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AP Statistics : How to use the multiplication rule

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Example Questions

Example Question #1 : Identifying Variables And Relationships

With a standard deck of cards, what is the probability of picking a spade then a red card if there is no replacement?

Possible Answers:

$\frac{11}{95}$

$\frac{19}{51}$

$\frac{13}{102}$

$\frac{23}{53}$

$\frac{17}{104}$

Correct answer:

$\frac{13}{102}$

Explanation:

In a standard deck of cards:

$52\; cards=26\; black + 26\; red$

$26\; black = 13\; spades + 13\; clubs$

$26\; red = 13\; hearts + 13\; diamonds$

$P(spade)\cdot P(red)=\frac{13}{52}\cdot \frac{26}{51}=\frac{13}{102}$

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Example Question #22 : Rules Of Probability

In a bag there are $\small 5$ red marbles, $\small 7$ green marbles, and $\small 3$ blue marbles. What is the probability of randomly selecting $\small 3$ marbles, one after the other without replacement, all the same color?

Possible Answers:

$\small p = \frac{3}{49}$

$\small p = \frac{9}{91}$

$\small p = \frac{1}{35}$

$\small p = \frac{46}{455}$

$\small p = \frac{1}{3}$

Correct answer:

$\small p = \frac{46}{455}$

Explanation:

$p(3\ red) = \frac{5}{15}\times \frac{4}{14}\times\frac{3}{13} = \frac{60}{2730}$

$p(3\ blue) = \frac{7}{15}\times\frac{6}{14}\times\frac{5}{13} = \frac{210}{2730}$

$p(3\ green) = \frac{3}{15}\times\frac{2}{14}\times\frac{1}{13} = \frac{6}{2370}$

$p(3\ same) =$ $\frac{\left ( 60+210+6\right )}{2730}$ $= \frac{276}{2370} = \frac{46}{455}$

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Example Question #30 : Probability

A child has a bag of marbles-- red, blue, and yellow. The child randomly selects one marble and then places it back in the bag. The child then selects a second marble. What is the probability that the first marble selected was blue and the second marble selected was yellow?

Possible Answers:

1/8

3/32

1/16

9/64

5/64

Correct answer:

3/32

Explanation:

To find the probability of possible outcomes for two separate events, multiply the probabilities of the two outcomes.

$2/8\cdot 3/8=6/64$

Then reduce the answer to the least common denominator.

6/64=3/32

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Example Question #31 : Probability

Automobile license plates in a certain area display three letters followed by three digits, and the letters Z and N are not used. How many plates are possible if neither repetition of letters nor of numbers is allowed?

Possible Answers:

None of the other answers

13,824,000

8,743,680

9,953,280

12,144,000

Correct answer:

8,743,680

Explanation:

If no letters are to be repeated and the letters Z and N are not used, then there are $\small 24\cdot 23\cdot 22$ possible 3-letter combinations. If no digits are to be repeated, then there are $\small 10 \cdot 9 \cdot 8$ possible 3-digit combinations. Multiply these two results together, and you get a total of 8,743,680 possible license plates $\small (24\cdot 23\cdot 22\cdot 10\cdot 9\cdot 8)$ .

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Example Question #11 : How To Use The Multiplication Rule

Tina and her 2 close friends are going to get together to see a movie Friday night, and Tina wants to determine the probability that they will all want to see the same one. There are 6 movies playing in the local theater, so the friends each wrote down the name of the movie they want to see. (They are all appealing movies that and have an equal chance of being chosen.)

What is the probability that all 3 girls will choose the same movie out of the 6 playing?

Possible Answers:

$\frac{1}{36}$

$\frac{1}{6}$

$\frac{1}{18}$

$\frac{1}{216}$

$3^{6}$

Correct answer:

$\frac{1}{216}$

Explanation:

When determining the probability of independent events, you multiply the probability of each event occurring.

The probability of a single girl choosing a specific movie is 1 out of 6, so to find the probability of this happening the same 3 times, you multiply 1/6 three times:

$\frac{1}{6}*\frac{1}{6}*\frac{1}{6}=\frac{1}{216}$

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Example Question #32 : Probability

Mike's five-person family is going out to dinner, and each person is planning to order a soda. The restaurant offers Soda1 and Soda2 only, and each family member likes both sodas equally.

