Recent questions in Binomial probability

High school probabilityAnswered question

ystyrixkzd 2023-03-31

Write formula for the sequence of -4, 0, 8, 20, 36, 56, 80, where the order of f(x) is 0, 1, 2, 3, 4, 5, 6 respectively

High school probabilityAnswered question

ballar9bod 2023-03-31

Assume that when adults with smartphones are randomly selected, 54% use them in meetings or classes (based on data from an LG Smartphone survey). If 8 adult smartphone users are randomly selected, find the probability that exactly 6 of them use their smartphones in meetings or classes?

High school probabilityAnswered question

Aydan Hardy 2023-03-31

A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n independent trials of the experiment.n=20,

p=0.7,

x=19

P(19)=

p=0.7,

x=19

P(19)=

High school probabilityAnswered question

Kelton Rogers 2023-03-30

In binomial probability distribution, the dependents of standard deviations must includes.

a) all of above.

b) probability of q.

c) probability of p.

d) trials.

a) all of above.

b) probability of q.

c) probability of p.

d) trials.

High school probabilityAnswered question

Tyrell Singleton 2023-02-17

Which of the following polynomials are binomials?

(1) p(x)=x+1

(2) q(x)=x^3+x

(3) r(x)=sqrt2+x+x^2

(4) r(u)=u+u^2−2

(1) p(x)=x+1

(2) q(x)=x^3+x

(3) r(x)=sqrt2+x+x^2

(4) r(u)=u+u^2−2

High school probabilityAnswered question

daliner4jl 2023-02-12

Find the remainder when ${5}^{99}$ is divided by 8 is

High school probabilityAnswered question

Harold Prince 2023-01-17

Use the summation formulas to rewrite the expression without the summation notation. Use the result to find the sums for $n=10,100,1000and10,000$.$\sum k=1n\frac{6k\left(k-1\right)}{{n}^{3}}$

High school probabilityAnswered question

Savion Cameron 2022-12-18

Which polynomial is a quintic binomial?

A. ${\mathrm{x}}^{4}-2{\mathrm{x}}^{3}-{\mathrm{x}}^{2}+7\mathrm{x}+11$

B. ${\mathrm{x}}^{2}+4\mathrm{x}$

C. $3{\mathrm{x}}^{5}+2$

D. $5{\mathrm{x}}^{2}-2\mathrm{x}+1$

A. ${\mathrm{x}}^{4}-2{\mathrm{x}}^{3}-{\mathrm{x}}^{2}+7\mathrm{x}+11$

B. ${\mathrm{x}}^{2}+4\mathrm{x}$

C. $3{\mathrm{x}}^{5}+2$

D. $5{\mathrm{x}}^{2}-2\mathrm{x}+1$

High school probabilityAnswered question

brojevnids4 2022-12-14

What is the expansion of ${\left(x-1\right)}^{4}$?

High school probabilityAnswered question

merodavandOU 2022-12-02

Use the binomial theorem to expand $(d-4b{)}^{3}$

High school probabilityAnswered question

Layla Fisher 2022-11-24

Expanding $(a+b{)}^{\frac{1}{2}}$

I was wondering if it's possible to expand $(a+b{)}^{\frac{1}{2}}$.For example, $(a+b{)}^{2}={a}^{2}+2ab+{b}^{2}$. But what is the expansion of $(a+b{)}^{\frac{1}{2}}$? I've learned about binomial theorem but I can't figure it out.

I was wondering if it's possible to expand $(a+b{)}^{\frac{1}{2}}$.For example, $(a+b{)}^{2}={a}^{2}+2ab+{b}^{2}$. But what is the expansion of $(a+b{)}^{\frac{1}{2}}$? I've learned about binomial theorem but I can't figure it out.

High school probabilityAnswered question

fabler107 2022-11-20

What is the coefficient of ${x}^{101}{y}^{99}$ in the expansion of $(2x-3y{)}^{200}$?

A. $C(200,99){2}^{101}(3{)}^{99}$

B. $C(200,99){2}^{101}(-3{)}^{99}$

C. $P(200,99){2}^{101}(3{)}^{99}$

D. $P(200,99){2}^{101}(-3{)}^{99}$

E. $C(200,2){2}^{101}(-3{)}^{99}$

A. $C(200,99){2}^{101}(3{)}^{99}$

B. $C(200,99){2}^{101}(-3{)}^{99}$

C. $P(200,99){2}^{101}(3{)}^{99}$

D. $P(200,99){2}^{101}(-3{)}^{99}$

E. $C(200,2){2}^{101}(-3{)}^{99}$

High school probabilityAnswered question

odcizit49o 2022-11-15

A card is drawn and replaced five times from an ordinary deck of 52 cards and the sequence of colors is observed. What is the probability that:

a) Five red cards were drawn?

b) Five black cards were drawn?

c) Three red and two black cards were drawn?

d) why is it necessary to replace the cards?

