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Algebra II : Algebra II

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Example Questions

Example Question #61 : Algebra Ii

What is the probability of flipping exactly one heads and exactly one tails in two coin tosses?

Possible Answers:

100%

Cannot be determined

50%

25%

Correct answer:

50%

Explanation:

Each of the two coin flips represents one event. The probability of obtaining exactly one heads and exactly one tails can be modeled as follows:

Event 1 * Event 2

The key to understanding this problem is to recognize that either heads followed by tails, or tails followed by heads, would satisfy the specific overall outcome asked for in the problem. Since Event 1 will produce either a heads or a tails, we have a 100% chance of obtaining an outcome from Event 1 that will satisfy one of the two requirements of the specific overall outcome (one heads and one tails). Event 2 will then have a 50% chance of producing an outcome that is the opposite, rather than the same, as the outcome of Event 1. We can therefore calculate the probability of our specific overall outcome as follows:

100% * 50%

1 * 0.5

0.5 = 50%

Therefore, the probability of obtaining one heads and one tails from two coin tosses is 50%.

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Example Question #1 : Binomial Theorem

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?

Possible Answers:

0.002

0.0002

0.000004

0.32

Correct answer:

0.002

Explanation:

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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Example Question #1 : Binomial Theorem

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

Possible Answers:

4,096

322,560

114, 688

448

Correct answer:

448

Explanation:

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 a = x, b = 4,n=8 , 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 i = 2 and evaluate:

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

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Example Question #1 : Binomial Theorem

What is the coefficient of x^4 in the polynomial (x+3)^7

Possible Answers:

315

945

Correct answer:

945

Explanation:

The binomial theorem says that we can represent the polynomial (x+a)^n as a sum

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

Thus, if we want to find the coefficient of x^4 ,

k=3 since 7-4=3 ,

a=3 and n=7 .

Therefore the coefficient of x^4 is

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

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Example Question #62 : Algebra Ii

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?

Possible Answers:

0.60

0.0055

0.040

0.045

0.0060

Correct answer:

0.0055

Explanation:

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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Example Question #4 : Binomial Theorem

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

Possible Answers:

Correct answer:

Explanation:

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 . n*p=50*.5=25

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Example Question #4 : Binomial Theorem

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.

Possible Answers:

0.00587

0.03694

0.08304

0.00213

0.00332

Correct answer:

0.00332

Explanation:

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 n,r,p,q .

: 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

n=12 , r=7 , $p= \frac{1}{5}$ , $q = \frac{4}{5}$

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: 0.00332

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Example Question #5 : Binomial Theorem

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?

Possible Answers:

$\frac{256}{625}$

$\frac{5}{1024}$

$\frac{4}{375}$

$\frac{4}{625}$

$\frac{7}{775}$

Correct answer:

$\frac{4}{625}$

Explanation:

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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Example Question #8 : Binomial Theorem

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?

Possible Answers:

$\frac{81}{8192}$

$\frac{135}{4096}$

$\frac{27}{2048}$

$\frac{45}{16384}$

$\frac{1215}{4096}$

Correct answer:

$\frac{135}{4096}$

Explanation:

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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Example Question #2 : Binomial Theorem

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

Possible Answers:

0.0625

0.5

0.117

0.000976

0.205

Correct answer:

0.205

Explanation:

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

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

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

=0.205

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #61 : Algebra Ii

Example Question #1 : Binomial Theorem

Example Question #1 : Binomial Theorem

Example Question #1 : Binomial Theorem

Example Question #62 : Algebra Ii

Example Question #4 : Binomial Theorem

Example Question #4 : Binomial Theorem

Example Question #5 : Binomial Theorem

Example Question #8 : Binomial Theorem

Example Question #2 : Binomial Theorem

All Algebra II Resources

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