Digit chosen at random to be odd, even or multiple of 3
Problem 1
| 5 |
| 9 |
| 4 |
| 9 |
| 1 |
| 3 |
Solution
Total number of digits
= 9 {1, 2, 3, .. , 9}
Experiment :
Choosing a digit from the digits 1,2 3, ... , 9.
Total Number of Possible Choices
= Number of ways in which one digit can be chosen from the total 9
| ⇒ n | = | 9C1 | ||
| = |
| |||
| = | 9 |
Let
A : the event of the digit chosen being odd.
B : the event of the digit chosen being even.
C : the event of the digit chosen being divisible by 3.
For Event A
Number of odd digits
= 5 {1, 3, 5, 7, 9}
| Favorable (Odd) | Unfavorable (Others) | Total | |
|---|---|---|---|
| Available | 5 | 4 | 9 |
| To Choose | 1 | 0 | 1 |
| Choices | 5C1 | 4C0 | 9C1 |
Number of Favorable Choices
= Number of ways in which one odd digit can be chosen from the total 5 favorable digits
| ⇒ mA | = | 5C1 | ||
| = |
| |||
| = | 5 |
Probability of the digit chosen being odd
⇒ Probability of occurrence of Event A
| = |
|
| ⇒ P(A) | = |
| ||
| = |
|
Odds
= Total Number of possible choices − Number of Favorable choices
| ⇒ mAc | = | n − mA |
| = | 9 − 5 | |
| = | 4 |
in favor
Odds in Favor of choosing an odd digit⇒ Odds in Favor of Event A
= Number of Favorable Choices : Number of Unfavorable Choices
= mA : mAc
= 5 : 4
against
Odds against choosing an odd digit⇒ Odds against Event A
= Number of Unfavorable Choices : Number of Favorable Choices
= mAc : mA
= 4 : 5
For Event B
Number of even digits
= 4 {2, 4, 6, 8}
| Favorable (Even) | Unfavorable (Others) | Total | |
|---|---|---|---|
| Available | 4 | 5 | 9 |
| To Choose | 1 | 0 | 1 |
| Choices | 4C1 | 5C0 | 9C1 |
Number of Favorable Choices
= Number of ways in which one even digit can be chosen from the total 4 favorable digits
| ⇒ mB | = | 4C1 | ||
| = |
| |||
| = | 4 |
Probability of the digit chosen being even
⇒ Probability of occurrence of Event B
| = |
|
| ⇒ P(B) | = |
| ||
| = |
|
For Event C
Number of digits which are multiples of 3
= 3 {3, 6, 9}
| Favorable (Multiples of 3) | Unfavorable (Others) | Total | |
|---|---|---|---|
| Available | 3 | 6 | 9 |
| To Choose | 1 | 0 | 1 |
| Choices | 3C1 | 6C0 | 9C1 |
Number of Favorable Choices
= Number of ways in which one digit which is a multiple of 3 can be chosen from the total 3 favorable digits
| ⇒ mC | = | 3C1 | ||
| = |
| |||
| = | 3 |
Probability of the digit chosen being a multiple of 3
⇒ Probability of occurrence of Event C
| = |
|
| ⇒ P(C) | = |
| ||
| = |
| |||
| = |
|