My thoughts:

a) ${}^{5}{P}_{1}{\left({\displaystyle \frac{26}{52}}\right)}^{1}(1-p{)}^{4}+...{+}^{5}{P}_{5}{\left({\displaystyle \frac{26}{52}}\right)}^{5}(1-p{)}^{0}$

b) isn't this the same as part (a) ?

c) isn't this the same as asking exactly 5 black or red cards were drawn ?

d) not sure about this one.

a) Five red cards were drawn?

b) Five black cards were drawn?

c) Three red and two black cards were drawn?

d) why is it necessary to replace the cards?

My thoughts:

a) ${}^{5}{P}_{1}{\left({\displaystyle \frac{26}{52}}\right)}^{1}(1-p{)}^{4}+...{+}^{5}{P}_{5}{\left({\displaystyle \frac{26}{52}}\right)}^{5}(1-p{)}^{0}$

b) isn't this the same as part (a) ?

c) isn't this the same as asking exactly 5 black or red cards were drawn ?

d) not sure about this one.

High school probabilityAnswered question

ajakanvao 2022-11-11

Differentiate between basic and binomial probability

So I saw a question under the topic for Binomial Distributions which asks that what is the probability of making 4 out of 7 free throws where the $P(makingafreethrow)=0.7$. Why can't the answer be a simple $(0.7{)}^{4}$? Why would it be $7C4\ast (0.7{)}^{4}\ast (0.3{)}^{3}$?

So I saw a question under the topic for Binomial Distributions which asks that what is the probability of making 4 out of 7 free throws where the $P(makingafreethrow)=0.7$. Why can't the answer be a simple $(0.7{)}^{4}$? Why would it be $7C4\ast (0.7{)}^{4}\ast (0.3{)}^{3}$?

High school probabilityAnswered question

Jonas Huff 2022-11-10

Probability using binomial distribution

A probability that a manufactured device has 3% or more deffects is $p=0.02$. If a company buys 5 devices, what is the probability that at least one has 3% or more defects.

I am thinking using binomial distribution to find probabilities that 1, 2, 3, 4 or all 5 devices are defective. So

$P(A)={\textstyle (}\genfrac{}{}{0ex}{}{5}{1}{\textstyle )}p(1-p{)}^{4}+{\textstyle (}\genfrac{}{}{0ex}{}{5}{2}{\textstyle )}{p}^{2}(1-p{)}^{3}+{\textstyle (}\genfrac{}{}{0ex}{}{5}{3}{\textstyle )}{p}^{3}(1-p{)}^{2}+{\textstyle (}\genfrac{}{}{0ex}{}{5}{4}{\textstyle )}{p}^{4}(1-p)+{\textstyle (}\genfrac{}{}{0ex}{}{5}{5}{\textstyle )}{p}^{5}$

Is this a correct approach or there is an easier way to solve the problem?

A probability that a manufactured device has 3% or more deffects is $p=0.02$. If a company buys 5 devices, what is the probability that at least one has 3% or more defects.

I am thinking using binomial distribution to find probabilities that 1, 2, 3, 4 or all 5 devices are defective. So

$P(A)={\textstyle (}\genfrac{}{}{0ex}{}{5}{1}{\textstyle )}p(1-p{)}^{4}+{\textstyle (}\genfrac{}{}{0ex}{}{5}{2}{\textstyle )}{p}^{2}(1-p{)}^{3}+{\textstyle (}\genfrac{}{}{0ex}{}{5}{3}{\textstyle )}{p}^{3}(1-p{)}^{2}+{\textstyle (}\genfrac{}{}{0ex}{}{5}{4}{\textstyle )}{p}^{4}(1-p)+{\textstyle (}\genfrac{}{}{0ex}{}{5}{5}{\textstyle )}{p}^{5}$

Is this a correct approach or there is an easier way to solve the problem?

