowing that they can be grouped into four groups by rank with cards within each group having the same rank.
The total number of permutations is 16!, but permutations that are different only by the order of cards within each group should be considered identical. Therefore, taking into account that we have 4 groups by 4 indistinguishable by rank cards in each group, we have to divide the total number of permutations by (4!)^4. So, the number of really different permutations is
16!/[(4!)^4]=63,063,000
Therefore, the probability of one specific permutation (when ranks are in order) is
1/63,063,000

Problem D

A square with a diagonal R is inscribed into a circle of a diameter R.
What is a probability that a point randomly placed inside a circle will be inside a square?

Solution
Obviously, the probability of a point to be inside a square equals to a ratio of an area of a square to an area of a circle.
The area of a square with a diagonal R equals to R^2/2.
The area of a circle with a diameter R is πR^2/4.
The ratio of the former to the latter, that is the probability we are calculating, equals to
(R^2/2)/(πR^2/4) = 2/π