How to use the multiplication rule - AP Statistics

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Question

Let us suppose that the probability of obtaining heads in a coin flip is 0.5 and the probability of the Earth being hit by an asteroid is 0.01.

The Pr(Obtaining Heads in a Coin Flip and Earth being hit by an Asteroid) .

Are these variables dependent or independent?

Answer

Recall that when two variables are independent, $P\left ( A\ and\ B \right )=P(A)\times P(B)$

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Question

In a standard deck of cards, without replacement, what is the probability of drawing three kings?

Answer

Start with 52 cards, probability of drawing first king: $\frac{4}{52}$

Now you have 51 cards. Probability of drawing second king: $\frac{3}{51}$

Now you have 50 cards. Probability of drawing third king: $\frac{2}{50}$

Multiply all probabilities:

$\frac{4}{52}\frac{3}{51}\frac{2}{50}=\frac{24}{132,600}=\frac{1}{5525}$

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Question

1 pencil and 7 pens are on a desk. One of the 8 items is randomly selected from the desk and then is replaced. Once again, an item is selected from the 8 items.

What is the probability that the pencil was selected both times?

Answer

To find the probability of outcomes for two separate events, multiply the probability of the two outcomes. Here, the two events are independent (the second event is not affected by the result of the first) and the probability is the same for both events.

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Question

1 pencil and 7 pens are on a desk. A pen was randomly selected from the 8 items and was set aside. A second item was then selected from the remaining 7 items.

What is the probability that the second item selected will be a pen?

Answer

The probability of the second event depends on the outcome of the first event. Since a pen was set aside as a result of the first event, there is a slightly lower probability of a pen being selected the second time. Six pens and one pencil remain. The probability of selecting a pen is:

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Question

If the probability of landing a heads in a coin flip is 0.5 and the probability of observing a meteor hit the earth is 0.03, and these events are independent, what is the probability of landing a heads AND observing a meteor hit the earth?

Answer

Since the two events are independent, multiply their probabilities to get their joint probability. Multiplying the probability of the coin flip, 0.5, by the probability of a meteor, 0.03, gives a probability of 0.015.

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Question

How many different combinations of 3 digit numbers can be formed using the numbers 1, 2, 3, 4, and 5, if repetitions are allowed?

Answer

The key to answering this question is noting that repetitions are allowed. This means that if a number is picked, it is replaced and may be picked again, thus allowing for duplicates or triplicates. Because there are 5 choices and after each number is picked there remain 5 choices (replacement), and the question is asking for 3 digit combinations, the answer is obtained by multiplying 5 * 5 * 5 = 125. In other words, there are 5 choices for the first digit, 5 choices for the second digit, and 5 choices for the third digit.

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Question

How many different combinations of 3 digit numbers can be formed using the numbers 1, 2, 3, 4, and 5, if repetitions are NOT allowed?

Answer

The important thing to note for this question is that there are no repetitions allowed. In other words, once a number had been chosen, it cannot be chosen for the second digit or the third digit. Thus, there are 5 choices for the first digit, 4 for the second, and 3 for the third. So, 5 * 4 * 3 = 60.

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Question

Given a fair coin, what is the probability of obtaining 5 heads and 3 tails from 8 tosses?

Answer

First, there are 8 trials and either choose 5 or 3 for heads or tails, respectively. Using this knowledge: $\frac{8!}{5!3!}$ . Next, the chance for either heads or tails is 0.5 and there are 5 heads and 3 tails. Thus: . Multiply: $\frac{8!}{5!3!}$ and and obtain 0.2188.

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Question

There are 52 total cards in a full deck of playing cards. If a card dealer chooses 4 cards from the deck at random and without replacement, what is the chance that the dealer draws four kings as the first four cards?

Answer

In a normal deck of playing cards, there are 4 kings. Thus, when the dealer draws the first card, the chance of the dealer obtaining a king is 4 out of 52. Because this card has been picked and is not replaced, the chance that the next card chosen is a king is 3 out of 52. The chance the third card is a king is 2 out of 52 and the fourth card is 1 out of 52. Each of these events is multiplied together, thus obtaining the correct answer, 0.0000037.

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Question

Research has found that the probability of having brown eyes is and the probability of having red hair is . Assuming these probabilities are independent, what is the probability of having brown eyes and red hair?

Answer

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Question

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

Answer

In a standard deck of cards:

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

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Question

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?

Answer

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.

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Question

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?

Answer

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

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

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

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

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Question

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?

Answer

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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Question

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?

Answer

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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Question

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?

Answer

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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Question

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?

Answer

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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Question

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

Answer

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.

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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Question

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 tutors from the agency, then the next year it hires more. How many of the tutors are expected to still be working with their assigned students at the end of the second year?

Answer

If the school starts with 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. .

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Question

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?

Answer

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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