### All GED Math Resources

## Example Questions

### Example Question #21 : Statistics

A card is selected from a standard deck of 52 cards. A student picks out a card, and does not replace the card. What is the probability that the student will pick out a club, and then another club?

**Possible Answers:**

**Correct answer:**

The deck is divided into four suits: Clubs, Diamonds, Hearts, and Spades. There are 13 of each suit in a standard deck.

The probability of picking out a club out of a plain deck is:

The student does not replace the card back into the deck, which means that there are 51 cards remaining, and there are only 12 clubs remaining in the deck.

The probability of picking out another club would be:

To identity the probability, simply multiply the first and second probabilities.

The answer is:

### Example Question #22 : Statistics

A closet contains the following:

- 10 shirts
- 2 jackets
- 3 pairs of jeans

Find the probability of grabbing a shirt.

**Possible Answers:**

**Correct answer:**

To find the probability of an event, we will use the following formula:

Now, given the event of grabbing a shirt from the closet, we can calculate the following:

because there are 10 shirts in the closet.

We can also calculate the following:

because there are 15 total things in the closet we could potentially grab:

- 10 shirts
- 2 jackets
- 3 pairs of jeans

Now, we can substitute. We get

Therefore, the probability of grabbing a shirt from the closet is .

### Example Question #23 : Statistics

Find the probability that we draw a 3 from a deck of cards.

**Possible Answers:**

**Correct answer:**

To find the probability of an event, we will use the following formula:

So, given the event of drawing a 3, we can calculate the following:

because there are 4 ways we can draw a 3 from a deck:

- 3 of clubs
- 3 of spades
- 3 of hearts
- 3 of diamonds

Now, we can calculate the following:

because there are 52 cards we could potentially draw from a deck of cards.

Knowing this, we can substitute into the formula. We get

Therefore, the probability of drawing a 3 from a deck of cards is .

### Example Question #24 : Statistics

A parking lot contains the following:

- 7 blue cars
- 1 white car
- 3 black cars

Find the probability the next car that leaves is a blue car.

**Possible Answers:**

**Correct answer:**

To find the probability of an event, we will use the following formula:

Now, given the event of a blue car leaves next, we can calculate the following:

\text{number of ways event can happen} = 7

because there are 7 blue cars in the lot

We can also calculate the following:

\text{total number of possible outcomes} = 11

because there are 11 total cars in the lot that could potentially leave:

- 7 blue cars
- 1 white car
- 3 black cars

Now, we can substitute. We get

Therefore, the probability that a blue car leaves the parking lot is .

### Example Question #25 : Statistics

The probability that it will rain tomorrow is , and the probability that a philosopher will say something strange is . What is the probability that it both does not rain and that a philosopher does not say something strange?

**Possible Answers:**

**Correct answer:**

The first thing you need to notice is that you need the probabilities for events *opposite* of those provided in your data. Luckily, since there is a total of for each discrete event, you know that there is a it will not rain and a chance that a philosopher will not say something strange.

Now, to calculate the chance of both these events occurring at the same time, you just need to multiply these two percentages together:

or

(You do not add them together, because you are talking about two independent events happening at the same time.)

### Example Question #26 : Statistics

In a small town, there is a chance that person owns a pig and a that a person owns a donkey. What is the chance that a person owns at least one of these animals?

**Possible Answers:**

**Correct answer:**

In a small town, there is a chance that person owns a pig and a that a person owns a donkey. What is the chance that a person owns at least one of these animals?

Based on our data, you can think of your information as meaning this:

- Has pig:
- Doesn't have pig:
- Has donkey:
- Doesn't have donkey:

Now, you are looking for cases of people owning both or just one. Given that these are independent events, you calculate that probability by multiplying the numbers together. Thus, you get:

- Pig and donkey:
- Pig, no donkey:
- Donkey, no pig:

You could add these up and get:

or

Another way you could do this is by finding out the percentage that holds for NEITHER donkey NOR pig. Then, you subtract this from . This will give you the left over (namely, NOT NONE or in other words, AT LEAST ONE):

or

Subtracted from , that is still

### Example Question #27 : Statistics

Find the probability of drawing a black card from a deck of cards.

**Possible Answers:**

**Correct answer:**

To find the probability of an event, we will use the following formula:

Given the event of drawing a black card from a deck of cards, we can calculate the following:

because there are 26 black cards in a deck

- 13 clubs
- 13 spades

Now, we can calculate the following:

because there are 52 total cards we could potentially draw.

So, we can substitute. We get

Therefore, the probability of drawing a black card from a deck of cards is .

### Example Question #28 : Statistics

Bob is playing a game that involves the buying, selling, and trading of properties. Above is a part of the board; Bob's token is represented by the black triangle.

It is now Bob's turn, and he must roll two fair six-sided dice and move clockwise the number of spaces rolled. He does not want to land on any of the purple or green spaces, as they represent properties owned by opponents; he does not want to land on the "Prison" space, either. What is the probability that he will *not* land on any of these spaces?

**Possible Answers:**

**Correct answer:**

For Bob, the desirable rolls would be 2, 5, 7, and 10, each of which would land him on a white space.

There are 36 ways to roll a pair of fair six-sided dice; the rolls are seen below, with the desirable rolls circled:

There are 14 desirable rolls out of 36 total, making the probability of rolling one of these

.

### Example Question #29 : Statistics

Julie is playing a game that involves the buying, selling, and trading of properties. Above is a part of the board; Julie's token is represented by the black triangle, on the space marked "Prison!!!"

In order to get off the Prison square, Julie must roll doubles - that is, she must roll the same number on both of two fair six-sided dice. She has three turns in which to do it, or she must pay a $1,000 fine. What is the probability that she will be able to avoid the fine by rolling doubles within three turns?

**Possible Answers:**

**Correct answer:**

Examine the diagram below, which shows the possible rolls of two fair six-sided dice. As seen, there are six ways to roll doubles:

The probability that Julie will roll doubles in any given turn is

,

so the probability she will *not *roll doubles in a given turn is 1 minus this, or

.

Apply the multiplication principle to find the probability that she will go three turns without rolling doubles:

.

Therefore, the probability that Julie will roll doubles within three turns is 1 minus this, or

.

### Example Question #30 : Statistics

A parking lot contains the following:

- 11 black cars
- 8 red cars
- 3 blue cars

Find the probability that the next car that leaves the parking lot is a black car.

**Possible Answers:**

**Correct answer:**

To find the probability of an event, we will use the following formula:

Now, given the event of a black car leaves next, we can calculate the following:

\text{number of ways event can happen} = 11

because there are 11 black cars in the lot.

We can also calculate the following:

\text{total number of possible outcomes} = 22

because there are 9 total cars in the lot that could potentially leave

- 11 black cars
- 8 red cars
- 3 blue cars

Now, we can substitute. We get

Therefore, the probability that a black car leaves the parking lot is .