Probability - AP Statistics

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Question

How many different football teams of 11 members may be created from 13 men, without regard to the position played by any of the members?

Answer

There are 13 men to choose from in order to make an 11 member football team. Thus, the answer is set up as: 13 choose 11: $\frac{13!}{11!2!}=78$ .

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Question

Number of Emails
Probability

The number of emails that Sam can respond to per hour is given by this probability distribution. What is the average number of emails that Sam respons to per hour?

Answer

The answer is found by multiplying by the probability :

emails

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Question

A child has a bag of marbles-- red, blue, and yellow. The child randomly selects one marble and gives it away. The child then selects a second marble. What is the probability that both marbles selected were yellow?

Answer

The two events in this case are dependent (the result of the first event will affect the outcome probabilities of the second event). Find the probability of a particular outcome for both events by multiplying the probability of event A by the probability of event B given the result of A. In other words, multiply the probability of both events, remembering to account for the result of event A in determining the probability of event B. Here, the probability of selecting a yellow marble the first time is 3/8, and the second time is 2/7.

$3/8\cdot 2/7=6/56$

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Question

There are 10 horses in a herd. 4 are males, 6 are females. 7 of the horses are brown while the remaining 3 are white. 5 female horses are brown. A horse is randomly selected from the herd. Given that the horse is brown, what is the probability that the horse is a male?

Answer

The probability of any event occurring is the number of outcomes in which case that event occurs divided by the number of total number of outcomes. Here you know that the total number of brown horses is 7, and if 5 of those brown horses are female then 2 must be male. So the probability that a brown horse is male is 2/7.

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Question

There are 10 horses in a herd. 4 are males, 6 are females. 7 of the horses are brown while the remaining 3 are white. At least 1 male horse is white and 5 female horses are brown.

One horse is selected at random from the herd. What is the probability of selecting a female horse given that the horse selected is white?

Answer

This question asks for the probability of an event given that one event has occurred, meaning it is a conditional probability. The equation for a conditional probability is:

$P(A|B)=\frac{P(A\cap B)}{P(A)}$

It is important to properly assign the A and B variables according to the question. It helps to read the equation out loud: "The probability of A given B equals the probability of A and B divided by the probability of A."

We need to find the probability of a female given that the horse is white. Therefore, female horse is A, white horse is B.

So, find each probability needed for the equation.

$P(A)=P(female)=\frac{6}{10}=0.6$

$P(B)=P(white)=\frac{3}{10}=0.3$

If 6 of the 7 females are brown, then 1 horse out of 10 is a white female horse.

$P(A\cap B)= P(female\cap white)= \frac{2}{10}=0.2$

Now, use the information to solve the equation.

$P(A|B)= \frac{0.2}{0.6}=0.33$

Alternatively you could simply count out the number of favorable outcomes: there are 6 total female horses and 5 are brown, so that leaves 1 white female horse (the "favorable outcome" here). And there are 3 total white horses. So if you know that you're selecting from the pool of 3 white horses and only 1 is female, the probability of choosing a female white horse from that set is 1/3.

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Question

There are 10 horses in a herd. 4 are males, 6 are females. 7 of the horses are brown while the remaining 3 are white. 1 male horse is white, 5 female horses are brown.

A horse is randomly selected from the herd. What is the probability that the horse chosen is brown or female?

Answer

This problem asks for the probability of one event or another, so the addition rule of probability is approriate. However, brown horse and female horse are not mutually exclusive events, because there are brown female horses. Therefore we must subtract the brown male horses to avoid double counting.

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$P(brown)=\frac{7}{10}$

$P(female)=\frac{6}{10}$

$P(female\cap brown)=\frac{5}{10}$

Now, fill in the equation.

$P(female \cup brown)=\frac{6}{10}+\frac{7}{10}-\frac{5}{10}=\frac{8}{10}=\frac{4}{5}$

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Question

The probability that there will be an accident on Highway each day is determined by the weather. If it's a dry day, there is a % chance of an accident, and a % chance if it's a wet day. The radio says that there is a % chance of rain today. What is the probability that there will be an accident on Highway today?

Answer

The chance of an accident is dependent on the weather conditions, so that is the first branch on the tree diagram.

The probabilities at the next branch are conditional and the last two branches have the chance of accident.

$P(wet \cap accident) = 0.2 \cdot0.010 = 0.0020$ and $P(dry\cap accident) = 0.80 \cdot 0.002 = 0.0016$ ,

so the total probability of an accident is .

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Question

There is a sample of 100 randomly selected children (60 of which are boys and 40 are girls). They were asked if they liked to play football. 55 boys answered yes to the question and 5 of the girls aswered yes to the question. What is the probability that a child said yes to the question give that they are a boy?

Answer

The conditional probability formula is the probability of both events happening divided by the probability of the given event. The probability of saying "yes" and also being a boy is $\frac{55}{100}$ . The probability of being a boy in the sample is $\frac{60}{100}$ . So $\frac{55}{100}\div\frac{60}{100}$ = $\frac{55}{60}$ = .

