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Question 195541:
Answer by jim_thompson5910(35256) (Show Source):
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# 1

The probability that the next call will be in person is simply the number of calls answered in person over the number of calls total. So

So the chances of a call answered in person are which in decimal form is 0.6 which gives a 60% chance

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# 2

Take note that there are 4 numbers that 5 on a single die namely: 1, 2, 3, and 4

So the relative frequency is which is 66.67%. So this means that the answer is A)

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# 3

First, we need to find the total number of marbles: 4 yellow + 12 green + 11 black = 27 total

So there are 27 marbles in total.

To find the chances of NOT yellow, we can do it two ways:

Method # 1:
Simply add up the number of marbles that are NOT yellow. So add up the 12 green and 11 black to get 23 marbles that are NOT yellow.

Now divide it by the total 27 to get:

So the chances of selecting a non yellow marble is which is 0.8519 in decimal form and gives a 85.19% chance of occurring

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Method # 2:

If we pick a marble, it is either yellow or NOT yellow (but not both). There are no other choices.

So this means that if we find the chances of picking a yellow marble, we can subtract this from 1 to find the chances of getting a non yellow marble.

So....

So the chances of selecting a non yellow marble is again which is 0.8519 in decimal form and gives a 85.19% chance of occurring

So we get the same answer using either method. This means that you can use whatever you find easier (to compute or understand)

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# 4

If we look through a deck of cards, we'll find that there are 4 jacks (one of each suit)

So

So there is a chance of picking a jack. This in decimal form is giving a 7.69% chance.

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# 5

Since the odds against Thunderbolt winning are 11:3, this means that the corresponding odds are

Winning: 3
Losing: 11
giving a total of 11+3=14

So ...

This means that the probability of winning is giving a decimal value of 0.2143 which means that there's a 21.43% of Thunderbolt winning.

So the answer is B)

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# 6

a)

First, list out the individual probabilities for every possible scenario:

P($300 Prize) = 1/500
P($200 Prize) = 1/500
P($50 Prize) = 3/500
P(Losing) = 996/500 = 249/125

Now simply multiply the probabilities listed above with their corresponding winning value and then add them all up like so:

Expected Value = P($300 Prize)*(Value) + P($200 Prize)*(Value) + P($50 Prize)*(Value) + P(Losing)*(Value)

Expected Value = (1/500)*(300) + (1/500)*(200) + (3/500)*(50) + (496/500)*(0)

Expected Value = 300/500 + 200/500 + 150/500 + 0

Expected Value = 3/5 + 2/5 + 3/10

Expected Value = 6/10 + 4/10 + 3/10

Expected Value = (6 + 4 + 3)/10

Expected Value = 13/10

Expected Value = 1.30

So he expects to earn $1.30 on average (without factoring in the cost)

However, we need to subtract the cost of playing a single game to get

1.30 - 2 = -0.70

So he expects to lose 70 cents

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b)

Since the expected winnings is $1.30, this means that the cost must equal the expected winnings to make the expected value equal to zero (since 1.30-1.30=0)

So the fair price for the ticket is $1.30

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# 7

Let
r = red ball
b = blue ball
g = green ball
y = yellow ball

Let's list out ALL the possible points in the sample space

{r,b}
{r,g}
{r,y}
{b,r}
{b,g}
{b,y}
{g,r}
{g,b}
{g,y}
{y,r}
{y,b}
{y,g}

So there are 12 different possible outcomes which means that there are 12 points in the sample space.

Take note that first the first ball there are 4 choices and then once we select a ball (and don't put it back) we'll have 3 choices. Multiply these values together to get

4 * 3 = 12

So we get the same answer.

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# 8

We could list out all the possible license plates, but that would take too long and there's a very high chance that we'll repeat values and/or forget to list some elements.

Since the first slot must be a vowel, this means that there are 5 choices for the first position. For the second slot, there are 25 choices (since repetition of letters is not allowed). Similarly, there are 24 choices for the third slot.

For the digits, there are 10 choices (for numbers 0 - 9). Since repetition is allowed, there are 10 choices for the last digit also.

Now simply multiply all of the choices to get

5 * 25 * 24 * 10 * 10 = 125 * 240 * 10 = 300,000

So there are 300,000 different license plates.

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# 9

Simply use the counting principle to find the number of possible outfits:

Number of Pants * Number of Shirts * Number of Ties = 7 * 9 * 4 = 63 * 4 = 252

So there are 252 different outfits

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# 10

a)

Number of Genders * Number of Colors * Number of Seat Choices = 2 * 3 * 3 = 6 * 3 = 18

So there are 18 different bike configurations.

b)

Here's the tree:

c)

Here are ALL of the possible configurations which form the sample space:

{male, red, soft}
{male, red, medium}
{male, red, hard}
{male, white, soft}
{male, white, medium}
{male, white, hard}
{male, blue, soft}
{male, blue, medium}
{male, blue, hard}
{female, red, soft}
{female, red, medium}
{female, red, hard}
{female, white, soft}
{female, white, medium}
{female, white, hard}
{female, blue, soft}
{female, blue, medium}
{female, blue, hard}

Note: an example point {female, red, soft} means that the bike has a female frame, is red, and has a soft seat.

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# 11

Let's use the table to answer these questions:

a)

So the chances are giving a decimal value of 0.4646 or a 46.46% chance
b)

So the chances are giving a decimal value of 0.4706 or a 47.06% chance
c)

So the probability is giving a decimal value of 0.5259 or a 52.59% chance

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# 12

For the first course, there are 6 choices. For the second there are 5 choices. For the third, there are 4 etc..

So multiplying all of this together, we get:

This means that there are 720 different ways for 6 instructors to teach six sections of a course in mathematics.

Note: the factorial button (!) is used as a shortcut. So

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# 13

a)

For first place, there are 7 choices. For second place, there are 6 choices (since it's not possible to place in two different positions). For third place, there are 5 choices, etc.. all the way down to 1.

So multiplying all of this together, we get

This means that there are 5,040 different arrangements.

b)

For first place, there are 7 choices. For second place, there are 6 choices and finally for third place there are 5 choices. Multiply these values together to get:

7*6*5 = 42*5 = 210

So there are 210 different ways to place in first, second and third place

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# 14

We could use the counting principle as we've done for earlier problems, but we'll have overlap and the sample space is far too large. So let's do it this way:

In this case, order does NOT matter since the candidates have no rank over one another (ie one isn't president or secretary).

Since order does not matter, we must use the combination formula:

Start with the combination formula.

Plug in and

Subtract to get 9

Expand 15!

Expand 9!

Cancel

Simplify

Expand 6!

Multiply 15*14*13*12*11*10 to get 3,603,600

Multiply 6*5*4*3*2*1 to get 720

Reduce.

So 15 choose 6 (where order doesn't matter) yields 5,005 unique combinations

This means that there are 5,005 different ways to select a group of 6 college candidates from a group of 15 applicants for an interview (where the order of the candidates doesn't matter).