High school probability Uncovered: Discover the Secrets with Plainmath's Expert Help

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?

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 $n>2$.

At the ceramic tableware factory, 10% of the produced plates are defective. During product quality control, 75% of defective plates are detected. The rest of the plates are for sale. Find the probability that a plate randomly selected at the time of purchase has no defects. . Round your answer to the nearest hundredth.

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
$$Var(X)=Var(X+r)=r{\displaystyle \frac{1-<!--\; -\; -->p}{p^2}}$$

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!

Classify the statement as an example of classical probability, emprical probability, or subjective probability. Explan your reasoning. The probability of winning a 500 - ticket reffle with 10 tickets is $\frac{10}{500}=0.02$
a) Empirical probability because the probability results from an estimate
b) Classical probability because each outcome in the sample space is equally likely
c) Subjective probability because the probability results from an estimete
d) Classical probability because the probability is based on observations obtained from a probability experiment
e) Subjective probability because each outcome in the sample space is equally likely
f) Empirical probability because the probability is based on observations obtained from a probability experiment

A die is thrown twice. Find the probability that a number greater than 3 is rolled both times.

In a family with 12 children the p = probability of a boy = 0.5. Find the binomial exact probability of less than 1 boy. And find the normal approximate probability of less then 1 boy.
a) Exact probability = 0.011, Approx probability = 0.013
b) Exact probability = 0.019, Approx probability = 0.022
c) Exact probability = 0.003, Approx probability = 0.005
d) Exact probability = 0.055, Approx probability = 0.057

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?

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:
$(\genfrac{}{}{0ex}{}{60}{20})$ $\ast <!--\; \ast \; -->{0.5}^{20}\ast <!--\; \ast \; -->(1-<!--\; -\; -->.5{)}^{40}$

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.

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?

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.

"Draw one card from a standard deck of playing cards. Let’s examine the independence of 3 events ‘the card is an ace’, ‘the card is a heart’ and ‘the card is red’. Define the events as $A{=}^{\prime }ac{e}^{\prime }$, $H=‘heart{s}^{\prime }$, $R=‘re{d}^{\prime }$."
With $P(A)=\frac{1}{13},P(H)=\frac{1}{4}$ and $P(R)=\frac{1}{21}$ I get $\frac{1}{4}$ for $P(H|R)$. So this basically says that P(H) and P(R) are independent, since H and P are independent iff $P(H|R)=P(H)$ . I know that that's not correct from the source, so now I am wondering how I can then calculate the conditional probability for dependent events and how I would know beforehand. Why is $P(H|R)$ actually $\frac{1}{2}$ ?

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?

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