Both the tickets drawn may have prime numbers or both composite numbers
Problem 3
Solution
Total number of tickets
= 50
Number of tickets drawn
= 2
Experiment :
Drawing 2 tickets from the 50 tickets
Total Number of Possible Choices
= Number of ways in which 2 tickets can be drawn from the 50 tickets
⇒ n | = | 50C2 | ||
= |
| |||
= | 25 × 49 | |||
= | 1,225 |
Let
A : the event of both the tickets having either a prime number or a composite number on it
For Event A
Prime Numbers :
= 15
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}
Numbers which have exactly two distinct factors, 1 and itself.
Composite Numbers :
= 34
{4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50}
Numbers which have more than two distinct factors.
Others : Neither Composite nor Prime
= 1
{1}
Event A can be accomplished in two alternative ways.
EA1 : Drawing both the tickets with prime numbers on them
EA2 : Drawing both the tickets with composite numbers on them
Prime | Composite | Others | Total | |
---|---|---|---|---|
Available | 15 | 34 | 1 | 50 |
To Choose | ||||
Alternative 1 | 2 | 0 | 0 | 2 |
Choices | 15C2 | 34C0 | 1C0 | 50C2 |
Alternative 2 | 0 | 2 | 0 | 2 |
Choices | 15C0 | 34C2 | 1C0 | 50C2 |
Number of favorable choices
= Number of ways in which the two tickets can be drawn having both prime numbers or both composite numbers on them
= Number of ways in which 2 tickets with prime numbers on them can be drawn + Number of ways in which 2 tickets with composite numbers on them can be drawn
= (Number of ways in which 2 tickets with prime numbers on them can be drawn from the available 15) + (Number of ways in which 2 tickets with composite numbers on them can be drawn from the available 34)
⇒ mA | = | mEA1 + mEA2 | ||||
= | (15C2) + (34C2) | |||||
= |
| |||||
= | (15 × 7) + (17 × 33) | |||||
= | 105 + 561 | |||||
= | 666 |
Probability of drawing both the tickets having prime numbers or composite numbers.
⇒ Probability of occurrence of Event A
= |
|
⇒ P(A) | = |
| ||
= |
|
Odds
Number of Unfavorable Choices= Total Number of possible choices − Number of Favorable choices
⇒ mAc | = | n − mA |
= | 1,225 − 666 | |
= | 559 |
in favor
Odds in Favor of drawing both the tickets having prime numbers or composite numbers.⇒ Odds in Favor of Event A
= Number of Favorable Choices : Number of Unfavorable Choices
= mA : mAc
= 666 : 559
against
Odds against drawing both the tickets having prime numbers or composite numbers.⇒ Odds against Event A
= Number of Unfavorable Choices : Number of Favorable Choices
= mAc : mA
= 559 : 666
Odds (alternative)
Probability of non-occurrence of Event A
⇒ P(Ac) | = | 1 − P(A) | ||
= | 1 −
| |||
= |
| |||
= |
|
in favor
Odds in Favor of the cards drawn belonging to the same suit⇒ Odds in Favor of Event A
= | Probability of occurrence of the event : Probability of non-occurrence of the event | ||||
= |
| ||||
= | 559 : 666 |
against
Odds against the cards drawn belonging to the same suit⇒ Odds against Event A
= | Probability of non-occurrence of the event : Probability of occurrence of the event | ||||
= |
| ||||
= | 559 : 666 |