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

Study concepts, example questions & explanations for Common Core: 7th Grade Math

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Example Questions

Example Question #14 : How To Find The Probability Of An Outcome

Set \(\displaystyle m\) consists of the numbers \(\displaystyle (3,4)\). Set n consists of the numbers \(\displaystyle (5,6)\). If a number is randomly selected from set \(\displaystyle m\) and multiplied by a number randomly selected from set \(\displaystyle n\), what is the chance the product will be \(\displaystyle 24\)?

Possible Answers:

\(\displaystyle \frac{1}{4}\)

\(\displaystyle \frac{1}{2}\)

\(\displaystyle \frac{1}{3}\)

\(\displaystyle \frac{2}{3}\)

Correct answer:

\(\displaystyle \frac{1}{4}\)

Explanation:

The only way that a product of \(\displaystyle 24\) will be acheived is if the number \(\displaystyle 4\) is selected from set \(\displaystyle m\), and the number \(\displaystyle 6\) is selected from set \(\displaystyle n\)

There is a one half chance of the number 4 being selected and a one half chance of the number 6 being selected, given that each set contains two numbers. 

Therefore, the probability of both these number being chosen is:

\(\displaystyle \frac{1}{2}* \frac{1}{2}=\frac{1}{4}\)

Example Question #522 : Data Analysis And Probability

If Mark flips a coin and then rolls a die, what are the odds that the coin will be heads and that the die will land on a multiple of 3?

Possible Answers:

\(\displaystyle \frac{1}{4}\)

\(\displaystyle \frac{1}{3}\)

\(\displaystyle \frac{1}{6}\)

\(\displaystyle \frac{1}{12}\)

Correct answer:

\(\displaystyle \frac{1}{6}\)

Explanation:

If Mark flips a coin, the chance that it will land on heads is \(\displaystyle \frac{1}{2}\). On a die, there are 2 out of 6 numbers that are a multiple of 3 (3 and 6); therefore, there is a \(\displaystyle \frac{1}{3}\) chance that the dice will be a multiple of 3. 

The probability that the coin will land on heads and that the dice will be a multiple of 3 is:

\(\displaystyle \frac{1}{2}*\frac{1}{3}=\frac{1}{6}\)

Example Question #21 : Probability

If Janet rolls a pair of six-sided dice, what is the probability that the dice will come up as snake eyes (meaning that both dice show one dot)?

Possible Answers:

\(\displaystyle \frac{1}{6}\)

\(\displaystyle \frac{1}{3}\)

\(\displaystyle \frac{1}{36}\)

\(\displaystyle \frac{1}{12}\)

Correct answer:

\(\displaystyle \frac{1}{36}\)

Explanation:

Given that there are six values that a die may show when rolled, there is a \(\displaystyle \frac{1}{6}\) chance that either dice will show a value of 1. Thus, the chance that both dice will show a value of 1 is equal to \(\displaystyle \frac{1}{6}* \frac{1}{6}\), which is equal to \(\displaystyle \frac{1}{36}\).

Thus, \(\displaystyle \frac{1}{36}\) is the correct answer. 

Example Question #82 : Statistics & Probability

If you are picking a random student from a bus with \(\displaystyle 9\) freshmen, \(\displaystyle 10\) sophomores, \(\displaystyle 11\) juniors, and \(\displaystyle 12\) seniors, what is the probability that you pick a senior?

Possible Answers:

\(\displaystyle 0.26\)

\(\displaystyle 0.21\)

\(\displaystyle 0.29\)

\(\displaystyle 0.71\)

\(\displaystyle 0.24\)

Correct answer:

\(\displaystyle 0.29\)

Explanation:

To find the probability of an event, you divide the possible outcomes of that event by total outcomes.  

The event outcomes would be the \(\displaystyle 12\) seniors on the bus.  

The total outcomes is the number of students on the bus which is 

\(\displaystyle 42 \left ( 9+10+11+12=42\right )\).  

So the probability of picking a senior would be 

\(\displaystyle \frac{12}{42}=0.29\).

Example Question #21 : Develop And Compare Probability Models And Find Probabilities Of Events: Ccss.Math.Content.7.Sp.C.7

What is the probability of drawing a card that is a spade from a deck of cards?

Possible Answers:

\(\displaystyle \frac{1}{13}\)

\(\displaystyle \frac{1}{2}\)

\(\displaystyle \frac{1}{52}\)

\(\displaystyle 1\)

\(\displaystyle \frac{1}{4}\)

Correct answer:

\(\displaystyle \frac{1}{4}\)

Explanation:

There are \(\displaystyle 52\) cards in a deck.  

There are four suits of cards including spades.  

They make up an even amount of the deck each.  

Since there are four of the suits, each represent \(\displaystyle \frac{1}{4}\) of the deck and that is the probability of drawing one.

Example Question #91 : Statistics & Probability

Find the probability of drawing a 2 from a deck of cards.

