a set of letters come together, do not come together in forming a word with letters which are all different
Problem 3
1 |
273 |
272 |
273 |
Solution
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 | = | (nL − ) + 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
= |
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⇒ P(A) | = |
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= |
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= |
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= |
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= |
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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)
Probability of non-occurrence of Event A
⇒ P(Ac) | = | 1 − P(A) | ||
= | 1 −
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= |
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= |
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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 | ||||
= |
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= | 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 | ||||
= |
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= | 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
= |
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⇒ P(B) | = |
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= |
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= |
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= |
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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 −
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= |
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= |
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