Probability - Algebra II

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Question

Assume that you guess on each question of a multiple choice test. There are 12 questions and each question has 4 possible answers. What is the probability of getting exactly 8 answers correct?

Answer

This problem can be solved using the binomial probability equation:

$\small P=\binom{n}{k}p^kq^{n-k}$

$\small n=total\ number\ of\ questions\ (trials)$

$\small k=number\ correct\ (successes)$

$\small n-k=number\ incorrect\ (failures)$

$\small p=probability\ of\ getting\ 1\ question\ correct\ (0.25)$

$\small q=1-p\ probability\ of\ getting\ 1\ question\ incorrect\ (0.75)$

$\small \small \binom{n}{k}=\frac{n!}{k!(n-k)!}$

$\small \small P=(\frac{n!}{k!(n-k)!})p^kq^{n-k}=\frac{12!}{8!(4)!}(0.25)^8(0.75)^4$

$\small P=\frac{12\times11\times10\times9}{4\times3\times2\times1}(0.25)^8(0.75)^4=0.002$

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Question

What is the coefficient of $x^{6}$ if the expression $(x + 4)^{8}$ is expanded?

Answer

By the Binomial Theorem, if the expression $(x+a) ^{n}$ is expanded, the result can be defined as

$(a+b) ^{n} = \sum_{i=0}^{n} C(n, i) a^{n-i} b^{i}$

If we set , then the above expression, with slight rearranging, becomes

$(x+4) ^{8} = \sum_{i=0}^{8} C(8, i) \cdot 4^{i} \cdot x^{8-i}$

The coefficient of the $x^{8-i}$ term is

$C(8, i) \cdot 4^{i}$

To find the the coefficient of $x^{6}$ , we set and evaluate:

$C(8, 2) \cdot 4^{2} = 28 \cdot 16 = 448$

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Question

What is the coefficient of in the polynomial

Answer

The binomial theorem says that we can represent the polynomial as a sum

$\sum_{k=0}^n\binom{n}{k} a^k x^{n-k}$

Thus, if we want to find the coefficient of ,

since ,

and .

Therefore the coefficient of is

$\binom{7}{4}3^3 = \frac{7!}{4!(7-4)!}3^3 = 945$

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Question

You're taking a multiple choice quiz that has questions. If each question has choices and you guess on all of them, what is the probability of getting exactly questions correct?

Answer

This question requires the application of the binomial theorem for probability. In order to determine the probability of getting exactly 6 questions right, we must remember the formula for this theorem:

$P=\begin{pmatrix} n\k \end{pmatrix}p^kq^{n-k}$

Where is the number of trials (total questions), is the number of successes (correct answers), is the probability of success in one trial (chance of answering a question correctly), is the probability of failure in one trial (chance of answering a question incorrectly), and is the probability of getting questions correct out of total questions. Because there are 5 choices for each question, the chances of answering a question correctly are 1/5, or 0.2, and therefore the chances of answering a question incorrectly are 4/5, or 0.8. Now we have all of our values and can plug them into the formula:

$P=\begin{pmatrix} 10\6 \end{pmatrix}(0.2)^6(0.8)^{10-6}$

The first part of the formula in which we have a 10 over a 6 in parentheses means we perform the following calculation:

$\begin{pmatrix} 10\6 \end{pmatrix}=\frac{10!}{(10-6)!6!}=\frac{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{(4\cdot3\cdot2\cdot1)(6\cdot5\cdot4\cdot3\cdot2\cdot1)}$

$=\frac{10\cdot9\cdot8\cdot7}{4\cdot3\cdot2\cdot1}=\frac{5040}{24}=210$

So now we can put this value into our formula, which gives us:

$P=(210)(0.2)^6(0.8)^{10-6}=0.0055$

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Question

Which is equivalent to $(a+b)^{4}$ ?

Answer

To answer this question, you could either use the binomial theorem or multiply . Both are detailed below.

Note: If you need help understanding Sigma Notation, please visit: https://www.varsitytutors.com/hotmath/hotmath_help/topics/sigma-notation-of-a-series; additionally, the notation is often read out loud as “n choose k” and is another way to write For more information on combinations, visit https://www.varsitytutors.com/hotmath/hotmath_help/topics/combinations

The Binomial Theorem is:

In our case, n=4. Plugging this in, we get:

Alternatively, we can solve this problem by multiplying to get the following:

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Question

A fair coin is tossed 10 times. What is the probability that four heads will be observed?

Answer

This is a binomial distribution with number of trials(n) equal to 10. The probability of success(p) is 0.5 because it is a fair coin. The number of success(r) is 4 because we want the probability of 4 heads.

The formula for a binomial distribution is

$(_nC_r)p^r(1-p)^{n-r}$

$(_{10}C_{4})*0.5^4+(1-0.5)^6$

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Question

A fair coin is tossed 50 times. What is the expected number of heads?

