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Calculate Binomial Probability for inequality.
How could the binomial probability be calculated for a case of $P(X\ge <!--\; \ge \; -->2)$ for a given value of n and p ?

Statistics Binomial Probability
Suppose that in a group of 10 people, each person has a 0.1 probability of having a certain disease, independently from person to person. Now consider a testing plan where the group is split into two groups of 5 people where in each group blood samples are taken from each person, then their blood samples are combined and tested together (so there is one test per group of 5). If a group tests positive for the disease then each person in that group is tested individually for the disease, if a group does not test positive, then the testing is concluded for that group. Let the random variable X represent the total number of tests required with this plan.
Find the probability distribution of X.
I know that there are 2, 7, 12 possible values of X but I don't know what formula to use to find P(X). I'm assuming its the binomial probability formula with $nCk(p^k)(q^n-<!--\; -\; -->k)$
but how do you use 2, 7 and 12 in that? what is n and k?

Use the Binomial Theorem to find the indicated coefficient or term
The coefficient of x in the expansion of $(2x+4{)}^5$
640
5120
2560
1280

How do I solve this question?
A school lab has sixteen computers. A teacher observes that, in the long run, in 80% of school days, at least 1 machine is not working properly. Assuming the probability of a computer not working properly is independent of the others, find the probability that:
a) a randomly chosen computer is not working in any school day
b) at least 2 computers are not working properly in any school day
Attempt: I’m really not sure how to approach this question. If for 80% of school days at least 1 computer doesnt work that means for 20% of school says at least 15 doesnt work. Now do you use the binomial expansion to find for all school days?

Binomial probability problem solving
The probability of sales representative making a sale with any one customer is $\frac{1}{3}$. The sales representative makes eight contacts a day. To find the probability of making four sales, evaluate the term
$$8C_4(\frac{1}{3}{)}^4(\frac{2}{4}{)}^4$$
In the expansion $$(\frac{1}{3}+\frac{2}{3}{)}^8$$
I know how to evaluate $$8C_4(\frac{1}{3}{)}^4(\frac{2}{4}{)}^4$$ but I don't understand what they mean by " in the expansion of $$(\frac{1}{3}+\frac{2}{3}{)}^8$$

Binomial probability for large n, small p
- I need to compute the probability of getting more than x "successes" in a large number of trials $\left({10}^{11}\right)$ of an event with a small probability $\left({10}^{-<!--\; -\; -->7}\right)$.
- Exact Binomial won't work, and the Poisson approximation does not seem appropriate.

Binomial probability formula
The binomial probability formula goes like this,
$$f(x\mid <!--\; \mid \; -->p)=p^{x_i}{(1-<!--\; -\; -->p)}^{1-<!--\; -\; -->x_i}$$
But I wonder why the success probability and failure probability are multiplied? Can somebody explain this for me, please?

Binomial or Joint Probability
The probability of getting a 6 when rolling a dice is 1/6.
Say I have 5 dices in front of me, the probability of 5 of them being a 6 is equal to:
$P[D_1=6]\cdot <!--\; \cdot \; -->P[D_2=6]\cdot <!--\; \cdot \; -->P[D_3=6]\cdot <!--\; \cdot \; -->P[D_4=6]\cdot <!--\; \cdot \; -->P[D_5=6]=(1/6{)}^5=0.000128601$
Where $P[D_i=6]$ is the probability of the $i^{th}$ dice being a 6
Now, why is it that I get the same answer by seeing this problem as 5 Bernoulli trials each with a probability of success equal to 1/6 and by calculating the desired probability via the binomial distribution, namely the probability of 5 successes out of 5 trials:
$P[5|5]=5!/(0!\cdot <!--\; \cdot \; -->5!)\cdot <!--\; \cdot \; -->(1/6{)}^5\cdot <!--\; \cdot \; -->(5/6{)}^0=0.000128601$
I'm trying to understand two things:
- Why I don't get the same result for 4 successes out of 5 trials compared to just doing $(1/6{)}^4$ if what I wan to know is the probability of 4 out of my 5 dices being a 6.
- Why are the results for 5 out of 5 the same? Is it a coincidence? Is one of them more or less valid than the other?

Binomial probabilities
In a city, the percentage of left-handed women is 16% and the percentage of left-handed men is 22%. A random sample of five men and five women are selected from the population of the city. What is the probability that the sample contains at least one left-handed woman and one left-handed man.
So yeah, using my awesome calculator I found probabilities of 0 men - 0.2289. 0 women - 0.418. It follows therefore that at least one man - 0.711, one woman - 0.582. My intuition told me that I now have to multiply last two values and even got me the right answer. Could anyone please explain why we multiply 0.711 and 0.582?

Binomial and probability
History indicates that 80% of all McDonald's customers order French fries with their meal purchases. During one afternoon at a slow restaurant, 15 people order meals. Assuming that all criteria for a binomial apply, what is the probability for each of the following
a) Exactly 13 of the customers order fries? I have 0.230
b) 14 or more customers order fries? I have 0.035
Is what I have right?

A fair coin is thrown ten times. Find the probability that more heads arise than tails?
Hi guys my question is a binomial theorem one I know that I'm just finding it hard to know what goes where. I get that in order to have more heads I must have 6 heads or more nut find it there to use the.

Simple binomial probability question
Var x has binomial distribution $n=3$ and $p=2/3$ Var y has binomial distribution $n2=4$ and $p2=1/2$
Var x and y are independent
Probability that $$P(X=Y)$$?

