# Drawing/Picking/Choosing a Single/One Letter/Character from a word

No. Problems & Solutions [Click Hide/Show to display the solutions below the question]
01.
A single letter is selected at random from the word “RIGHTEOUSLY”. The probability and odds that it is a consonant is:

 Solution
Number of letters in the word "RIGHTEOUSLY" = 11 {R, I, G, H, T, E, O, U, S, L, Y}

In the experiment of selecting a letter from the letters of the word "RIGHTEOUSLY"

Total No. of Possible Choices = Number of ways in which a unique letter can be drawn from the total 11
⇒ n = 11C1
=
 11 1
= 11
 Where there are repetitions, drawing either of the repeated letters would mean the same. Thus only the unique letters are considered for finding the number of possible choices.

Let "A" be the event of the letter selected being a vowel

# For Event "A"

Number of consonants in the letters of the word "RIGHTEOUSLY" = 7 {R, G, H, T, S, L, Y}

Number of Favourable/Favorable Choices = The number of ways in which one letter which is a consonant can be
selected from the total 7 consonants
mA = 7C1
=
 7 1
= 7

Probability of the letter drawn being a vowel
⇒ Probability of occurrence of Event "A" =
 Number of Favourable/Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(A) =
 mA n
=
 7 11

# • Odds

 ⇒ mAc n − mA No. of UnFavorable Choices = Total No. of possible choices − No. of Favorable choices = = 11 − 7 = 4

## » in favor/favour

Odds in Favour of the letter selected being a consonant
 ⇒ Odds in Favor/Favour of Event "A" = No. of Favourable Choices : No. of UnFavorable Choices = mA : mAc = 7 : 4

## » against

Odds against the letter selected being a consonant
 ⇒ Odds against Event "A" = No. of UnFavourable Choices : No. of Favorable Choices = mAc : mA = 4 : 7

No. Problems & Solutions [Click Hide/Show to display the solutions below the question]
01.
A single letter is selected at random from the word “PROBABILITY”. The probability and odds that it is a vowel is:

 Solution
Number of letters in the word "PROBABILITY" = 11 {P, R, O, B, A, B, I, L, I, T, Y}

Number of unique letters in the word "PROBABILITY" = 9 {P, R, O, B, A, I, L, T, Y}

In the experiment of selecting a letter from the letters of the word "PROBABILITY"

Total No. of Possible Choices = Number of ways in which a unique letter can be drawn from the total 9
⇒ n = 9C1
=
 9 1
= 9
 Where there are repetitions, drawing either of the repeated letters would mean the same. Thus only the unique letters are considered for finding the number of possible choices.

Let "A" be the event of the letter selected being a vowel

# For Event "A"

Number of vowels in the unique letters = 3 {O, A, I}

Number of Favourable/Favorable Choices = The number of ways in which one letter which is a vowel can be
selected from the total 3 vowels
mA = 3C1
=
 3 1
= 3

Probability of the letter drawn being a vowel
⇒ Probability of occurrence of Event "A" =
 Number of Favourable/Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(A) =
 mA n
=
 3 9
=
 1 3

# • Odds

 ⇒ mAc n − mA No. of UnFavorable Choices = Total No. of possible choices − No. of Favorable choices = = 9 − 3 = 6

## » in favor/favour

Odds in Favour of the letter selected being a vowel
 ⇒ Odds in Favor/Favour of Event "A" = No. of Favourable Choices : No. of UnFavorable Choices = mA : mAc = 3 : 6 = 1 : 2

## » against

Odds against the letter selected being a vowel
 ⇒ Odds against Event "A" = No. of UnFavourable Choices : No. of Favorable Choices = mAc : mA = 6 : 3 = 2 : 1

No. Problems for Practice
01.

 Author Credit : The Edifier