Question 1197443: Consider an experiment involving a random arrangement of the letters of the word
LEXICOGRAPHY.
a. Determine the cardinality of the sample space S, i.e., the number of possible
arrangements of the letters of the word LEXICOGRAPHY.
Define the random variable B to be the number of vowels in the first six (6) letters of the
the random arrangement of the letters of the word LEXICOGRAPHY.
b. What are the possible values of B? Determine the probability mass function of the random variable B. Your solution must be consistent with the sample space in part (a).
Show complete solutions for the PMF, following the set standard indicated in the
instructions written above. Express your final answer in a table and using the fractions.
c. Find the mean and variance of the random variable B.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
There are 12 letters in the word LEXICOGRAPHY.
There are 12 selections for the first slot, then 11 for the second, 10 for the third, and so on until we get to 1 selection for the last slot.
12! = 12*11*10*9*...*3*2*1 = 479,001,600
The exclamation mark indicates a factorial.
Since there are no repeated letters, we don't have to worry about overcounting.
Answer: 479,001,600
This value is a slightly larger than 479 million.
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Part (b)
I'll use X in place of the variable B.
X = number of vowels in the first 6 letters of the arrangement of LEXICOGRAPHY
The four vowels in that word are E, I, O, A. They are recorded in the order they come up in the word LEXICOGRAPHY.
This means we could have the following vowel count in the first 6 letters
X = 0 vowels
X = 1 vowel
X = 2 vowels
X = 3 vowels
X = 4 vowels
We'll use both the combination formula nCr and the permutation formula nPr to count how many sub-cases there are for X = 0 through X = 4.
The notation I'll use is C(n,r) for combinations.
I'll also use P(n,r) for permutations.
As a refresher, here are the formulas for each
C(n,r) = (n!)/(r!*(n-r)!)
P(n,r) = (n!)/((n-r)!)
An interesting connection is that
C(n,r) = P(n,r)/(r!)
which rearranges to
P(n,r) = (r!)*C(n,r)
which is handy if you know exactly one of the nPr or nCr values and want to find the other one.
On spreadsheet software, the commands for nPr and nCr are PERMUT (not PERMUTE, I don't know why they decided to leave the last E off) and COMBIN (the CHOOSE function is not what we want since that deals with pulling elements out of a set).
nPr = PERMUT
nCr = COMBIN
Example usage
Type in =COMBIN(10,3) to find the value of C(10,3) = 120.
Use the nCr combination formula mentioned above to verify this result.
Don't forget about the equal sign up front when typing in the COMBIN function.
Another example:
Type in =PERMUT(9,2) to compute P(9,2); the result of which is 72.
Don't forget about the equal sign up front.
Use the nPr formula mentioned above to confirm this result.
Let's get back to the problem at hand.
X goes from x = 0 to x = 4.- P(4,x) represents the number of ways to pick the x vowels to be somewhere in the first 6 slots.
- C(6,x) has us figure out where those x vowels go in those first six slots.
- P(8,6-x) then computes the number of ways to pick the consonants for the remaining 6-x slots (somewhere in the first 6 slots).
- P(6,6) = 720 is the number of ways to select the remaining 6 letters (slot7,slot8,...,slot12)
We'll multiply out those items.
Here's what the spreadsheet looks like before each item is calculated
I'm using LibreOffice spreadsheet, which is free software. Those commands work in Excel and similar spreadsheet programs.
Calculating each item gets us this
Use the command SUM to add up all the values in column F.
The command is =SUM(F2:F6) to sum the values from F2 to F6.
You should get 479,001,600 as that sum.
Which if you recall, this is the result of part (a).
This makes sense because X = 0 through X = 4 represent all of the possible ways to partition the sample space.
This is what your teacher meant when s/he wrote "Your solution must be consistent with the sample space in part (a)". This is a checksum of sorts.
Column F represents the frequency for cases X = 0 through X = 4.
For example, there are exactly 14,515,200 permutations in which there aren't any vowels in the first six slots.
What we'll do is form a new column where we divide values in column F over the grand total 479001600. This is to determine the relative frequency for each case X = 0 through X = 4.
For example, X = 0 has a frequency of 14,515,200 and relative frequency of 14515200/479001600 which reduces to 1/33.
Do this for the other values of X as well.
Here's how to quickly reduce fractions
In cells G1 and G2 type the numerator and denominator. Both of which need to be integers.
In cell H1, we'll have this command =G1/GCD(G1,G2)
In cell H2, we'll have this command =G2/GCD(G1,G2)
Therefore, the fraction (G1)/(G2) reduces fully to (H1)/(H2).
GCD = greatest common divisor, it's the same as the GCF
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Answers:
What are the possible values of B? 0, 1, 2, 3, and 4
The PMF is this table
| B | P(B) | | 0 | 1/33 | | 1 | 8/33 | | 2 | 5/11 | | 3 | 8/33 | | 4 | 1/33 |
PMF = probability mass function
Side notes:
As a check, the P(B) values need to add up to 1.
5/11 = 15/33
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Part (c)
I'll use X in place of the variable B.
Here's the PMF from part (b) earlier.
| X | P(X) | | 0 | 1/33 | | 1 | 8/33 | | 2 | 5/11 | | 3 | 8/33 | | 4 | 1/33 |
Form a third column labeled X*P(X)
| X | P(X) | X*P(X) | | 0 | 1/33 | 0/33 | | 1 | 8/33 | 8/33 | | 2 | 5/11 | 30/33 | | 3 | 8/33 | 24/33 | | 4 | 1/33 | 4/33 |
Add up the values in the X*P(X) column and you should get 2 exactly.
This is the mean of X
mu is the greek letter that represents the mean.
Then form a fourth column labeled (X-mu)^2*P(X)
| X | P(X) | X*P(X) | (X-mu)^2*P(X) | | 0 | 1/33 | 0/33 | 4/33 | | 1 | 8/33 | 8/33 | 8/33 | | 2 | 5/11 | 30/33 | 0/33 | | 3 | 8/33 | 24/33 | 8/33 | | 4 | 1/33 | 4/33 | 4/33 |
Add up everything in the fourth column to get the fraction 24/33, which reduces to 8/11. This is the variance of X.
Side note: Apply the square root to the variance to get the standard deviation.
Answers:
Mean = 2
Variance = 8/11
Convert the variance to decimal form if your teacher requires it.
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