Answer to Question #210136 in Electrical Engineering for Bahawal Tahir

Question #210136

Question1

b) The English alphabet contains 21 consonants and 5 vowels. How many strings of

six lower case letters of the English alphabet contain:i) Exactly one vowel?

i) Exactly 2 vowels

III) At least 1 vowel

IV) At least 2 vowels)

c).If A and B are independent events, show that A and B are also independent. Are A And B independent?

d) Find Population Mean, median, mode and Sample Standard Deviation for the following

data set: 5, 10, 15, 20, 25, 30


1
Expert's answer
2021-06-28T08:53:16-0400

Part b)

i) Exactly 2 vowels

There are 26 letters in a total of which 21 are consonants and 5 are vowels. We are interested in strings contains 6 letters. 

Position vowel: 6 ways (as there are 6 letters in the string) 

Vowel: 5 ways

Second letter: 21 ways (needs to a consonant)

Third letter: 21 ways (needs to be a consonant)

Fourth letter: 21 ways (needs to be a consonant.)

Fifth letter: 21 ways (needs to be a consonant)

Sixth letter: 21 ways (needs to be a consonant) 

Use the product rule: 

"6* 5*21*21*21*21*21= 6*5*21^5= 122,523,030"

Thus there are 122,523,030 strings containing exactly one vowel. 


III) At least 1 vowel

There are 26 letters in a total of which 21 are consonants and 5 are vowels. We are interested in strings contains 6 letters. 

There are 26 possible letters for each letter in the string. By the product rule: 

Number of strings = "26* 26* 26* 26* 26 *26 = 26^6 = 308,915,776"

When the string contains no vowels, then there are 21 possible letters for each letter in the string. By the product rule: 

Number of strings with no vowels = "21 *21 *21* 21* 21* 21 = 21^6 = 85,766,121"

Strings that do not have any vowels will have at least one vowel. 

Number of strings with at least one vowel = Number of strings — Number of strings with no vowels = "26^6 - 21^6 = 308,915,776 - 85,766,121 = 223,149,655"


IV) At least 2 vowels

Number of strings with at least two Nrowels

= Number of strings — Number of strings with no Nrowels — Number of strings with at least one vowel

="26^6 \u2014 21^6 \u2014 6 * 5 * 21^5"

"= 308,915,776\u2014 85,766,121 \u2014 122,523,030 = 100,626,625"

c) The events A and B are independent, so, "P(A \u2229 B) = P(A) P(B)" .

"A = ( A \u2229 B) \u222a (A \u2229 B\u2019)."

Also, "P(A) = P[(A \u2229 B) \u222a (A \u2229 B\u2019)]."

or,"P(A) = P(A \u2229 B) + P(A \u2229 B\u2019)."

or, "P(A) = P(A) P(B) + P(A \u2229 B\u2019)"

or, "P(A n B\u2019) = P(A) \u2212 P(A) P(B) = P(A) (1 \u2013 P(B)) = P(A) P(B\u2019)"

d)Mean

"\\frac{ 5+ 10+ 15+ 20+ 25+ 30}{6}=17.5"

median

15, 20

"\\frac{ 15+ 20}{2}=17.5"

mode

None

Sample Standard

"s= \\sqrt{ \\frac{ \\sum(x_i- \\bar{x})^2}{N-1}}= \\sqrt{ \\frac{437.5}{6-1}}= \\sqrt{87.5}=9.35"


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