Ace Your Binomial Probability Tests with Plainmath's Expert Help and Detailed Resources

I saw the following claim in some book without a proof and couldn't prove it myself.
${\displaystyle \frac{d}{dp}}\mathbb{P}(\text{Bin}(n,p)\le <!--\; \le \; -->d)=-<!--\; -\; -->n\cdot <!--\; \cdot \; -->\mathbb{P}(\text{Bin}(n-<!--\; -\; -->1,p)=d)$
So far I got:
$\begin{array}{l}{\displaystyle \frac{d}{dp}}\mathbb{P}(\text{Bin}(n,p)\le <!--\; \le \; -->d)=\\ {\displaystyle \frac{d}{dp}}\underset{i=0}{\overset{d}{\sum <!--\; \sum \; -->}}\left(\begin{array}{c}n\\ i\end{array}\right)p^i{(1-<!--\; -\; -->p)}^{n-<!--\; -\; -->i}=\\ -<!--\; -\; -->n\cdot <!--\; \cdot \; -->{(1-<!--\; -\; -->p)}^{n-<!--\; -\; -->1}+\underset{i=1}{\overset{d}{\sum <!--\; \sum \; -->}}\left(\begin{array}{c}n\\ i\end{array}\right)[i{p}^{i-<!--\; -\; -->1}{(1-<!--\; -\; -->p)}^{n-<!--\; -\; -->i}-<!--\; -\; -->p^i(n-<!--\; -\; -->i){(1-<!--\; -\; -->p)}^{n-<!--\; -\; -->i-<!--\; -\; -->1}]\end{array}$
But I am not very good playing with binomial coefficients and don't know how to proceed.

Intuitive way to arrive at the maximizing argument for the binomial probability
The binomial probability term $q^n(1-<!--\; -\; -->q{)}^{N-<!--\; -\; -->n}$ is maximized when $q=n/N$. This can be easily arrived at by differentiating the given probability term with respect to q. Is there a more intuitive way to arrive at this value of q that maximizes the probability ?

Probability question regarding binomial distribution
At a party, it is discovered that a renegade guest has drawn a controversial image on the bathroom wall. There are 40 people at the party, and in the absence of other evidence, each is equally likely to have drawn the controversial image. The drawing style of the controversial image is analysed and found to have characteristics unique to 8% of the general (innocent) population. A suspect whose drawing style is analysed is found to match the controversial image. Assuming the guests are representative of the general population, what is the probability that the suspect is guilty given the drawing evidence?

Probability distribution binomial
An assembly system is composed of n independent and identical parts. During any given “run” of the system, all parts have a probability of p of working. Suppose the random variable Y represents the number of working parts in any given “run”. You are told that $P(Y=3)=2P(Y=2)$ and $V[Y]=p$. What is the probability distribution of Y?
I know thus is a binomial wherein the mean is np and the variance is np(1-p) how would you solve this?

Negative Binomial: $Y\sim <!--\; \sim \; -->NB(r,p)$ for $r=$ number of success and p is the probability of each success
$$p(k)=p(Y=k)={(}_{r-<!--\; -\; -->1}^{k-<!--\; -\; -->1})p^r(1-<!--\; -\; -->p{)}^{k-<!--\; -\; -->r},r\ge <!--\; \ge \; -->k$$
then
$$Pr(l\le <!--\; \le \; -->Y\le <!--\; \le \; -->k)=Pr(Y\le <!--\; \le \; -->k)-<!--\; -\; -->P(Y\le <!--\; \le \; -->l-<!--\; -\; -->1)=Pr(N_k\ge <!--\; \ge \; -->r)-<!--\; -\; -->Pr(N_{l-<!--\; -\; -->1}\ge <!--\; \ge \; -->r)$$
where $N_k$ is the number of successes in the first k trials and $N_{l-<!--\; -\; -->1}$ the number of successes in the first l-1 trials.
I'm not understanding how theses equate to each other, why is it greater than equal to?
$$Pr(Y\le <!--\; \le \; -->k)-<!--\; -\; -->P(Y\le <!--\; \le \; -->l-<!--\; -\; -->1)=Pr(N_k\ge <!--\; \ge \; -->r)-<!--\; -\; -->Pr(N_{l-<!--\; -\; -->1}\ge <!--\; \ge \; -->r)$$

The probability of Jack hitting a target is 1/4. How many times must he fire so that the probability of hitting the target at least once is greater than 2/3?
using the formula $(nCx)(p^x)((p^{\prime }{)}^{n-<!--\; -\; -->x})$
I identified $p=1/4$, $p^{\prime }=3/4$, $x=?$, $n=?$
in order to solve this que, I have to solve x but i couldn't identified it.

