7th Grade Math - Develop and Compare Probability Models and Find Probabilities of Events: CCSS.Math.Content.7.SP.C.7

A bag contains $\small 2$ red marbles, $\small 7$ pink marbles, and $\small 6$ purple marbles. What is the probability of not choosing a purple marble?

$\small \frac{3}{5}$

$\small \frac{2}{5}$

$\small \frac{2}{3}$

$\small \frac{3}2$

Explanation

The probability (p) is equal to the number of a specific event (purple marbles) divided by the total number of events. The probability of NOT p (called q) is equal to $\small 1-q.$ In this example:

$\small P_{purple}=\frac{6}{2+7+6}=\frac{6}{15}=\frac{2}{5}$

$\small Q_{not\ purple}=1-\frac{2}{5}=\frac{3}{5}$

In a dice game, what is the probability of rolling a factor of 5 on a six-sided die?

$\frac{1}{3}$

$\frac{5}{6}$

$\frac{1}{6}$

Explanation

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

$\text{probability of event} = \frac{\text{number of ways event can happen}}{\text{total number of possible outcomes}}$

Now, given the event of rolling a factor of 5, we can calculate the following.

$\text{number of ways event can happen} = 2$

because there are 2 factors of 5 on a dice:

We can also calculate the following.

$\text{total number of possible outcomes} = 6$

because there are 6 different possible numbers we can roll on a dice:

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

$\text{probability of rolling a factor of 5} = \frac{2}{6}$

$\text{probability of rolling a factor of 5} = \frac{1}{3}$

Therefore, the probability of rolling a factor of 5 on a dice is $\frac{1}{3}$ .

Find the probability of drawing a 3 from a deck of cards.

$\frac{1}{13}$

$\frac{1}{4}$

$\frac{1}{52}$

$\frac{4}{13}$

Explanation

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

$\text{probability of event} = \frac{\text{number of ways event can happen}}{\text{total number of possible outcomes}}$

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

$\text{number of ways event can happen} = 4$

Beause there are four 3s we could draw from a deck of cards

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

Now we can calculate the following:

$\text{total number of possible outcomes} = 52$

Because there are 52 cards in the deck we could potentially draw.

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

$\text{probability of drawing a 3} = \frac{4}{52}$

$\text{probability of drawing a 3} = \frac{2}{26}$

$\text{probability of drawing a 3} = \frac{1}{13}$

In a dice game, what is the probability of rolling a factor of 4 on a six-sided die?

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{5}{6}$

Explanation

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

$\text{probability of an event} = \frac{\text{number of times the event can happen}}{\text{total number of possible outcomes}}$

So, in the event of rolling a factor of 4 on a dice, we can determine the number of times that event can happen. So,

$\text{number of times the event can happen} = 3$

because there are 3 factors of 4 on a dice:

Now, we can determine the total number of possible outcomes. We get

$\text{total number of possible outcomes} = 6$

because there are 6 different outcomes we can get when rolling the dice:

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

$\text{probability of rolling an odd number} = \frac{3}{6}$

and we simplify.

$\text{probability of rolling an odd number} = \frac{1}{2}$

Therefore, the probability of rolling a factor of 4 on a dice is $\frac{1}{2}$ .

A box contains the following:

9 blue crayons
3 green crayons
4 red crayons
1 yellow crayon

Find the probability of grabbing a red crayon from the box.

$\frac{4}{17}$

$\frac{4}{13}$

$\frac{1}{4}$

$\frac{1}{17}$

Explanation

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

$\text{probability of event} = \frac{\text{number of ways event can happen}}{\text{total number of possible outcomes}}$

Now, given the event of grabbing a red crayon from the box, we can calculate:

$\text{number of ways event can happen} = 4$

because there are 4 red crayons in the box we could grab.

We can also calculate:

$\text{total number of possible outcomes} = 17$

because there are 17 total crayons we could potentially grab:

9 blue crayons
3 green crayons
4 red crayons
1 yellow crayon

So, we get

$\text{probability of grabbing a red crayon} = \frac{4}{17}$

Therefore, the probability of grabbing a red crayon from the box is $\frac{4}{17}$ .

In a dice game, what is the probability of rolling a factor of 2 on a six-sided die?

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{1}{6}$

Explanation

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

$\text{probability of event} = \frac{\text{number of ways event can happen}}{\text{total number of possible outcomes}}$

Now, given the event of rolling a factor of 2 on a dice, we can calculate the following.

$\text{number of ways event can happen} = 2$

because there are 2 factors of 2 on a dice:

Now, we can calculate he following.

$\text{total number of possible outcomes} = 6$

because there are 6 possible numbers we could potentially roll on a dice:

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

$\text{probability of rolling a factor of 2} = \frac{2}{6}$

$\text{probability of rolling a factor of 2} = \frac{1}{3}$

Therefore, the probability of rolling a factor of 2 on a dice is $\frac{1}{3}$ .

In a dice game, what is the probability of rolling a factor of 6 on a six-sided die?

$\frac{2}{3}$

$\frac{1}{4}$

$\frac{1}{6}$

$\frac{5}{6}$

Explanation

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

$\text{probability of event} = \frac{\text{number of ways event can happen}}{\text{total number of possible outcomes}}$

Now, given the event of rolling a factor of 6 on a dice, we can calculate the following:

$\text{number of ways event can happen} = 4$

because there are 4 factors of 6 on a dice:

Now, we can calculate the following:

$\text{total number of possible outcomes} = 6$

because there are 6 possible outcomes we could potentially get when rolling a dice:

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

$\text{probability of rolling a factor of 6} = \frac{4}{6}$

$\text{probability of rolling a factor of 6} = \frac{2}{3}$

Therefore, the probability of rolling a factor of 6 on a dice is $\frac{2}{3}$ .

Two fair six-sided dice are thrown. What is the probability that the product of the two numbers rolled is between and inclusive?

$\frac{11}{36}$

$\frac{1}{3}$

$\frac{1}{4}$

$\frac{5}{18}$

$\frac{1}{2}$

Explanation

The rolls that yield a product between and inclusive are:

$\begin{matrix} \textrm{Roll} & \textrm{Prod} \ (2,6) &12 \ (3,4) &12 \ (3,5) &15 \ (3,6) & 18\ (4,3) & 12\ (4,4) &16 \ (4,5) &20 \ (5,3) &15 \(5,4) & 20 \ (6,2) & 12 \(6,3) & 18 \end{matrix}$

Therefore there are rolls that fit our criteria out of a total of possible rolls, so the probability of this outcome is $\frac{11}{36}$ .

If John were to roll a die times, roughly how many times would he roll a

Explanation

A die has sides, with each side displaying a number between .

Let's first determine the probability of rolling a after John rolls the die a single time.

There is a total of sides on a die and only one value of on one side; thus, our probability is:

$\frac{1}{6}$

This means that roughly $\frac{1}{6}$ of John's rolls will be a ; therefore, in order to calculate the probability we can multiply $\frac{1}{6}$ by —the number of times John rolls the die.

$\frac{1}{6}\times\frac{120}{1}=\frac{120}{6}=20$

If John rolls a die times, then he will roll a roughly times.

If John were to roll a die $1\textup,200$ times, roughly how many times would he roll a

Explanation

A die has sides, with each side displaying a number between .

Let's first determine the probability of rolling a after John rolls the die a single time.

There is a total of sides on a die and only one value of on one side; thus, our probability is:

$\frac{1}{6}$

This means that roughly $\frac{1}{6}$ of John's rolls will be a ; therefore, in order to calculate the probability we can multiply $\frac{1}{6}$ by $1\textup,200$ —the number of times John rolls the die.