Possible Answers:

\(\displaystyle \frac{1}{52}\)

\(\displaystyle \frac{2}{52}\)

\(\displaystyle 4\)

\(\displaystyle \frac{1}{4}\)

\(\displaystyle \frac{1}{13}\)

Correct answer:

\(\displaystyle \frac{1}{13}\)

Explanation:

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

\(\displaystyle \text{probability of an event} = \frac{\text{number of ways the event can happen}}{\text{number of total possible outcomes}}\)

So, when looking at the event of drawing a 2 from a deck of cards, we get

\(\displaystyle \text{number of ways the event can happen} = 4\)

It is 4 because we can draw a 2 from a deck of cards 4 different ways:

  • 2 of hearts
  • 2 of diamonds
  • 2 of clubs
  • 2 of spades


Now, 

\(\displaystyle \text{number of total possible outcomes} = 52\)

because there are 52 cards in the entire deck.  Now,

\(\displaystyle \text{probability of drawing a 2} = \frac{4}{52}\)

We can simplify, we get

\(\displaystyle \frac{4}{52} = \frac{2}{26} = \frac{1}{13}\)

 

Therefore, the probability of drawing a 2 from a deck of cards is \(\displaystyle \frac{1}{13}\)

Example Question #23 : Develop And Compare Probability Models And Find Probabilities Of Events: Ccss.Math.Content.7.Sp.C.7

Find the probability of drawing a 9 of hearts from a deck of cards.

Possible Answers:

\(\displaystyle 4\)

\(\displaystyle \frac{1}{13}\)

\(\displaystyle \frac{1}{4}\)

\(\displaystyle \frac{1}{52}\)

\(\displaystyle \frac{2}{52}\)

Correct answer:

\(\displaystyle \frac{1}{52}\)

Explanation:

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

\(\displaystyle \text{probability of an event} = \frac{\text{number of ways the event can happen}}{\text{number of total possible outcomes}}\)

So, when looking at the event of drawing a 9 of hearts from a deck of cards, we get

\(\displaystyle \text{number of ways the event can happen} = 1\)

It is 1 because we can draw a 9 of hearts from a deck of cards in only 1 way:

  • 9 of hearts


Now, 

\(\displaystyle \text{number of total possible outcomes} = 52\)

because there are 52 cards in the entire deck.  Now,

\(\displaystyle \text{probability of drawing a 9 of hearts} = \frac{1}{52}\)

Therefore, the probability of drawing a 2 from a deck of cards is \(\displaystyle \frac{1}{52}\)

Example Question #531 : Data Analysis And Probability

At Lisa's school, \(\displaystyle 52\%\) of her class consists of girls. There are \(\displaystyle 50\) students in her class. On the first day of school, \(\displaystyle 12\) girls wore skirts. How many girls did not wear skirts?

Possible Answers:

\(\displaystyle 13\)

\(\displaystyle 12\)

\(\displaystyle 14\)

\(\displaystyle 16\)

\(\displaystyle 26\)

Correct answer:

\(\displaystyle 14\)

Explanation:

Given that \(\displaystyle 52\%\) of the class is girls, and there are \(\displaystyle 50\) students, we must first determine how many girls are in the class. 

\(\displaystyle 52\%=\frac{52}{100}\)

\(\displaystyle \frac{52}{100}=\frac{x}{50}\)

Because \(\displaystyle 100\) is double the value of \(\displaystyle 50\), \(\displaystyle 52\) is double the value of \(\displaystyle x\)\(\displaystyle x\) should be half of \(\displaystyle 52\).

\(\displaystyle \frac{52}{100}=\frac{2\times x}{2\times50}=\frac{2x}{100}\)

\(\displaystyle 52=2x\)

\(\displaystyle x=\frac{52}{2}=26\)

If \(\displaystyle 12\) of the \(\displaystyle 26\) girls wear skirts, then we can subtract to find the number of girls not wearing skirts.

\(\displaystyle 26-12=14\)

Example Question #532 : Data Analysis And Probability

Billy's mom baked a pizza with eight slices. Half the slices have pepperoni only. Two of the slices have both pepperoni and onions. One slice has onions only. One slice has only cheese. 

 

If Billy is allergic to pepperoni, and takes a slice of pizza with his eyes blindfolded, what is the percentage chance that he will select a piece he is not allergic to?

Possible Answers:

\(\displaystyle 50\%\)

\(\displaystyle 0\%\)

\(\displaystyle 80\%\)

\(\displaystyle 25\%\) 

Correct answer:

\(\displaystyle 25\%\) 

Explanation:

The pizza that Billy's mom baked was composed of these types of slices:

Half the slices have pepperoni only. Two of the slices have both pepperoni and onoins. One slice has onions only. One slice has only cheese. 

Therefore, 

\(\displaystyle 4\) slices had pepperoni because \(\displaystyle 8\div2=4\)

\(\displaystyle 2\) slices had pepperoni and onions 

\(\displaystyle 1\) slice had onion only

\(\displaystyle 1\) slice had only cheese. 

Billy could eat either the onion only slice, or the cheese only slice. This means that out of \(\displaystyle 8\) pieces of pizza, he could eat \(\displaystyle 2\)

\(\displaystyle \frac{2}{8}=\frac{1}{4}\)

Given that \(\displaystyle 25\%\) is the percent equivalent of \(\displaystyle \frac{1}{4}\), that is the correct answer. 

Example Question #533 : Data Analysis And Probability

Two fair dice are thrown. What is the probability that the difference of the numbers that show on the dice will be exactly \(\displaystyle 1\)?

Possible Answers:

\(\displaystyle \frac{1}{3}\)

\(\displaystyle \frac{7}{18}\)

\(\displaystyle \frac{1}{4}\)

\(\displaystyle \frac{5}{18}\)

Correct answer:

\(\displaystyle \frac{5}{18}\)

Explanation:

The following rolls result in the difference of the dice being \(\displaystyle 1\):

\(\displaystyle (1,2), (2,3), (3,4), (4,5), (5,6)\) and the reverse of each of these.

These are \(\displaystyle 10\) rolls out of a possible \(\displaystyle 36\), so the probability of this event is \(\displaystyle \frac{10}{36} = \frac{5}{18}\)

 

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