Probability and Variance-Bernoulli, Binomial, Negative Binomial
I am taking a course in probability and I have trouble computing the variance of a random variable.
There are 2 cases we saw in class that I would like to understand:
First I will state the definition of variance of expected value:
- If X is a random variable who's values are in N, then
$$\mathbb{E}(X)={\mathrm{\Sigma <!--\; \Sigma \; -->}}_{n\ge 0}\mathbb{P}(X>n)$$
- If X is a random variable and E(X) exists, then the variance of X is:
$$Var(X)=\mathbb{E}((X-<!--\; -\; -->\mathbb{E}(X){)}^2)$$
Now here are the examples I'd like to understand:
1. binomial distribution: If X is a random variable that follows a binomial distribution of parametres n and p then we can write $X=X_1+X_2+X_3+...+X_n$ where the $X_i$'s are bernoulli variables of parametre p. Then
$$Var(X)=np(1-<!--\; -\; -->p)$$
2. negative binomial distribution (Pascal law): If X is a random variable that follows a pascal law of parameters r and p the $X+r=X_1+...+X_r$ where $X_i$'s are independant geometric variables. Then
$$Var(X)=Var(X+r)=r{\displaystyle \frac{1-<!--\; -\; -->p}{p^2}}$$

Convex combination with binomial probabilities
Suppose that we have some zk>0, k={0,1,⋯,n}. I want to compare weighted averages of zk's when the weights are defined by binomial probabilities.
More specifically, for p and q, where p,q∈(0,1), and for some λ∈(0,1), let x=λp+(1−λ)q.
In this case, should the following be true for all λ∈(0,1)?
$$max\{\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{k})p^k(1-<!--\; -\; -->p{)}^{n-<!--\; -\; -->k}{z}_k,\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{k})q^k(1-<!--\; -\; -->q{)}^{n-<!--\; -\; -->k}{z}_k\}\ge <!--\; \ge \; -->\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{k})x^k(1-<!--\; -\; -->x{)}^{n-<!--\; -\; -->k}{z}_k.$$
I could see that it's true for $n=2$ but I can't show that it still holds for any $n>2$.

Binomial probability
Suppose that the probability of a company supplying a defective product is a and the probability that the supplied product is not defective is b. Before each product supplied is released for further use, it must be screened through a system which detects the potential defectiveness of the product. The probability that this process successfully detects a defective product is x, and the probability that it does not is y. What is the probability that at least one defective product will pass through the system undetected after 20 products have been screened?

Binomial probability doesn't give the correct number?
If you roll five dice your chances of getting all five matching is $\frac{1}{6^5}=\frac{1}{7776}\approx <!--\; \approx \; -->0.08\mathrm{\%<!--\; \%\; -->}$.
I recently learned the binomial probability formula: $P=nCr\times <!--\; \times \; -->p^r\times <!--\; \times \; -->(1-<!--\; -\; -->p{)}^{n-<!--\; -\; -->r}$.
- $nCr=\frac{n!}{r!(n-<!--\; -\; -->r)!}$ << The different possible combinations
- r is the number of dice you need to match
- n is the number of dice you're rolling
I use $\frac{1}{6}$ in the formula because that's dice probability
Because I am rolling 5 dice and need 5 matches, I set $r=5$ and $n=5$ and came up with this equation: $P=\frac{5!}{(5-<!--\; -\; -->5)!5!}\times <!--\; \times \; -->(\frac{1}{6}{)}^5\times <!--\; \times \; -->(1-<!--\; -\; -->\frac{1}{6}{)}^{5-<!--\; -\; -->5}\approx <!--\; \approx \; -->0.012\mathrm{\%<!--\; \%\; -->}$
But when I set $r=4$, $n=4$ with the formula below I do get the desired result. $P=\frac{4!}{(4-<!--\; -\; -->4)!4!}\times <!--\; \times \; -->(\frac{1}{6}{)}^4\times <!--\; \times \; -->(1-<!--\; -\; -->\frac{1}{6}{)}^{4-<!--\; -\; -->4}\approx <!--\; \approx \; -->0.08\mathrm{\%<!--\; \%\; -->}$
Why is this? Shouldn't r and n be 5 because I'm rolling 5 dice and need 5 matches?

A company has 500 employees, and 60% of them have children.
Suppose that we randomly select 4 of these employees.
What is the probability that exactly 3 of the 4 employees selected have children?
We know that 300 of these employees have children.
I tried to figure it out, how to work with the binomial equation by the given's

Suppose a health insurance company can resolve 60% of claims using a computerised system, the remaining needing work by humans. On a particular day, 10 claims arrived, assuming claims are independent, what is the probability that:
Q2.1) Either 3 or 4 (inclusive) claims require work by a human?
Q2.2) No more than 9 claims require work by a human?
I have identified that:
$n=10,p=0.6,q=0.4$
How would I go about these questions? Any help would be appreciated!

Binomial Probability Word Problem
This problem appears very simple, but I am almost positive that it should not be so simple. For the question below I reason that the probability for part a is 0.6 and part b is 0.6 as well. What am I missing?
Hirsbrunner produces tubas and ships them in lots of twenty. Suppose that 60% of all such lots contain no defective tubas, 30% contain one defective, and 10% contain two defectives. Now suppose that a lot is inspected, with two tubas being selected from it at random, and neither is found to be defective.
a) What is the probability that there are no defectives in that lot?
b) Suppose that the inspected lot is from a shipping container that contains 10 lots, and the other 9 lots were not inspected. What is the probability that there are no defectives in that container?