High school probabilityAnswered question

Aden Lambert 2022-11-03

Relationship between binomial and negative binomial probabilities

Let X be a negative binomial random variable with parameters r and p, and let Y be a binomial random variable with parameters n and p. Show that $\mathbb{P}(X>n)=\mathbb{P}(Y<r).$

I would like to get an analytic solution. Basically I want to show the following equality mathematically:

$$\sum _{i=n+1}^{\mathrm{\infty}}{\textstyle (}\genfrac{}{}{0ex}{}{i-1}{r-1}{\textstyle )}{p}^{r}(1-p{)}^{r}=\sum _{i=0}^{r-1}{\textstyle (}\genfrac{}{}{0ex}{}{n}{i}{\textstyle )}{p}^{i}(1-p{)}^{n-i}$$

Let X be a negative binomial random variable with parameters r and p, and let Y be a binomial random variable with parameters n and p. Show that $\mathbb{P}(X>n)=\mathbb{P}(Y<r).$

I would like to get an analytic solution. Basically I want to show the following equality mathematically:

$$\sum _{i=n+1}^{\mathrm{\infty}}{\textstyle (}\genfrac{}{}{0ex}{}{i-1}{r-1}{\textstyle )}{p}^{r}(1-p{)}^{r}=\sum _{i=0}^{r-1}{\textstyle (}\genfrac{}{}{0ex}{}{n}{i}{\textstyle )}{p}^{i}(1-p{)}^{n-i}$$

High school probabilityAnswered question

Kamila Frye 2022-10-26

Binomial probabilities in a production process

Screws are made in a production process where the probability of any one screw being defective is constant at $p=0.1$ i.e. 10% of screws produced are defective.

Screws are placed in bags of 15 at the end of the process. At various intervals a bag is checked and the process is stopped if more than 3 screws in that bag are found to be defective.

I am trying to answer the question "What is the probability that there will be sufficient defective screws to stop the process?".

I am using a sum of binomial probabilities and have calculated the answer to be 0.0555. Does this appear correct?

Screws are made in a production process where the probability of any one screw being defective is constant at $p=0.1$ i.e. 10% of screws produced are defective.

Screws are placed in bags of 15 at the end of the process. At various intervals a bag is checked and the process is stopped if more than 3 screws in that bag are found to be defective.

I am trying to answer the question "What is the probability that there will be sufficient defective screws to stop the process?".

I am using a sum of binomial probabilities and have calculated the answer to be 0.0555. Does this appear correct?

High school probabilityAnswered question

JetssheetaDumcb 2022-10-22

Prove that $\sum _{k=0}^{m}\frac{{\textstyle (}\genfrac{}{}{0ex}{}{m}{k}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}}=\frac{n+1}{n+1-m}$

High school probabilityAnswered question

rancuri5a 2022-10-02

What is Cumulative Binomial probabilities?

I am new to this so don't know if I am asking the right question as I just read about its usage but didn't know what exactly a Cumulative Binomial probability is.

So my question is, What is Cumulative Binomial probabilities ? any example will be of great help.

I am new to this so don't know if I am asking the right question as I just read about its usage but didn't know what exactly a Cumulative Binomial probability is.

So my question is, What is Cumulative Binomial probabilities ? any example will be of great help.

High school probabilityAnswered question

Elena Marquez 2022-10-02

Why is Binomial Probability used here?

A test consists of 10 multiple choice questions with five choices for each question. As an experiment, you GUESS on each and every answer without even reading the questions.What is the probability of getting exactly 6 questions correct on this test?

The answer says:

$$P={\textstyle (}\genfrac{}{}{0ex}{}{10}{6}{\textstyle )}(0.2{)}^{6}(0.8{)}^{4}$$

I was thinking though:

$$P(A\text{and}B)=P(A)\cdot P(B)$$

Probability of getting 6 right and 4 wrong would be just:

$$(0.2{)}^{6}(0.8{)}^{4}$$

Why is the binomial coefficient used there?

A test consists of 10 multiple choice questions with five choices for each question. As an experiment, you GUESS on each and every answer without even reading the questions.What is the probability of getting exactly 6 questions correct on this test?

The answer says:

$$P={\textstyle (}\genfrac{}{}{0ex}{}{10}{6}{\textstyle )}(0.2{)}^{6}(0.8{)}^{4}$$

I was thinking though:

$$P(A\text{and}B)=P(A)\cdot P(B)$$

Probability of getting 6 right and 4 wrong would be just:

$$(0.2{)}^{6}(0.8{)}^{4}$$

Why is the binomial coefficient used there?

Binomial probability is one of those subjects that we learn during high school studies, yet good binomial probability examples are not mentioned in most cases, which doesn’t help to solve various questions or find the answers without turning to formulas that do not fit. It’s all about being able to calculate the probability as binomial probability problems can be compared to purchasing a lottery ticket where one can either win or leave without any luck. Addressing the lack of examples, we provide binomial questions and explain how certain answers have been found to address this or that binomial probability equation.