Problem 3
If 7 cards are drawn at random. From a pack of 52 cards then find the probability that 3 of them are red and 4 are black.
Solution
Total number of cards in the pack
= 52
Number of cards drawn
= 7
Experiment :
Drawing 7 cards from the pack of cards
Total Number of Possible Choices
= Number of ways in which 7 cards can be drawn from the 52 cards
⇒ n | = | 52C7 When simplification leads to large calculations leaving the term as it is may help. Simplify when needed in the steps while deriving the required answer. |
| = | |
| = | 52 × 51 × 50 × 49 × 48 × 47 × 46 | 7 × 6 × 5 × 4 × 3 × 2 × 1 |
|
| = | 52 × 17 × 10 × 7 × 47 × 46 |
Let
A : the event of drawing 3 red 4 black cards
For Event A
| Red | Black | Total |
---|
Available | 26 | 26 | 52 |
To Choose | 3 | 4 | 7 |
Choices | 26C3 | 26C4 | 52C7 |
Number of Favorable Choices
= Number of ways in which 3 red and 4 black cards can be drawn from the total 52
= Number of ways in which 3 red cards can be drawn from the available 26 × Number of ways in which 4 black cards can be drawn from the available 26
⇒ mA | = | 26C3 × 26C4 |
| = | × 26 × 25 × 24 × 23 | 4 × 3 × 2 × 1 |
|
| = | (26 × 25 × 4) × (26 × 25 × 23) |
Probability of drawing 3 red 4 black cards
⇒ Probability of occurrence of Event A
= | Number of Favorable Choices for the Event | Total Number of Possible Choices for the Experiment |
|
⇒ P(A) | = | |
| = | |
| = | × × 7 × 6 × 5 × 4! | 52 × 51 × 50 × 49 × 48 × 47 × 46 |
|
| = | |
| = | |
Odds
Probability of non-occurrence of Event A
⇒ P(Ac) | = | 1 − P(A) |
| = | 1 − |
| = | |
| = | |
in favor
Odds in Favor of drawing 3 red 4 black cards
⇒ Odds in Favor of Event A
= | Probability of occurrence of the event : Probability of non-occurrence of the event |
= | : |
= | 2,875 : 2,718 |
against
Odds against drawing 3 red 4 black cards
⇒ Odds against Event A
= | Probability of non-occurrence of the event : Probability of occurrence of the event |
= | : |
= | 2,718 : 2,875 |