Tenth Grade (Grade 10) Statistics and Probability Concepts Questions for Tests and Worksheets

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Grade 10 Combinations and Permutations CCSS: HSS-CP.B.9

What is the formula for finding the probability of event A, given that both events A and B have already occurred?

P(A | B)
P(A | B and A)
P(A and B) / P(B)
P(A and B) / P(A)

Grade 10 Combinations and Permutations CCSS: HSS-CP.B.9

What is the formula for finding the probability of the intersection of two events A and B, given that event A has already occurred?

P(A and B)
P(B | A)
P(A | B)
P(B)

Grade 10 Combinations and Permutations CCSS: HSS-CP.B.9

If event A is independent of event B, what is the value of P(A and B)?

P(A) * P(B)
P(A) + P(B)
P(A) / P(B)
0

Grade 10 Combinations and Permutations CCSS: HSS-CP.B.9

What is the formula for finding the probability of event A, given that event B has not occurred?

P(A | complement of B)
P(A) * P(complement of B)
P(A) / P(complement of B)
P(A) + P(complement of B)

Grade 10 Represent and Determine Probability CCSS: HSS-CP.A.4

95 people were asked what type of soft serve ice cream they were most likely to buy. Their choices were vanilla, chocolate, and swirl (a combination of vanilla and chocolate). The chart shows the results. Which of the following statements are correct? There may be more than one correct answer.

	$"Vanilla"$	$"Chocolate"$	$"Swirl"$	$\mathbf{ Total}$
$"Male"$	$6$	$22$	$9$	$37$
$"Female"$	$10$	$27$	$21$	$58$
$\mathbf{Total}$	$16$	$49$	$30$	$95$

There is almost no difference in probability in choosing someone who likes vanilla, whether choosing from all people, only men, or only women.
It is nearly equally likely that you will choose someone at random who likes swirl, whether choosing from all people, only men, or only women.
It is more likely that you will choose a man from people who like swirl, than if you choose from people who like vanilla.
Randomly choosing a woman from people who like vanilla is more likely than choosing a woman from people who like chocolate.

Grade 10 Combinations and Permutations CCSS: HSS-CP.B.9

What is the formula for finding the probability of two events A and B occurring together, given that event A has already occurred?

P(A) * P(B)
P(B | A)
P(A | B)
P(A and B)

Grade 10 Combinations and Permutations CCSS: HSS-CP.B.9

What is the formula for finding the joint probability of two events A and B?

P(A | B)
P(A) * P(B | A)
P(A and B)
P(A) + P(B)

Grade 10 Combinations and Permutations CCSS: HSS-CP.B.9

What is the formula for finding the probability of the union of two events A and B?

P(A) + P(B)
P(A) * P(B)
P(A) / P(B)
P(A) + P(B) - P(A and B)

Grade 10 Combinations and Permutations CCSS: HSS-CP.B.9

What is the formula for finding the marginal probability of event A?

P(A)
P(A | B)
P(B | A)
P(A and B)

Grade 10 Combinations and Permutations CCSS: HSS-CP.B.9

If two events are independent, what is the formula for their joint probability?

P(A) * P(B)
P(A and B)
P(A) + P(B)
P(A | B)

Grade 10 Combinations and Permutations CCSS: HSS-CP.B.9

What is the formula for finding the probability of event A given that event B has already occurred?

P(A | B)
P(B | A)
P(A)
P(B)

Grade 10 Represent and Determine Probability CCSS: HSS-CP.A.3

If P(A) = 0.36, P(B) = 0.44, and P(A|B) = 0.44, then events A and B are independent.

True
False

Grade 10 Collecting and Interpreting Data CCSS: HSS-CP.A.1

Given

$A = {1,2,3,4,5}, \ \ B = {4,5,6,7,8}, \ \ C = {6,7,8,9,10}$ , find

$A uu B uu C$ .

${1,2,3,4,5,6,7,8,9,10}$
$emptyset$
${4,5,6,7,8}$
${1,2,3,9,10}$

Grade 10 Combinations and Permutations

If 15 runners are in a race, how many possible ways are there to award 1st, 2nd, and 3rd place trophies?

455
2730
217,945,728,000
1,307,674,368,000

Grade 10 Collecting and Interpreting Data CCSS: HSS-CP.A.2

P(A) = 0.4 that it will rain today. P(B) = 0.2 that you will get your allowance today. P(A and B) = 0.08. The two events are

independent.
not independent.
there is not enough information.

Grade 10 Represent and Determine Probability CCSS: HSS-CP.B.8

At Monarch Collegiate Institute, a survey was conducted about the number of students who play rugby and the number of students who read science fiction books. Let R be the event that a student plays rugby, and S be the event that a student reads science fiction books. If there were 60 students surveyed, P(R|S) = 5/7, P(S|R) = 1/3, and P(S) = 7/60, which of the following expressions gives the number of students who said they play rugby and why? Choose all correct answers.

$60 ( (7/60 * 5/7) / (1/3) )$ , by applying the general Multiplication Rule to find P(R), then multiplying by 60.
$60 (1 - 7/60 + (7/60 * 5/7))$ , since it can assumed that everyone who doesn't read science fiction plays rugby. Therefore, subtract P(S) from one, but add back in P(S and R) (found using the general Multiplication Rule), and then multiply by 60.
$7/60 * 1/3 = 7/180 = "P(S and R)", \ \ \ 60 ( (7/180) / (5/7) )$ , by applying the general Multiplication Rule to find P(S and R), then using it again to find P(R) and then multiplying by 60.
Not possible. If 60 is multiplied by 5/7 (the conditional probability of choosing someone who plays rugby, given that they read science fiction), the number is not a whole number. This means that either the total number of participants or this probability is incorrect.