Solving question by binomial probability
If, over a given period in Brisbane, rain falls at random on 4 days of every 10, find the probability that the first 2 days of a given week will be wet and the remainder of the week fine.
using the formula $(nCx)(p^x)((p^{\prime }{)}^{n-<!--\; -\; -->x})$
I put $n=10,x=2,p=0.4,p^{\prime }=0.6$
and I got 0.12, but the actual answer is 0.012.

Suppose Player A and Player B are flipping a coin. Player A flips the coin 20 times, and Player B 40 times. What would be the probability that Player A gets 10 heads from his 20 flips, given that 20 heads were found in total.
I know we can use the binomial distribution to find the denominator(20 heads in 60 flips), but I'm wondering what exactly would b in the numerator. Would I use the binomial distribution as well?
This is my denominator:
$(\genfrac{}{}{0ex}{}{60}{20})$ $\ast <!--\; \ast \; -->{0.5}^{20}\ast <!--\; \ast \; -->(1-<!--\; -\; -->.5{)}^{40}$

Having difficulty with Binomial Probabilities question
I've never worked with binomial probabilities before and I found this question below a little hard. If anyone could help that would be great.
Use Binomial Probabilities to find probability that for the certain number of trials N and probability of success in each trial p number of successes will be:
1) exactly k 2) less than k 3) more than k
N: 10 k 6 P: 0.5
Use numbers assigned to N, K, P

Intuitively: binomial probability as summation?
The expected number of successes in n independent trials with success probability p each is given by the expectation np of the binomial probability $X\sim <!--\; \sim \; -->Binom(n,p)$. Yet you could also write:
$E[X]=\underset{i=0}{\overset{n}{\sum <!--\; \sum \; -->}}iP(i)=\underset{i=0}{\overset{n}{\sum <!--\; \sum \; -->}}i(\genfrac{}{}{0ex}{}{n}{i})p^i(1-<!--\; -\; -->p{)}^{n-<!--\; -\; -->i}=np$
Why, intuitively, are these are the same quantity? I'm having a bit of trouble equating these two things and would really appreciate a clean explanation as to why you don't have to think about summing over all possible # of successes that you could get for n trials.

Binomial probability for sporting event where the order of the first 3 outcomes does not matter
There is a sporting event where team A has 1/3 chance of winning and team B has a 2/3 chance of winning. In order to win a team needs to be the first to win 4 matches. There are no ties.
The question: What is the probability that team A will win?
My thoughts:
It will take no more than 7 matches for team A or B to win.
Neither team will win within 3 matches, so the order for the first 3 matches does not matter.
The possible scores where A wins are $(4-<!--\; -\; -->0)$ $(4-<!--\; -\; -->1)$ $(4-<!--\; -\; -->2)$ $(4-<!--\; -\; -->3)$
So I need to find the binomial probability for each possibility and sum them. However, the order of the first 3 matches does not affect the outcome, so they should not be included in the nCr part.
That's where I'm stuck as to how I need to write out nCr for each possibility whilst taking into account that the order of the first 3 matches does not matter.

When you toss a coin 2n times, the expected number of heads is n. Given the probability density function for a standard normal random variable $\mathrm{\varphi <!--\; \varphi \; -->}(x)={\displaystyle \frac{1}{\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}}}{e}^{-<!--\; -\; -->\frac{x^2}{2}}$, approximate the probability that you observe $n+j$ heads when you toss the coin 2n times using the density function of a standard random variable $\mathrm{\varphi <!--\; \varphi \; -->}$ and the standard deviation $\mathrm{\sigma <!--\; \sigma \; -->}$. Hint: use Stirling's formula: $n!\sim <!--\; \sim \; -->\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}{n}^{n+\frac{1}{2}}{e}^{-<!--\; -\; -->n}$.
I know that the exact probability that I get $n+j$ heads is $(\genfrac{}{}{0ex}{}{2n}{n+j})\cdot <!--\; \cdot \; -->{\displaystyle \frac{1}{2^n}}$ and I tried to use Stirling's formula on the binomial but it just made it more complicated.

Problem 1. 10 identical bulbs, of which 6 will be yellow. He randomly plants 6 of the bulbs. Prob that exactly 4 are yellow.
Isn't this binomial prob?
I did as follows $6C4(.6{)}^4(.4{)}^2$ to get 0.311. Answer is $\frac{3}{7}$.
Problem 2. Bag of bulbs of which 40% produce red tulips. Plant 15 of them. Prob that 6 will be red? I did similar: $15C6(.4{)}^6(.6{)}^9$ to get 20.7% which I think is correct.
Are these the same type of problems, and why isn't Problem 1 working?