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Question

Variable has a Poisson distribution with a mean of . What is the variance of Variable ?

Answer

Because has a Poisson distribution we know that:

$E(A)=\mu$ and

$V(A)=\sigma^2=\mu$ .

Therefore, since we are given the mean of 25, we can find its variance to also be 25.

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Question

Answer

No explanation available

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Question

Answer

No explanation available

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Question

Let us suppose you are a waiter. You work your first four shifts and receive the following in tips: (1) 20, (2) 30, (3) 15, (4) 5. What is the mean amount of tips you will receive in a given day?

Answer

The answer is 17.5. Simply take the values for each day, add them, and divide by the total number of days to obtain the mean: $\frac{20+30+15+5}{4}$

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Question

Robert's work schedule for next week will be released today. Robert will work either 45, 40, 25, or 12 hours. The probabilities for each possibility are listed below:

45 hours: 0.3

40 hours: 0.2

25 hours: 0.4

12 hours: 0.1

What is the mean outcome for the number of hours that Robert will work?

Answer

We are required to find the mean outcome where the probability of each possible result varies--the random/weighted mean. First, multiply each possible outcome by the probability of that outcome occurring. Second, add these results together.

$(45\cdot 0.3) + (40\cdot 0.2) + (25\cdot 0.4) + (12\cdot 0.1) = 32.7$

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Question

There are collectable coins in a bag. are ounces, are ounces, are ounces, and are ounces. If one coin is randomly selected, what is the mean possible weight in ounces?

Answer

We are required to find the mean outcome where the probability of each possible result varies--the random/weighted mean.

First, multiply each possible outcome by the probability of that outcome occurring.

Second, add these results together.

$(0.4\cdot 3)+(0.1\cdot 5)+(0.3\cdot 6)+(0.2\cdot 7)=4.9$

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Question

A basketball player makes of his three-point shots. If he takes three-point shots each game, how many points per game does he score from three-point range?

Answer

First convert $40% \rightarrow 0.40$ .

The player's three-point shooting follows a binomial distribution with and .

On average, he thus makes $np = 10\cdot 0.4 = 4$ three-point shots per game.

This means he averages 12 points per game from three-point range if he tries to make 10 three-pointers per game.

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Question

Tim samples the average plant height of potato plants for his science class and finds the following distribution (in inches):

Which of the following is/are true about the data?

i: the mode is

ii: the mean is

iii: the median is

iv: the range is

Answer

Analyzing the data, there are more 6s than anything else (mode), the median is between and , the mean is , and the range is

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Question

During a week's worth of soccer practice, a player practices total free kicks and has a chance of scoring. What is the probability that he or she scored at least times? Assume each shot is independent.

Answer

Two steps are crucial here.

First, we need to recognize this is a binomial distribution with and .

Second, we need to realize we can use a normal approximation of the binomial since $np = 50\cdot 0.25 = 12.5$ and $n(1 - p) = 50\cdot 0.75 = 37.5$ , which are both larger than 5.

With that said, we can calculate a -score and its -value, keeping in mind that our mean will be and our standard deviation will be $\sqrt{np(1-p)}$ , which is about .

$Z = \frac{18 - 12.5}{3.062} = 1.80$

$P(Z \geq 1.80) = 0.036$

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Question

Which of the following would be considered a binomial experiment?

Answer

There are four conditions that need to be satisfied for a binomial experiment:

Each trial must have two outcomes.

Each trial must be independent.

All trials must be identical.

The probabilities of the outcomes remain constant must not change with each trial.

The only choice that satisfies all four of these conditions (and is therefore a binomial experiment) is the rock-paper-scissors scenario.

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Question

Suppose you are throwing three darts and you have a one third chance of hitting the bull's eye. Each throw is independent of one another. What is the chance of hitting the bull's eye at least once?

Answer

To calculate Prob(at least one bull's eye), we can instead compute one minus the complementary probability, P(no bull's eye).

So we have P(at least one bull's eye)=1-P(no bull's eye).

The chance of getting no bull's eyes is $\left (\frac{2}{3}\right)^3$ .

This means the probability of getting at least one bull's eye is

$1-\left(\frac{2}{3}\right)^3=\frac{27}{27}-\frac{8}{27}=\frac{19}{27}$

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Question

A particle travels left with probability one sixth and right with probability five sixths. Each movement is independent of the others. What is the chance that after three movements, the particle ends up one unit to the right?

Answer

The movements that this particle can make include: RRL, RLR, LRR.

The chance of getting RRL is $\small \left( \frac{5}{6} \right )^2\left(\frac{1}{6} \right )$ . This is also the chance of getting any of those movements.

To get the total probability, we can add up the individual probabilities since the events are all mutually exclusive.

Thus, we get the following as the solution.

$\small 3\left( \frac{5}{6} \right )^2\left(\frac{1}{6} \right )$

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