Binomial or Uniform Probability?
What is the probability of rolling exactly two sixes in 7 rolls of a die?
I know this is a binomial probability.
$P(X=2)=(\genfrac{}{}{0ex}{}{7}{2})(1/6{)}^2(5/6{)}^5$.
By the definition of probability formula "Probability formula is the ratio of number of favorable outcomes to the total number of possible outcomes." Then why probability of rolling exactly two sixes in 7 rolls of a die isn't $=2/42$? The denominator is 42 because in one die there are 6 faces, and in 7 rolls of a die, there are $7\times <!--\; \times \; -->6=42$ faces.

Binomial distribution and finding the probability
I want to know how to use the binomial distribution table, given that the mean of a binomial distribution is 5 and the standard deviation is 2. What is the actual probability of 5 successes? The(mean) is found by number of trials *probability of success in any trial; while the S.D. is found by square root of (number of trials) * (probability of success) * (probability of success-1). How do i use the binomial distribution table to find the probability or even without it, but without the use of permutations and combinations.

Fast(er) way of computing the cumulative binomial probability?
While revising for my probability test, I saw this question from one of the previous exams:
You flip a fair coin 100 times. What is the probability that you have less than 45 heads?
My question is a simple one. I know that to calculate the probability of a specific amount of n heads (or tails), we can use the binomial distribution with formula $P(X=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-<!--\; -\; -->p{)}^{n-<!--\; -\; -->k}$, with $n=100$ and $p=0.5$ in this case. I also know that we can compute the chance of $P(X<k)$ as either $P(X=0)+P(X=1)+...+P(X=k-<!--\; -\; -->10)$ or $1-<!--\; -\; -->(P(X=k)+P(X=k+1)+...+P(X=n))$
However this is a simple question only worth two points out of over 60 total points. I cannot imagine that you need to compute $P(X=n)$ 44 separate times and sum them together for such a small amount of points.
Is there any way to rewrite the formula or apply a different trick to drastically lower the amount of computations you need to do?

Comparing Binomial Probability to Poisson Random Variable Probability
A text file contains 6000 characters. When the file is sent by e-mail from one machine to another, each character (independently of all other characters) has probability 0.001 of being corrupted. Use a Poisson random variable to estimate the probability that the file is transferred without error.
Compare this to the answer obtained when you model the number of errors as a binomial random variable.
For the binomial probability I got 0.2471%(to 4 significant figs).
For the Poisson probability I got 0.2478%(to 4 significant figs).
However I'm not sure how I'm supposed to compare them, clearly I can see that the binomial probability is slightly lower, but I don't understand why this is the case?

Binomial probability majority question
Here $p=0.25$, so $q=0.75$, $X=0,1,2,3,4,5$
The problem says what is the probability that the majority will accept; is it asking me to find $P(X=3)$ or $P(X\ge <!--\; \ge \; -->3)$?

Total probability and binomial distribution
This is taken from Ross book: when coin 1 is flipped, it lands on heads with probability .4; when coin 2 is flipped, it lands on heads with probability .7. One of these coins is randomly chosen and flipped 10 times. Compute the probability that the coin lands on heads on exactly 7 of the 10 flips.
What is incorrect behind this reasoning? let C be the event that the coin lands on heads. By total probability we have $P(C)=\frac{1}{2}.7+\frac{1}{2}.4=.55$. Then we apply a binomial distribution with $n=10$, $k=7$ and $p=.55$. However, this is incorrect. Why?

A problem on probability involving binomial distribution
I'm trying to figure out the method required to solve this problem, so I stripped out the actual values to keep from getting a direct answer.
A random sample of n people who walk to work are chosen, what is the probability that at least r of them are injured, given that the probability of being injured while walking to work is p.
I don't know where to go. It feels like a binomial probability problem, but it covers a range of trials and not just one value exactly. My guess was to calculate $1-<!--\; -\; -->BinomCDF(n,p,r-<!--\; -\; -->1)$. Does this seem accurate? For example, if $n=15$, $r=7$, $p=0.5$.
I would have $1-<!--\; -\; -->BinomCDF(15,0.5,6)$ or
$$1-<!--\; -\; -->\underset{i=1}{\overset{6}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{15}{i}){0.5}^i(1-<!--\; -\; -->0.5{)}^{15-<!--\; -\; -->i}.$$

What is the probability of having exactly k successes from n coins if all n coins are flipped and there are x successes, then n - x coins are re-flipped to give the additional $k-<!--\; -\; -->x$ successes?
Example: $$n=3,k=3,p=1/2$$, and let:
$$f(k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-<!--\; -\; -->p{)}^{n-<!--\; -\; -->k}$$
I tried adding the success probabilities for the following scenarios:
All 3 success on the first attempt $f(3)=0.125$
2 on the first attempt, then 1 on the second $f(2)\cdot <!--\; \cdot \; -->f(1)=0.140625$
1 on the first attempt, then 2 on the second $$f(1)\cdot <!--\; \cdot \; -->f(2)=0.140625$$
0 on the first attempt, then 3 on the second $f(0)\cdot <!--\; \cdot \; -->f(3)=0.015625$
I get $P(3)=f(3)+f(2)\cdot <!--\; \cdot \; -->f(1)+f(1)\cdot <!--\; \cdot \; -->f(2)+f(0)\cdot <!--\; \cdot \; -->f(3)=0.421875$.
However, doing the same for $k=2$ and $k=1$, I get:
$$\begin{gathered} P(2)=f(2)+f(1)\cdot <!--\; \cdot \; -->f(1)+f(0)\cdot <!--\; \cdot \; -->f(2)=0.5625 \\ P(1)=f(1)+f(0)\cdot <!--\; \cdot \; -->f(1)=0.421875 \end{gathered}$$
But I know this is wrong because $P(3)+P(2)+P(1)>1$