probability of forming words with letters which are all not different and specified letters do not come together
Problem 3
42 |
143 |
Solution
Number of letters
= 13
{A, S, S, A, S, S, I, N, A, T, I, O, N}
⇒ nL = 13
Number of Letters
of a first kind
Number of A's = 3
⇒ nLa = 3
of a second kind
Number of S's = 4
⇒ nLb = 4
of a third kind
Number of I's = 2
⇒ nLc = 2
of a fourth kind
Number of N's = 2
⇒ nLd = 2
Number of places in the word to be formed,⇒ Number of Letters in the word to be formed
= 13
⇒ nP = 13
Experiment :
Forming a word using the 13 letters
Total Number of Possible Choices
= Number of words that can be formed using the 13 letters of which nLa are of one kind, nLb are of another kind, nLc are of a third kind and nLd are of another kind
⇒ n | = |
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= |
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= |
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= | 13 × 11 × 5 × 3 × 7! | |||
= | 2,145 × 7! |
Let
A : the event of no two S's coming together in the word formed
For Event A
Number of Letters to stay separate
⇒ Number of S's
= 4
⇒ nDL = 4
Number of other letters
= Total number of letters − Number of letters to stay separate
⇒ nOL | = | nL − nDL |
= | 13 − 4 | |
= | 9 |
Number of places to place the letters to stay separate
= Number of other letters + 1
⇒ nDP | = | nOL + 1 |
= | 9 + 1 | |
= | 10 |
In other letters
Number of Letters
of a first kind
Number of A's = 3
⇒ nOLa = 3
of a second kind
Number of I's = 2
⇒ nOLb = 2
of a third kind
Number of N's = 2
⇒ nOLc = 2
In letters to be separated
In arranging S's in the specified places, nDP > nDL and there are repeating letters. Consider space to be a letter and assume that there is word formed with the letters and spaces.
S | S | S | S | _ | _ | _ | _ | _ | _ |
Number of Letters
of a first kind
Number of S's = 4
⇒ nDLa = 4
of a second kind
Number of spaces = 6
⇒ nDLb = 6
Number of Favorable Choices
= Number of words that can be formed using the letters of the word ASSASSINATION with no two S's together
= Number of ways in which the nOL Other letters can be arranged in as many places creating the nDP spaces on either side for placing the nDL letters to be kept separated × Number of ways in which the nDL letters to be kept separated can be arranged in the nDP spaces created
⇒ mA | = |
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= |
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= |
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= | (3 × 7!) × (10 × 3 × 7) | |||||
= | 630 × 7! |
Probability of forming words using all the letters of the word ASSASSINATION such that no two S's come 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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Odds
Number of Unfavorable Choices= Total Number of possible choices − Number of Favorable choices
⇒ mAc | = | n − mA |
= | 2,145 × 7! − 630 × 7! | |
= | (2,145 − 630) × 7! | |
= | 1,515 × 7! |
in favor
Odds in Favor of forming the words with no two S's together⇒ Odds in Favor of Event A
= Number of Favorable Choices : Number of Unfavorable Choices
= mA : mAc
= 630 × 7! : 1,515 × 7!
= 42 : 101
against
Odds against forming the words with no two S's together⇒ Odds against Event A
= Number of Unfavorable Choices : Number of Favorable Choices
= mAc : mA
= 1,515 × 7! : 630 × 7!
= 101 : 42
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 no two S's together⇒ Odds in Favor of Event A
= | Probability of occurrence of the event : Probability of non-occurrence of the event | ||||
= | P(A) : P(Ac) | ||||
= |
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= | 43 : 101 |
against
Odds against forming the words with no two S's together⇒ Odds against Event A
= | Probability of non-occurrence of the event : Probability of occurrence of the event | ||||
= | P(Ac) : P(A) | ||||
= |
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= | 101 : 43 |