Statistics - GED Math

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 .

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 .

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.)

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

If the weighting is for heads, this means that it is for tails. Therefore, the probability of flipping tails will be:

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

.

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 .

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

.

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 .

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

Now, given the event of grabbing a marker, we can calculate the following:

because there is 1 marker we can grab.

We can also calculate the following:

because there are 7 objects we could potentially grab:

• 2 pencils
• 4 pens
• 1 marker

Now, we can substitute. We get

Therefore, the probability of grabbing a marker from the bag is .

Every result for a fair coin is independent of the previous trials.

The probability of landing heads or tails is .

For the coin to land on tails six consecutive times,

The answer is:

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

Given the event of the teacher calling on a girl, we can calculate the following:

because there are 14 girl in the class.

We can also calculate the following:

because there are 26 total students in the class the teacher could potentially call on:

• 12 boys
• 14 girls

So, we substitute. We get

Therefore, the probability that the teacher calls on a girl is

Start by figuring out the probability of rolling a or . Since a six sided die has an equal chance of landing on each number, the probability would be , or .

Next, figure out the probability of spinning green. Since the spinner is divided equally into quarters, the probability of spinning green is .

Because the question asks about the probability of both events happening, you will need to multiply the two probabilities together.

Start by finding the probability of spinning an even number. We have total possibilities for the spinner. Out of those six, will give us even numbers: . Thus, we can write the following probability:

Next, determine the probability of rolling a on the die. Since we are using a fair, sided die, the probability of rolling a will be .

Since the question asks about the probability of both events happening, we will need to multiply the two probabilities.

Recall the definition of a probability:

Now, the desired outcome for this scenario is the beanbag landing in the circle. Thus, we can use the area of the circle as the desired outcome. In addition, the total possible outcomes can then be represented by the area of the entire square playing field.

Start by finding the area of the playing field. Recall how to find the area of a square:

Plug in the given side length of the square.

Next, find the area of the circle.

Since the question informs us that the diameter of the circle is one-fourth that of the side of the square playing field, the diameter of the circle must be . It also follows that the radius of the circle must be .

Recall how to find the area of the circle.

Plug in the given radius to find the area of the circle.

Now that we have both areas, we can write probability:

Recall how to find the probability of an event:

Since there are students out of the total who have two pets, the probability for that occurring is .

Now, since we are selecting students without replacement, this means that the total number of students to choose from when it's time to select the second student is . There are also students who have no pets, so we can write as the probability that the second student selected will have no pets.

Since the question wants to know the probability of both events happening, we will multiply the probabilities together:

If you have a standard, six-sided die, what is the probability of rolling two numbers greater than 3 on two separate rolls?

We need to find the probability of two independent events. To do so, we need to find the probability of each event and multiply them.

There are three numbers on a six-sided die that are greater than 3 (4,5,and 6).

Thus, our probability of each event is

But, we want the probability of doing this twice, so we need to multiply our probabilities.

So, our answer is

Each of the larger regions is one-eighth of the ring; each of the smaller regions is one-sixteenth of the ring. There are three small green regions and no large green regions, so the green portion of the ring is

of the ring. This is the probability that the arrow will land in a green region.

Recall what a probability is:

For the die, we can find the probability of rolling an even number:

Next, for the spinner, we can find the probability of landing on blue:

Since we want the probability of both evens happening, we will need to multiply the individual probabilities together: