Statistics & Probability - 7th Grade Math

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Question

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

Answer

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.

$\frac{1}{6}\times\frac{300}{1}=\frac{300}{6}=50$

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

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Question

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

Answer

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.

$\frac{1}{6}\times\frac{600}{1}=\frac{600}{6}=100$

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

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Question

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

Answer

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

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

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

$\frac{2}{6}$

This means that roughly $\frac{2}{6}$ of John's rolls will be a or a ; therefore, in order to calculate the probability we can multiply $\frac{2}{6}$ by —the number of times John rolls the die.

$\frac{2}{6}\times\frac{300}{1}=\frac{600}{6}=100$

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

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Question

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

Answer

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

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

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

$\frac{2}{6}$

This means that roughly $\frac{2}{6}$ of John's rolls will be a or a ; therefore, in order to calculate the probability we can multiply $\frac{2}{6}$ by —the number of times John rolls the die.

$\frac{2}{6}\times\frac{600}{1}=\frac{1200}{6}=200$

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

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Question

If John were to roll a die times, roughly how many times would he roll an even number?

Answer

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

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

There is a total of sides on a die and even numbers: ; thus, our probability is:

$\frac{3}{6}$

This means that roughly $\frac{3}{6}$ of John's rolls will be an even number; therefore, in order to calculate the probability we can multiply $\frac{3}{6}$ by —the number of times John rolls the die.

$\frac{3}{6}\times\frac{400}{1}=\frac{1200}{6}=200$

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

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Question

If John were to roll a die times, roughly how many times would he roll an odd number?

Answer

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

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

There is a total of sides on a die and odd numbers: ; thus, our probability is:

$\frac{3}{6}$

This means that roughly $\frac{3}{6}$ of John's rolls will be an odd number; therefore, in order to calculate the probability we can multiply $\frac{3}{6}$ by —the number of times John rolls the die.

$\frac{3}{6}\times\frac{400}{1}=\frac{1200}{6}=200$

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

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Question

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

Answer

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

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

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

$\frac{3}{6}$

This means that roughly $\frac{3}{6}$ of John's rolls will be a , , or a ; therefore, in order to calculate the probability we can multiply $\frac{3}{6}$ by —the number of times John rolls the die.

$\frac{3}{6}\times\frac{600}{1}=\frac{1800}{6}=300$

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

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Question

If John were to roll a die times, roughly how many times would he roll an odd number or a

Answer

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

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

There is a total of sides on a die and odd numbers: and one ; thus, our probability is:

$\frac{4}{6}$

This means that roughly $\frac{4}{6}$ of John's rolls will be an odd number or a ; therefore, in order to calculate the probability we can multiply $\frac{4}{6}$ by —the number of times John rolls the die.

$\frac{4}{6}\times\frac{300}{1}=\frac{1200}{6}=200$

If John rolls a die times, then he will roll an odd number or a roughly times.

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Question

If John were to roll a die times, roughly how many times would he roll an even number or a

Answer

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

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

There is a total of sides on a die and even numbers: and one ; thus, our probability is:

$\frac{4}{6}$

This means that roughly $\frac{4}{6}$ of John's rolls will be an even number or a ; therefore, in order to calculate the probability we can multiply $\frac{4}{6}$ by —the number of times John rolls the die.

$\frac{4}{6}\times\frac{600}{1}=\frac{2400}{6}=400$

If John rolls a die times, then he will roll an even number or a roughly times.

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Question

A student flips a coin four times and rolls a six-sided die. What is the probability that the coin will land on heads all four times and the die will show a ?

Answer

There are two sides to a coin: heads and tails. The probability of the coin landing on heads is $\frac{1}{2}$

To determine the probability of the coin landing on heads four times in a row, we take the probability of the coin landing on heads and multiply it four times.

$\frac{1}{2}\times \frac{1}{2} \times\frac{1}{2}\times\frac{1}{2}=\frac{1}{16}$

Based on the question, we want to combine the probability of flipping a coin four times, with the coin landing on heads all four times, with the probability of rolling a on a die. There are sides to a die, and only one of those sides has the number ; thus, the probability of rolling a on a die is $\frac{1}{6}$

We want the probability of all of these events occurring, so we need to multiply:

$\frac{1}{16}\times\frac{1}{6}=\frac{1}{96}$

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Question

What is the probability of flipping a coin three times, with the coin landing on heads all three times, and rolling a on a die?

Answer

There are two sides to a coin: heads and tails. The probability of the coin landing on heads is $\frac{1}{2}$

To determine the probability of the coin landing on heads three times in a row, we take the probability of the coin landing on heads and multiply it three times.

$\frac{1}{2}\times \frac{1}{2} \times\frac{1}{2}=\frac{1}{8}$

Based on the question, we want to combine the probability of flipping a coin three times, with the coin landing on heads all three times, with the probability of rolling a on a die. There are sides to a die, and only one of those sides has the number ; thus, the probability of rolling a on a die is $\frac{1}{6}$

We want the probability of all of these events occurring, so we need to multiply:

$\frac{1}{8}\times\frac{1}{6}=\frac{1}{48}$

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Question

What is the probability of flipping a coin two times, with the coin landing on heads all two times, and rolling a on a die?

Answer

There are two sides to a coin: heads and tails. The probability of the coin landing on heads is $\frac{1}{2}$

To determine the probability of the coin landing on heads two times in a row, we take the probability of the coin landing on heads and multiply it two times.

$\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

Based on the question, we want to combine the probability of flipping a coin two times, with the coin landing on heads all two times, with the probability of rolling a on a die. There are sides to a die, and only one of those sides has the number ; thus, the probability of rolling a on a die is $\frac{1}{6}$

We want the probability of all of these events occurring, so we need to multiply:

$\frac{1}{4}\times\frac{1}{6}=\frac{1}{24}$

Compare your answer with the correct one above

Question

What is the probability of flipping a coin four times, with the coin landing on heads all four times, and rolling a or a on a die?

Answer

There are two sides to a coin: heads and tails. The probability of the coin landing on heads is $\frac{1}{2}$

To determine the probability of the coin landing on heads four times in a row, we take the probability of the coin landing on heads and multiply it four times.

$\frac{1}{2}\times \frac{1}{2} \times\frac{1}{2}\times\frac{1}{2}=\frac{1}{16}$

Based on the question, we want to combine the probability of flipping a coin four times, with the coin landing on heads all four times, with the probability of rolling a or a on a die. There are sides to a die, and one of those sides has the number and one has the number ; thus, the probability of rolling a or a on a die is $\frac{2}{6}$

We want the probability of all of these events occurring, so we need to multiply:

$\frac{1}{16}\times\frac{2}{6}=\frac{2}{96}=\frac{1}{48}$

Compare your answer with the correct one above

Question

What is the probability of flipping a coin three times, with the coin landing on heads all three times, and rolling a or a on a die?

Answer

There are two sides to a coin: heads and tails. The probability of the coin landing on heads is $\frac{1}{2}$

To determine the probability of the coin landing on heads three times in a row, we take the probability of the coin landing on heads and multiply it three times.

$\frac{1}{2}\times \frac{1}{2} \times\frac{1}{2}=\frac{1}{8}$

Based on the question, we want to combine the probability of flipping a coin three times, with the coin landing on heads all three times, with the probability of rolling a or a on a die. There are sides to a die, and one of those sides has the number and one has the number ; thus, the probability of rolling a or a on a die is $\frac{2}{6}$

We want the probability of all of these events occurring, so we need to multiply:

$\frac{1}{8}\times\frac{2}{6}=\frac{2}{48}=\frac{1}{24}$

Compare your answer with the correct one above

Question

What is the probability of flipping a coin two times, with the coin landing on heads all two times, and rolling a or a on a die?

Answer

There are two sides to a coin: heads and tails. The probability of the coin landing on heads is $\frac{1}{2}$

To determine the probability of the coin landing on heads two times in a row, we take the probability of the coin landing on heads and multiply it two times.

$\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

Based on the question, we want to combine the probability of flipping a coin two times, with the coin landing on heads all two times, with the probability of rolling a or a on a die. There are sides to a die, and one of those sides has the number and one has the number ; thus, the probability of rolling a or a on a die is $\frac{2}{6}$

We want the probability of all of these events occurring, so we need to multiply:

$\frac{1}{4}\times\frac{2}{6}=\frac{2}{24}=\frac{1}{12}$

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Question

What is the probability of flipping a coin four times, with the coin landing on heads all four times, and rolling a , , or a on a die?

Answer

There are two sides to a coin: heads and tails. The probability of the coin landing on heads is $\frac{1}{2}$

To determine the probability of the coin landing on heads four times in a row, we take the probability of the coin landing on heads and multiply it four times.

$\frac{1}{2}\times \frac{1}{2} \times\frac{1}{2}\times\frac{1}{2}=\frac{1}{16}$

Based on the question, we want to combine the probability of flipping a coin four times, with the coin landing on heads all four times, with the probability of rolling a , , or a on a die. There are sides to a die, and one of those sides has the number , one side has a , and one has the number ; thus, the probability of rolling a , , or a on a die is $\frac{3}{6}$

We want the probability of all of these events occurring, so we need to multiply:

$\frac{1}{16}\times\frac{3}{6}=\frac{3}{96}=\frac{1}{32}$

Compare your answer with the correct one above

Question

What is the probability of flipping a coin four times, with the coin landing on heads all three times, and rolling a , , or a on a die?

Answer

There are two sides to a coin: heads and tails. The probability of the coin landing on heads is $\frac{1}{2}$

To determine the probability of the coin landing on heads three times in a row, we take the probability of the coin landing on heads and multiply it three times.

$\frac{1}{2}\times \frac{1}{2} \times\frac{1}{2}=\frac{1}{8}$

Based on the question, we want to combine the probability of flipping a coin three times, with the coin landing on heads all three times, with the probability of rolling a , , or a on a die. There are sides to a die, and one of those sides has the number , one side has a , and one has the number ; thus, the probability of rolling a , , or a on a die is $\frac{3}{6}$

We want the probability of all of these events occurring, so we need to multiply:

$\frac{1}{8}\times\frac{3}{6}=\frac{3}{48}=\frac{1}{16}$

Compare your answer with the correct one above

Question

What is the probability of flipping a coin two times, with the coin landing on heads all two times, and rolling a , , or a on a die?

Answer

There are two sides to a coin: heads and tails. The probability of the coin landing on heads is $\frac{1}{2}$

To determine the probability of the coin landing on heads two times in a row, we take the probability of the coin landing on heads and multiply it two times.

$\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

Based on the question, we want to combine the probability of flipping a coin two times, with the coin landing on heads all two times, with the probability of rolling a , , or a on a die. There are sides to a die, and one of those sides has the number , one side has a , and one has the number ; thus, the probability of rolling a , , or a on a die is $\frac{3}{6}$

We want the probability of all of these events occurring, so we need to multiply:

$\frac{1}{4}\times\frac{3}{6}=\frac{3}{24}=\frac{1}{8}$

Compare your answer with the correct one above

Question

What is the probability of flipping a coin four times, with the coin landing on heads all four times, and rolling any number on a die except

Answer

There are two sides to a coin: heads and tails. The probability of the coin landing on heads is $\frac{1}{2}$

To determine the probability of the coin landing on heads four times in a row, we take the probability of the coin landing on heads and multiply it four times.

$\frac{1}{2}\times \frac{1}{2} \times\frac{1}{2}\times\frac{1}{2}=\frac{1}{16}$

Based on the question, we want to combine the probability of flipping a coin four times, with the coin landing on heads all four times, with the probability of rolling any number on a die except . There are sides to a die, and five sides that aren't the number ; thus, the probability of rolling any number on a die except is $\frac{5}{6}$

We want the probability of all of these events occurring, so we need to multiply:

$\frac{1}{16}\times\frac{5}{6}=\frac{5}{96}$

Compare your answer with the correct one above

Question

What is the probability of flipping a coin three times, with the coin landing on heads all three times, and rolling a number on a die other than

Answer

There are two sides to a coin: heads and tails. The probability of the coin landing on heads is $\frac{1}{2}$

To determine the probability of the coin landing on heads three times in a row, we take the probability of the coin landing on heads and multiply it three times.

$\frac{1}{2}\times \frac{1}{2} \times\frac{1}{2}=\frac{1}{8}$

Based on the question, we want to combine the probability of flipping a coin three times, with the coin landing on heads all three times, with the probability of rolling any number on a die except . There are sides to a die, and five sides that aren't the number ; thus, the probability of rolling any number on a die except is $\frac{5}{6}$

We want the probability of all of these events occurring, so we need to multiply:

$\frac{1}{8}\times\frac{5}{6}=\frac{5}{48}$

Compare your answer with the correct one above

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