What is the probability that all 5 family members will order Soda1?

Possible Answers:

$\frac{1}{64}$

$\frac{1}{32}$

$\frac{1}{2}$

$\frac{1}{16}$

$\frac{1}{10}$

Correct answer:

$\frac{1}{32}$

Explanation:

When determining the probability of several events occuring independently, you use the multiplication rule, meaning that you multiply the probability of each individual event occuring.

In this problem, the events are independent, meaning that each person's soda order does not affect the probabilty of someone else's order.

The probability of one person choosing Soda1 is 1 out of 2, or $\frac{1}{2}$ .

The probability of all five people ordering Soda1 is:

$\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2} = \frac{1}{32}$

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Example Question #34 : Statistical Patterns And Random Phenomena

Jenna and her two sisters are all picking a random number from 1 to 10.

What is the probability that they will all choose the same number?

Possible Answers:

$\frac{1}{10}$

$\frac{1}{9}{}$

$\frac{1}{10000}$

$\frac{1}{100}$

$\frac{1}{1000}$

Correct answer:

$\frac{1}{1000}$

Explanation:

When determining the probability of several events occuring independently, you use the multiplication rule, meaning that you multiply the probability of each individual event occuring.

In this problem, the events are independent, meaning that each person's number choice doesn't impact the other person's number choice

The probability of one person choosing a specific number is $\frac{1}{10}$ .

The probability of all 3 people choosing the same specific number is:

$\frac{1}{10}*\frac{1}{10}*\frac{1}{10}= \frac{1}{1000}$

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Example Question #12 : How To Use The Multiplication Rule

A person rolls a single 6 sided dice. What is the probability they will roll a 2 or a 4?

Possible Answers:

$\frac{5}{6}$

$\frac{1}{3}$

$\frac{3}{6}$

$\frac{4}{6}$

Correct answer:

$\frac{1}{3}$

Explanation:

In a single roll of a dice, rolling a 4 is mutually exclsuve of getting a 2. Therefore we will use the addition rule to find the probability of getting a 2 or a 4.

P(AorB)=P(A)+P(B)

First, find the probability of each seperate event.

$P(2)=\frac{1}{6}$

$P(4)=\frac{1}{6}$

Because the problem asks for the probability of a 2 "or" a 4, we will add the probability of each event.

$P(2or4)=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}$

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Example Question #11 : How To Use The Multiplication Rule

A tutoring agency helps match tutors with students. The agency knows that percent of all the tutors it places with students will leave the position within a year, but after the first year, only percent of the tutors who stay on will leave. At the start of the school year, an elementary school hires 100 tutors from the agency, then the next year it hires 100 more. How many of the 200 tutors are expected to still be working with their assigned students at the end of the second year?

Possible Answers:

156

172

154

112

140

Correct answer:

156

Explanation:

If the school starts with 100 tutors, it can expect to still be tutoring at the end of the first year. After that, we expect only percent or tutors to leave. So at the end of the second year, there will be the original tutors hired the at the start of the first year and the remaining tutors who were hired at the start of the second year. 80 + 76 = 156 .

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Example Question #33 : Probability

If you have a deck of cards, what is the probability that you draw a spade after you drew a non-spade on the first draw without replacement?

Possible Answers:

Correct answer:

Explanation:

You must use the multiplication rule which is the probability of one event happening after one has already taken place is the product of both probabilities. The probability of drawing a non-spade on the first draw is $\frac{39}{52}$ . Since there is no replacment, there are now 51 cards in the deck. The probability of drawing the spade on the second draw is $\frac{13}{51}$ . The probability of both happening after one another is then $\frac{39}{52}\ast \frac{13}{51}$ =

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AP Statistics : How to use the multiplication rule

Study concepts, example questions & explanations for AP Statistics

All AP Statistics Resources

Example Questions

Example Question #1 : Identifying Variables And Relationships

Example Question #22 : Rules Of Probability

Example Question #30 : Probability

Example Question #31 : Probability

Example Question #11 : How To Use The Multiplication Rule

Example Question #32 : Probability

Example Question #34 : Statistical Patterns And Random Phenomena

Example Question #12 : How To Use The Multiplication Rule

Example Question #11 : How To Use The Multiplication Rule

Example Question #33 : Probability

All AP Statistics Resources

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