Answer

The expected probably for the binomial distribution is n*p. N is the number of trials, in this case the 50 coin tosses. p is the probability of heads. Since the coin is a fair coin the probability is .

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Question

A student takes a 12 question multiple choice test. There are 5 answer choices, and the student guesses on all the questions. What is the probability the student will get exactly 7 questions correct in order to pass? Round to 5 decimal places.

Answer

In order to determine the probability, we will need to use the binomial theorem.

The equation can be written in two ways:

$\binom{n}{r}\cdot p^r\cdot (1-p)^{n-r}$

Or:

$_nC_r \cdot p^r \cdot q^{n-r}$

Identify the definition and values for .

: represents the total number of trials
: represents the number of events
: represents the probability of occurrence per trial
: is the probability that the occurrence will not happen per trial

, , $p= \frac{1}{5}$ , $q = \frac{4}{5}$

Substitute the values into the formula.

$_{12}C_{7} \cdot (\frac{1}{5})^7 \cdot (\frac{4}{5})^{12-7}$

Recall that:

$C(n,r) =\frac{n!}{r! (n - r)!}$

$_{12}C_{7} = \frac{12!}{7!(12-7)!} = 792$

The terms become:

$792 \cdot (\frac{1}{5})^7 \cdot (\frac{4}{5})^{12-7} = \frac{811008}{244140625}$

Simplify the terms with calculator.

The answer is:

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Question

Suppose Billy takes a 5 question multiple choice test that has 5 answer choices per question. What is the probability that Billy will get exactly four correct to pass the test?

Answer

Write the binomial formula.

$P = _nC_r\cdot p^r\cdot q^{n-r} = \frac{n!}{r!(n-r)!}\cdot p^r\cdot q^{n-r}$

$n=\textup{number of trials }= 5$

$r= \textup{number of success} =4$

$p=\textup{probability of success per trial} = \frac{1}{5}$

$q=\textup{probability of failure per trail} = \frac{4}{5}$

Evaluate the probability.

$P=\frac{5!}{4!(5-4)!}\cdot (\frac{1}{5})^4\cdot (\frac{4}{5})^{5-4}$

$P = 5\cdot(\frac{1}{5})\cdot(\frac{1}{5})\cdot(\frac{1}{5})\cdot(\frac{1}{5})\cdot(\frac{4}{5})$

The answer is: $\frac{4}{625}$

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Question

Suppose a competitor has to answer four out of six multiple choice questions correctly to win a prize. There are four answer choices per question. What is the probability that this competitor will succeed answering exactly four correct questions if he or she guessed on all the questions?

Answer

This problem requires the binomial theorem. Write the formula.

$P = _nC_r\cdot p^r\cdot q^{n-r}$

This formula can also be rewritten as:

$P= \frac{n!}{r!(n-r)!}\cdot p^r\cdot q^{n-r}$

Identify all the terms.

$n=\textup{number of trials } =6$

$r= \textup{number of success} =4$

There are four answer choices per question, which means there is only one correct answer.

$p=\textup{probability of success per trial} = \frac{1}{4}$

The failure rate would be three out of the four questions.

$q=\textup{probability of failure per trail} = \frac{3}{4}$

Substitute the terms into the formula and simplify the terms.

$P= \frac{6!}{4!(6-4)!}\cdot (\frac{1}{4})^4\cdot (\frac{3}{4})^{6-4}$

$P= \frac{6\times 5}{2}\cdot (\frac{1}{4})(\frac{1}{4})(\frac{1}{4})(\frac{1}{4})\cdot (\frac{3}{4})(\frac{3}{4})$

$P= 15\cdot \frac{9}{4^6} = \frac{135}{4096}$

The answer is: $\frac{135}{4096}$

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Question

A biased coin is tossed. The probability of landing heads is , and the probability of landing tails is . What is the probability of tossing three tails in a row, followed by one head?

Answer

Each coin toss is an independent event, meaning the outcome of one toss does not affect the outcome of the other coin tosses. To find the overall probability, multiply the probabilities of each toss together:

$P = \frac{1}{3}\frac{1}{3}\frac{1}{3}*\frac{2}{3} = \frac{2}{81}$

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Question

What is the probability of drawing 2 aces from a standard deck of playing cards when the first card is not replaced?

Answer

There are 4 aces in a deck of 52 playing cards.

$\small \frac{4}{52}=\frac{1}{13}$

After 1 ace is removed, there are 3 aces and 51 cards remaining:

$\small \frac{3}{51}=\frac{1}{17}$

To determined the probability of both events, multiply the probability of drawing each card:

$\small P=\frac{1}{13}\times \frac{1}{17}=\frac{1}{221}$

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Question

A bag of marbles contains 4 gold and 6 blue marbles. What is the probability that you will draw first a blue marble then a gold marble? Assume the first marble you draw is not placed back in the bag.

