Exercise 1 Page 783

In a standard deck of

52

cards there are

4

aces. We want to find the probability that a randomly dealt

5 -

card hand contains a pair of aces. To find it, we will use the theoretical probability.

P = \frac{Favorable Outcomes}{Possible Outcomes}

We start by finding the number of possible outcomes. This will be the number of combinations in which we can choose

5

out of

52

cards in a standard deck. The order in which we choose them is not important, since we consider them as a whole

5 -

card hand. Let's recall the formula for the number of combinations of

n

objects taken

r

at a time.

_{n} C_{r} = \frac{n !}{( n - r ) ! \cdot r !}

There are

52

cards in a standard deck, so

n = 52 .

Out of them,

5

cards are randomly selected. Therefore, we know that

r = 5 .

Let's substitute these values and find the number of possible combinations.

_{n} C_{r} = \frac{n !}{( n - r ) ! \cdot r !}

SubstituteII

$n = 52$ , $r = 5$

_{52} C_{5} = \frac{5 2 !}{( 5 2 - 5 ) ! \cdot 5 !}

▼

Evaluate right-hand side

SubTerm

Subtract term

_{52} C_{5} = \frac{5 2 !}{4 7 ! \cdot 5 !}

Write as a product

_{52} C_{5} = \frac{5 2 \cdot 5 1 \cdot 5 0 \cdot 4 9 \cdot 4 8 \cdot 4 7 !}{4 7 ! \cdot 5 !}

CancelCommonFac

Cancel out common factors

_{52} C_{5} = \frac{5 2 \cdot 5 1 \cdot 5 0 \cdot 4 9 \cdot 4 8 \cdot 4 7 !}{4 7 ! \cdot 5 !}

SimpQuot

Simplify quotient

_{52} C_{5} = \frac{5 2 \cdot 5 1 \cdot 5 0 \cdot 4 9 \cdot 4 8}{5 !}

Multiply

_{52} C_{5} = \frac{3 1 1 8 7 5 2 0 0}{5 !}

Write as a product

_{52} C_{5} = \frac{3 1 1 8 7 5 2 0 0}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}

Multiply

_{52} C_{5} = \frac{3 1 1 8 7 5 2 0 0}{1 2 0}

CalcQuot

Calculate quotient

_{52} C_{5} = 2598960

Therefore, the number of

possible

outcomes

is

2598960 .

Next, we will look for the number of

favorable

outcomes .

We want to have a pair of aces in a dealt

5 -

card hand. Therefore, we need to look for the number of combinations of

2

aces taken out of

4,

so that our

5 -

card hand contains a pair of aces.

_{n} C_{r} = \frac{n !}{( n - r ) ! \cdot r !}

SubstituteII

$n = 4$ , $r = 2$

_{4} C_{2} = \frac{4 !}{( 4 - 2 ) ! \cdot 2 !}

▼

Evaluate right-hand side

SubTerm

Subtract term

_{4} C_{2} = \frac{4 !}{2 ! \cdot 2 !}

$2! = 2$

_{4} C_{2} = \frac{4 !}{2 \cdot 2}

Multiply

_{4} C_{2} = \frac{4 !}{4}

Write as a product

_{4} C_{2} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{4}

Multiply

_{4} C_{2} = \frac{2 4}{4}

CalcQuot

Calculate quotient

_{4} C_{2} = 6

We found there are

6

different combinations of a pair of aces that can be chosen. Notice that we want to have a

5 -

card hand. Let's find out how many more cards we need to draw.

5 - 2 = 3 cards remaining

We want to have exactly two aces in the dealt hand, therefore when choosing the

3

remaining cards we need to pick from cards other than aces.

52 - 4 = 48 cards to choose from

Therefore, we need to find the number of combinations of

3

cards chosen out of

48

cards other than aces. Therefore, let's evaluate

_{48} C_{3} .

_{n} C_{r} = \frac{n !}{( n - r ) ! \cdot r !}

SubstituteII

$n = 48$ , $r = 3$

_{48} C_{3} = \frac{4 8 !}{( 4 8 - 3 ) ! \cdot 3 !}

▼

Evaluate right-hand side

SubTerm

Subtract term

_{48} C_{3} = \frac{4 8 !}{4 5 ! \cdot 3 !}

Write as a product

_{48} C_{3} = \frac{4 8 \cdot 4 7 \cdot 4 6 \cdot 4 5 !}{4 5 ! \cdot 3 !}

CancelCommonFac

Cancel out common factors

_{48} C_{3} = \frac{4 8 \cdot 4 7 \cdot 4 6 \cdot 4 5 !}{4 5 ! \cdot 3 !}

SimpQuot

Simplify quotient

_{48} C_{3} = \frac{4 8 \cdot 4 7 \cdot 4 6}{3 !}

Write as a product

_{48} C_{3} = \frac{4 8 \cdot 4 7 \cdot 4 6}{3 \cdot 2 \cdot 1}

Multiply

_{48} C_{3} = \frac{1 0 3 7 7 6}{6}

CalcQuot

Calculate quotient

_{48} C_{3} = 17296

Notice that the choice of aces is independent from the choice of other cards. That enables us to use the Fundamental Counting Principle, since this principle is used to find the number of possible outcomes for a combination of independent events.

Fundamental Counting Principle
If an event $A$ has $n$ possible outcomes and an event $B$ has $m$ possible outcomes, then the total number of different outcomes for $A$ and $B$ combined is $n \times m .$