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| 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:
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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 |
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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.
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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 |
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) |
= |
|
|
= |
|
• Odds
|
No. of UnFavorable Choices
|
= |
Total No. of possible choices − No. of Favorable choices |
| ⇒ mAc |
= |
n − mA
|
|
= |
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 |
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| 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:
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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 |
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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 |
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) |
= |
|
|
= |
|
|
= |
|
• Odds
|
No. of UnFavorable Choices
|
= |
Total No. of possible choices − No. of Favorable choices |
| ⇒ mAc |
= |
n − mA
|
|
= |
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 |
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| No. |
Problems for Practice |
| 01. |
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