Total number of numbered balls in the box
= 6
Number of balls drawn
= 2
Experiment :
Drawing 2 numbered balls from the 6 balls
Total Number of Possible Choices
= Number of ways in which 2 balls can be drawn from the 6 balls
Let
A : the event of drawing balls numbered 1 or 2
For Event A
Event A can be accomplished in two alternative ways.
EA1 : Drawing two balls numbered 1 and 2
EA2 : Drawing one of the balls numbered 1 or 2 and another ball not numbered 1 or 2
| 1 or 2 | Others | Total |
|---|
| Available | 2 | 4 | 6 |
| To Choose | | | |
| Alternative 1 | 2 | 0 | 2 |
| Choices | 2C2 | 4C0 | 6C2 |
| Alternative 2 | 1 | 1 | 2 |
| Choices | 2C1 | 4C1 | 6C2 |
Number of ways in which the two numbered balls with 1 or 2 marked on it can be selected
= Number of ways in which 2 balls numbered 1 or 2 can be drawn + Number of ways in which 1 ball numbered 1 or 2 and another ball not numbered 1 or 2 can be drawn
= (Number of ways in which 2 balls numbered 1 or 2 can be drawn from the available 2) + (Number of ways in which 1 ball numbered 1 or 2 can be drawn from the available 2 × Number of ways in which 1 ball not numbered 1 or 2 can be drawn from the available 4)
| ⇒ mA | = | mEA1 + mEA2 |
| = | (2C2) + (2C1 × 4C1) |
| = | 1 + (2 × 4) |
| = | 1 + 8 |
| = | 9 |
Probability of drawing the balls numbered 1 or 2
⇒ Probability of occurrence of Event A
| = | | Number of Favorable Choices for the Event | | Total Number of Possible Choices for the Experiment |
|
Odds
Probability of non-occurrence of Event A
| ⇒ P(Ac) | = | 1 − P(A) |
| = | 1 − |
| = | |
| = | |
in favor
Odds in Favor of drawing the balls numbered 1 or 2
⇒ Odds in Favor of Event A
| = | Probability of occurrence of the event : Probability of non-occurrence of the event |
| = | : |
| = | 3 : 2 |
against
Odds against drawing the balls numbered 1 or 2
⇒ Odds against Event A
| = | Probability of non-occurrence of the event : Probability of occurrence of the event |
| = | : |
| = | 2 : 3 |
The event of drawing either 1 or 2 is a complementary of the event of not drawing either 1 or 2. Thus the probability can also be found from the probability of its complimentary event.
Let
B : the event of not drawing balls numbered 1 or 2
For Event B
| 1 0r 2 | Others | Total |
|---|
| Available | 2 | 4 | 6 |
| To Choose | 0 | 2 | 2 |
| Choices | 2C0 | 4C2 | 6C2 |
Number of Favorable Choices
= Number of ways in which 2 balls with numbers other than 1 or 2 can be drawn from the total 6
= Number of ways in which 2 balls with numbers other than 1 or 2 can be drawn from the available 4
Probability of drawing balls with numbers other than 1 or 2
⇒ Probability of occurrence of Event B
| = | | Number of Favorable Choices for the Event | | Total Number of Possible Choices for the Experiment |
|
For Event A [Alternative]
Probability of drawing the balls numbered 1 or 2
⇒ Probability of occurrence of Event A
= Probability of non-occurrence of Event B
| ⇒ P(A) | = | P(Bc) |
| = | 1 − P(B) |
| = | 1 − |
| = | |
| = | |
