Exercise 34 Page 936

Practice makes perfect

a To find the probability that Meleah will have to wait

5

minutes or less to see each van, we will use geometric probability. Let's model this situation using a coordinate grid. The

x -

axis will represent the time until Meleah sees the van for company

B,

and the

y -

axis will represent the time until she sees the van for company

A .

Since the van for company $A$ arrives every $7$ minutes, it will arrive in at most $7$ minutes. Likewise, since the van for company $B$ arrives every $12$ minutes, it will arrive in at most $12$ minutes.

The sample space in our model is represented by a rectangle with width

12

and length

7 .

Let's calculate its area

A .

A = ℓ w

SubstituteII

$ℓ = 7$ , $w = 12$

A = (7) (12)

Multiply

A = 84

The outcomes where both vans appear in

5

minutes or less are represented by points with both coordinates less than or equal to

5 .

The event that Meleah will have to wait for

5

minutes or less to see each van is represented by a square with side length

5 .

Let's calculate the area

A_{S}

of this square.

A_{S} = 5^{2} \Rightarrow A_{S} = 25

The probability that Meleah will have to wait

5

minutes or less is the ratio of the area

A_{S}

of the region representing the outcomes where Meleah has to wait

5

minutes or less to see each van to the area

A

of the region representing the sample space.

P (5 minutes or less to each van) = \frac{A _{S}}{A}

SubstituteII

$A_{S} = 25$ , $A = 84$

P (5 minutes or less to each van) = \frac{2 5}{8 4}

CalcQuot

Calculate quotient

P (5 minutes or less to each van) = 0.2976619 \dots

RoundDec

Round to $2$ decimal place(s)

P (5 minutes or less to each van) \approx 0.30

WritePercent

Convert to percent

P (5 minutes or less to each van) \approx 30 %

The probability that Meleah will have to wait for

5

minutes or less to see each van is

\frac{2 5}{8 4},

which can be written as a decimal as approximately

0.30,

or approximately

30 % .

b To find the probability that Meleah will have to wait

5

minutes or less to see one of the vans, we will use geometric probability. Let's model this situation using a coordinate grid, like we did in Part A.

The outcomes where at least one of the vans appear in $5$ minutes or less are represented by points with either coordinate less than or equal to $5 .$

The event that Meleah will have to wait $5$ minutes or less for either van is represented by a region that consists of two rectangles and a square. The square has side length $5,$ one of the rectangles has length $5$ and width $2,$ and the other rectangle has length $7$ and width $5 .$ Let's calculate the areas of these figures.

	Length	Width	Area
Square	$5$	$5$	$25$
Rectangle $1$	$5$	$2$	$10$
Rectangle $2$	$7$	$5$	$35$

The area

A_{R}

of the entire region is the sum of the areas of the square and the two rectangles.

A_{R} = 25 + 10 + 35 \Rightarrow A_{R} = 70

Now, let's use geometric probability. The probability that Meleah will have to wait

5

minutes or less to see one of the vans is the ratio of the area

A_{R}

of the region representing this event to the area

A

of the region representing the sample space. We know from Part A that

A

is equal to

84 .

P (5 minutes or less to any van) = \frac{A _{R}}{A}

SubstituteII

$A_{R} = 70$ , $A = 84$

P (5 minutes or less to any van) = \frac{7 0}{8 4}

CalcQuot

Calculate quotient

P (5 minutes or less to any van) = 0.833333 \dots

RoundDec

Round to $2$ decimal place(s)

P (5 minutes or less to any van) \approx 0.83

WritePercent

Convert to percent

P (5 minutes or less to any van) \approx 83 %

The probability that Meleah will have to wait

5

minutes or less for one of the vans is

\frac{7 0}{8 4},

which can be written as a decimal as approximately

0.83,

or approximately

83 % .

c We want to consider a situation where the van for company

B

arrives before the van for company

A

and decide whether Meleah should wait for the van for company

A .

Let's model our situation using a coordinate grid, like we did in Part A.

Since in this case the van for company $B$ arrives before the van for company $A,$ we will only consider the points where the $x -$ coordinate is less than the $y -$ coordinate.

Here, the sample space is represented by a right triangle with legs of length

a = 7

and

b = 7 .

Let's calculate the area

A

of this region.

A = \frac{1}{2} a b

SubstituteII

$a = 7$ , $b = 7$

A = \frac{1}{2} (7) (7)

Multiply

A = 24.5

Now let's consider the situations where Meleah should wait for the van for company

A .

If the van for company

A

will arrive in less than

5

minutes Meleah will not be late, so she should wait. This situation is represented by the points with the

y -

coordinate less than or equal to

5 .

If the van for company $B$ arrives in more than $5$ minutes, Meleah runs the risk of being late even if she does not wait, so she should wait as well. This situation is represented by the points with the $x -$ coordinate greater than $5 .$

Now we will find the area of the shaded region. Note that the shaded region consists of two right triangles: one with both legs of length $5,$ and another with both legs of length $2 .$ Let's calculate their areas!

Leg Length	Area	Simplify
$5$	$\frac{1}{2} (5) (5)$	$12.5$
$2$	$\frac{1}{2} (2) (2)$	$2$

The area

A_{S}

of the shaded region is the sum of the areas of the triangles.

A_{S} = 12.5 + 2 \Rightarrow A_{S} = 14.5

Now we can use geometric probability. The probability that Meleah should wait for the van for company

A

is the ratio of the area

A_{S}

of the shaded region to the area

A

of the region that represents the sample space.

P (Meleah should wait) = \frac{A _{S}}{A}

SubstituteII

$A_{S} = 14.5$ , $A = 24.5$

P (Meleah should wait) = \frac{1 4 . 5}{2 4 . 5}

CalcQuot

Calculate quotient

P (Meleah should wait) = 0.591837 \dots

RoundDec

Round to $2$ decimal place(s)

P (Meleah should wait) \approx 0.59

The probability that Meleah should wait for the van for company

A

is approximately

0.59 .

Since this leaves a large probability that Meleah will risk being late for the competition, she should not wait for the van for company

A .

Subchapter links

Check Your Understanding p.934

Practice and Problem Solving p.934-936

H.O.T. Problems p.936

Standardized Test Practice p.937

Spiral Review p.937

Skills Review p.937

Exercise 34 Page 936

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