Drawing two cards of the same suite
Problem 7
Solution
Total number of cards in the pack
= 52
Number of cards drawn
= 2
Experiment :
Drawing 2 cards from the pack of cards
Total Number of Possible Choices
= Number of ways in which 2 cards can be drawn from the 52 cards
⇒ n | = | 52C2 | ||
= |
| |||
= | 26 × 51 | |||
= | 1,326 |
Let
A : the event of the two cards drawn being of of the same suite
For Event A
Event A can be accomplished in 4 alternative ways, the cards drawn being
- 2 spades
- 2 clubs
- 2 hears
- 2 diamonds
Alternative 1
Spades | Others | Total | |
---|---|---|---|
Available | 13 | 39 | 52 |
To Choose | 2 | 0 | 2 |
Choices | 13C2 | 39C0 | 52C0 |
Alternative 2
Clubs | Others | Total | |
---|---|---|---|
Available | 13 | 39 | 52 |
To Choose | 2 | 0 | 2 |
Choices | 13C2 | 39C0 | 52C0 |
Alternative 3
Hearts | Others | Total | |
---|---|---|---|
Available | 13 | 39 | 52 |
To Choose | 2 | 0 | 2 |
Choices | 13C2 | 39C0 | 52C0 |
Alternative 4
Diamonds | Others | Total | |
---|---|---|---|
Available | 13 | 39 | 52 |
To Choose | 2 | 0 | 2 |
Choices | 13C2 | 39C0 | 52C0 |
Number of favorable choices
= Number of ways in which 2 cards are drawn such that they belong to the same suite
= Number of ways in which 2 spades can be selected + Number of ways in which 2 clubs can be selected + Number of ways in which 2 hearts can be selected + Number of ways in which 2 diamonds can be selected
= (Number of ways in which 2 spades can be selected from the available 13) + (Number of ways in which 2 clubs can be selected from the available 13) + (Number of ways in which 2 hearts can be selected from the available 13) + (Number of ways in which 2 diamonds can be selected from the available 13)
⇒ mA | = | mEA1 + mEA2 + mEA3 + mEA4 | ||||||||
= | 13C2 + 13C2 + 13C2 + 13C2 | |||||||||
= |
| |||||||||
= | (13 × 6) + (13 × 6) + (13 × 6) + (13 × 6) | |||||||||
= | 78 + 78 + 78 + 78 | |||||||||
= | 312 |
Probability of the cards drawn belonging to the same suit
⇒ 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 |
= | 1,326 − 312 | |
= | 1,014 |
in favor
Odds in Favor of the cards drawn belonging to the same suit⇒ Odds in Favor of Event A
= Number of Favorable Choices : Number of Unfavorable Choices
= mA : mAc
= 312 : 1,014
= 4 : 13
against
Odds against the cards drawn belonging to the same suit⇒ Odds against Event A
= Number of Unfavorable Choices : Number of Favorable Choices
= mAc : mA
= 1,014 : 312
= 13 : 4
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 | ||||
= |
| ||||
= | 4 : 13 |
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 | ||||
= |
| ||||
= | 13 : 4 |