Drawing two cards of the same suite

Problem 7

If two cards are drawn from a pack of 52 cards, then the probability that they belong to the same suit is

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
=
52 × 51
2!
= 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 13C239C052C0

Alternative 2

Clubs Others Total
Available 13 39 52
To Choose 2 0 2
Choices 13C239C052C0

Alternative 3

Hearts Others Total
Available 13 39 52
To Choose 2 0 2
Choices 13C239C052C0

Alternative 4

Diamonds Others Total
Available 13 39 52
To Choose 2 0 2
Choices 13C239C052C0

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 × 12
2 × 1
+
13 × 12
2 × 1
+
13 × 12
2 × 1
+
13 × 12
2 × 1
= (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

=
Number of Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(A) =
mA
n
=
312
1,326
=
4
17

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 −
4
17
=
17 − 4
17
=
13
17

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
17
:
13
17
= 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
17
:
4
17
= 13 : 4