a set of letters come together, do not come together in forming a word with letters which are all different

Problem 3

In a random arrangement of the letters of the word uncopyrightable. Find (a) the probability and odds that all the vowels come together and (b) the probability that the vowels do not come together
Ans :
1
273
, 1 : 272, 272 : 1,
272
273

Solution

In the word uncopyrightable,

Number of Letters/Characters

= 15

{U, N, C, O, P, Y, R, I, G, H, T, A, B, L, E}

⇒ nL = 8

Number of Vowels

= 5

{U, O, I, A, E}

Number of Consonants

= 10

{N, C, P, Y, R, G, H, T, B, L}

Experiment :

Forming a word using the 15 letters

Total Number of Possible Choices

= Number of words that can be formed using the 15 letters

⇒ n = 15P15
= 15!

When there are large factorial values, using the factorial form would reduce the burden of calculations

Let

A : the event of the word formed with all the vowels together

B : the event of the word formed with all the vowels not coming together

For Event A

Number of groups of letters

= 1

⇒ nG = 1

Number of letters in the first group

⇒ Number of Vowels

= 5

⇒ nGL1 = 5

Total Number of letters considering the groups of letters as a unit

⇒ Number of Vowels

= Total number of letters − Number of letters grouped + Number of groups of letters

⇒ nGL = (nLi=1nGnGLi) + nG
= nL − (nGL1) + nG
= 15 − (5) + 1
= 11

{N, C, (U, O, I, A, E), P, Y, R, G, H, T, B, L}

Number of Favorable Choices

= Number of words that can be formed with the letters of the word uncopyrightable with the vowels coming together

= Number of ways in which the letters with the grouped letters as a unit can be arranged × Number of ways in which the letters of the first group (vowels) can be inter arranged between themselves

⇒ mA = nGLPnGL × nGL1PnGL1
= nGL! × nGL1!
= 11! × 5!

Probability of forming the word using all the letters of the word uncopyrightable with the vowels coming together

⇒ Probability of occurrence of Event A

=
Number of Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(A) =
mA
n
=
11! × 5!
15!
=
11! × 5 × 4 × 3 × 2 × 1
15 × 14 × 13 × 12 × 11!
=
1
3 × 7 × 13
=
1
273

Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices

⇒ mAc = n − mA
= 15! − (11! × 5!)
= (15 × 14 × 13 × 12 × 11!) − (11! × 120)
= (32,760 − 120) × 11!
= 32,640 × 11!

in favor

Odds in Favor of forming the words with the vowels together

⇒ Odds in Favor of Event A

= Number of Favorable Choices : Number of Unfavorable Choices

= mA : mAc

= 11! × 5! : 32,640 × 11!

= 120 : 32,640

= 1 : 272

against

Odds against forming the words with the vowels together

⇒ Odds against Event A

= Number of Unfavorable Choices : Number of Favorable Choices

= mAc : mA

= 32,640 × 11! : 11! × 5!

= 32,640 : 120

= 272 : 1

Odds (alternative)

Once we find the probability for an event, it would be easier finding the odds using probabilities. This would eliminate the need to find the number of unfavorable choices and simplification.

Probability of non-occurrence of Event A

⇒ P(Ac) = 1 − P(A)
= 1 −
1
273
=
273 − 1
273
=
272
273

in favor

Odds in Favor of forming the words with the vowels together

⇒ Odds in Favor of Event A

= Probability of occurrence of the event : Probability of non-occurrence of the event
=
1
273
:
272
273
= 1 : 272

against

Odds against forming the words with the vowels together

⇒ Odds against Event A

= Probability of non-occurrence of the event : Probability of occurrence of the event
=
272
273
:
1
273
= 272 : 1

For Event B

Number of Favorable Choices

= Number of words that can be formed with the letters of the word uncopyrightable with the vowels not coming together

= Total number of words that can be formed with the 15 letters of the word uncopyrightable − Number of words that can be formed with all the vowels together

⇒ mB = mAc
= 32,640 × 11!

Probability of forming the word using all the letters of the word uncopyrightable with the vowels not coming together

⇒ Probability of occurrence of Event B

=
Number of Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(B) =
mB
n
=
32,640 × 11!
15!
=
32,640 × 11!
15 × 14 × 13 × 12 × 11!
=
272
273

For Event B [Alternative]

Probability of forming the word using all the letters of the word uncopyrightable with the vowels not coming together

⇒ Probability of occurrence of Event B

= Probability of non-occurrence of Event A

⇒ P(B) = P(Ac)
= 1 − P(A)
= 1 −
1
273
=
273 − 1
273
=
272
273