3
$\begingroup$

"A sample space is the set of all possible outcomes of a random process or experiment. An event is a subset of a sample space" enter image description here For any finite set A, N(A) denotes the number of elements in A. enter image description here enter image description here

Suppose $A[1], A[2], A[3], . . . , A[n]$ is a one-dimensional array and $n ≥ 50$.

a. How many elements are in the array?
b. How many elements are in the subarray
$A[4], A[5], . . . , A[39]$?
c. If $3 ≤ m ≤ n$, what is the probability that a randomly chosen array element is in the subarray $A[3], A[4], . . . , A[m]$?
d. What is the probability that a randomly chosen array element is in the subarray shown below if n = 39?
$A[\lfloor \frac n2 \rfloor], A[\lfloor \frac n2 \rfloor + 1], . . . , A[n]$

Source: Discrete Mathematics with Applications Susanna S. Epp I'm not sure about my answer for d. Can you show how to solve the question d?

My answer
a. $n$
b. $36 $
c. $\frac {m-2}n$
d(edited).

$A[\lfloor \frac {39}{2} \rfloor]=A[19], A[\lfloor \frac {39}{2} \rfloor + 1]=A[20], . . . , A[39]$ has 39-18=21 elements

So $\frac {21}{39}$

$\endgroup$

1 Answer 1

1
$\begingroup$

I too think you might have made a mistake. Firstly, there is no $m$ or infact, no variable in the question.. unless the question was -

What is the probability that a randomly chosen array element is in the subarray shown below if $m = 39$? $A[\lfloor \frac m2], A[⌊\frac m2⌋ + 1], . . . , A[m]$

If so, then the answer would be-
$⌊\frac m2⌋= 19$ and so, there are $21$ elements.

The probability would then be $\frac {21}n$

If the original question is right though, then the answer would be -

$\frac {21}{39}$ putting $n=39$ above.

$\endgroup$
2
  • $\begingroup$ There are 39-18=21 elements in an array A[39], ..., A[$\lfloor \frac {m}{2} \rfloor$ + 1 = 20] A[$\lfloor \frac {m}{2} \rfloor$ = 19] $\endgroup$
    – buzzee
    Mar 7, 2016 at 8:34
  • $\begingroup$ @buzzee Thank you. Edited. Anyway, you already got it. $\endgroup$ Mar 7, 2016 at 12:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.