Answer

Selecting a specific sequence of marbles from the bag without replacement is a dependent or conditional probabiliy. If you successfully draw a blue marble, it affects the chances of subsequently drawing a gold marble.

In the notation of probability we can call the probability of drawing a blue marble and the probability of drawing a gold marble .

If we replaced the first marble we drew, we would simply multiply times , but since we aren't we want to know or, the probability of B, given A.

So the probability of A and B is equal to:

$\ast$

$\frac{6}{10}\ast\frac{4}{9}=\frac{24}{90}=\frac{4}{15}$

In words, you have a six in ten chance of drawing a blue marble. If you successfully do that, you then have a four in nine chance of drawing a gold marble. Multiply those odds together, and reduce the fraction and you get a four in fifteen probability.

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Question

50 high school seniors were asked about their plans following graduation. 20 planned to attend an in state college. 12 planned to attend an out of state college. 8 did not plan to attend college and 10 were undecided.

What percent of this group of seniors is not planning to attend an in state college?

Answer

First, calculate the number of students not planning on attending an in state college.

Either subtract those planning on attending an in state college from the total...

50 - 20 = 30

...or add up all the other categories.

12 + 8 + 10 = 30

Then divide by the total number of students.

30/50 = 60%

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Question

In a sample of 100 high school students, 65 students played an organized sport. 48 students have a dog as a pet. What is the probability that a randomly selected student has a dog and plays a sport?

Answer

Playing a sport, which we can call P(A), and owning a dog, which we can call P(B), are traits that we can assume are independent. The probability of one does not affect the probability of the other.

So, P(A and B) = P(A) * P(B)

Fill in the probabilities given into this equation.

P(A and B) = (0.65) * (0.48) = 0.312 = 31.2%

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Question

In a sample of cars at a parking garage, the probability of a randomly selected car being blue was found to be 0.22. The probability of a car being blue OR green was found to be 0.33. What is the probability of car from this garage being green? (Assume a car can be only one color.)

Answer

The probability of finding either a blue car (A) or a green car (B) is written
P(A or B).

This type of probability can be found using the equation:

P(A or B) = P(A) + P(B) - P(A and B)

We need to solve this equation for P(B).

Two of the terms are given in the prompt:

P(A or B) = 0.33

P(A) = 0.22

The third, P(A and B) can be inferred. Since a car can be only one color,
P(A and B) = 0

Thus,

0.33 = 0.22 + P(B) - 0

P(B) = 0.11

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Question

In a standard deck of card, what is the probality of drawing either a face card or a red card?

Answer

The probability of drawing a face card P(A) or a red card P(B) can be written as P(A or B), and be calculated using:

P(A or B) = P(A) + P(B) - P(A and B)

The first two terms on the right side of the equation are fairly straightforward. Since we'll be adding and subtracting fractions, I will keep the denominator the same and won't reduce the fractions until the end.

P(A) = 12/52

P(B) = 26/52 = 1/2

The third, P(A and B), can be calculated using:

P(A and B) = P(A) * P(B)

P(A and B) = 12/52 * 1/2 = 12/104 = 6/52

(Conversely, think there are 12 face cards in a deck of 52, half of which \[6\] are red.)

Back to our original equation!

P(A or B) = 12/52 + 26/52 - 6/52

= 32/52 = 8/13

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Question

If a coin is tossed in the air and a 6-sided die is rolled, what is the probability of getting a tail on the coin and a 3 on the die?

Answer

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

$P(3)=\frac{1}{6}$

Multiply these two probabilities together:

$P(x)=\frac{1}{2}\cdot \frac{1}{6}=\frac{1}{12}$

Therefore the probability of flipping a tail on a coin and rolling a 3 is $\frac{1}{12}$ .

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Question

A restaurant serves apple pie and cherry pie. On average, 60% of the customers choose apple pie. If the next 4 customers order pie, what is the probability 3 of them order the cherry pie?

Answer

Binomial probability

$_{4}C_{3}\ast.6\ast.4^{3}$

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Question

Suppose a loaded coin lands on heads $\frac{4}{5}$ times.

What is the probability that the coin will land HTTH?

Answer

The probability of flipping the coin in this particular order is calculated by multiplying the probabilities of each turn by each other. Because we have a "loaded" coin the probabilities are as follows:

$\frac{4}{5}\cdot\frac{1}{5}\cdot\frac{1}{5}\cdot\frac{4}{5}$ .

Then simplify:

$\frac{4\cdot1\cdot1\cdot4}{5\cdot5\cdot5\cdot5}=\frac{16}{625}$ .

This fraction is in its simplest form.

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