MATHEMATICS ARRANGEMENTS QUESTION ANSWER 2022

Running head: MATHEMATICS
MATHEMATICS
Name of the Student:
Name of the University:
Author Note

1
MATHEMATICS
Answer to question number a
For solving the problem, the following information is required to be considered:
For a.b to be odd both the numbers a and b should be odd.
That is, they should end with odd numbers.
Therefore, fifth digit and the tenth digit in the arrangement should be odd so that X = 0.
Total number of cases = 10! = 3628800
Number of case for X = 0, is 5x4. 8!. = 806400
Now, for the cases where the product of the numbers are even, the following criteria can be
considered:
Therefore, number of case for X = 1, 3628800 – 806400 = 2822400.
Therefore, E (X) = Sum of X(i) * f(i)/ f(i) = 2822400/3628800 = 0.7778
Answer to question number 1
Given that X1 and X2 are independent Bernoulli random variables with the parameter p1 and p2.
P [X1 < X2] = 1/6 and P [X1 + X2 <= 1] = 5/6
Therefore, P [X1 + X2 <= 1] = 5/6
Or, P [X1 < X1 + X2 <= 1] = 5/6
Therefore, p1 = 1/3 and p2 = 1/2
Answer to question number 2
Given that, Val(X)=Val(Y) = [0..99].

2
MATHEMATICS
Part a
For m the number of possibilities = 98
For n the number of possibilities = 98
Therefore, ∫
0
98
∫
1
99
f ( x , y ) .dx . dy
= 99/98.98 = 0.0103.
Part b
Pr [X + Y < 10]
Or, Pr [2X < 10]
Or, Pr [X < 5] = 4/100 = 1/25 = 0.04
Answer to question number 3
All the permutations are to identified for this system. The combinations are required to be
rejected as there is no requirement of arrangements for this particular set of problem.
The number of letter generated by the computer is 8.
The set of letter for selection is, let x = {A, B, C, D, E}
Therefore, the total sample size = 5^8 = 390625.
Part a
Sample size for all the letters to be same = 5
This this because the n(x) or the cardinality of the given set of objects are 5.