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

What are the applications of Archimedes' principle ? AIt is used in designing of ships and submarines. BIt is used in lactometers to determine the purity of milk. CIt is used in hydrometers to determine density of fluids. DIt is used in hydraulic lifts.

An airline claims that there is a 0.10 probability that a coach-class ticket holder who flies frequently will be upgraded to first class on any flight. This outcome is independent from flight to flight. Sam is a frequent flier who always purchases coach-class tickets.
(a) What is the probability that Sam's first upgrade will occur after the third flight?
(b) What is the probability that Sam will be upgraded exactly 2 times in his next 20 flights?
(c) Sam will take 104 flights next year. Would you be surprised if Sam receives more than 20 upgrades to first class during the year?

To win a particular lottery game, a player chooses 4 numbers from 1 to 50. Each number can only be chosen once.
1. How many different number combinations are possible =
2. What is the probability that the chosen number is 2500=

A(n) _______ distribution has a​ "bell" shape.
A) normal
B) outlier
C) relative frequency
D) polygon

What does a cross product of 0 mean?

Which of the following statements about probabilities and components is not true?
Choose the correc answer below:
a) $P(E)-<!--\; -\; -->P(E^{\prime })=1$
b) $P(E)=1-<!--\; -\; -->P(E^{\prime })$
c) $P(E)+P(E^{\prime })=1$
d) $P(E^{\prime })=1-<!--\; -\; -->P(E)$

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.

If randomly picked from m symbols, each symbol being equally likely, how long on average would it take to pick m/2 different symbols?
Lets take the specific example of the English alphabet. In this case, I have 26 letters (symbols). Lets say these letters are plastic toys in a box. At random, I take one toy letter from the box, and write it down on a piece of paper. I then put that toy letter back into the box, and repeat the process. How many times, on average, would I have to repeat this process for my piece of paper to contain 13 of the 26 letters of the alphabet?
Is there a way to generalize the answer to m symbols instead of 26? What happens when m is odd?

Determine the number of different ways that the letters of "success" can be arranged

Which option best completes the diagram?
A. More prosperous citizens have money to spend.
B. The country’s government discovers a new natural resource.
C. More educated citizens are more likely to create a democratic government.
D. The country creates alliances with others in the region

Relationship among Exponential, Gamma, and Normal distribution
I am studying stochastic processes where I stumbled upon the theorem that says the sum of exponential distributions is gamma distribution. However, from the central limit theorem, we know that sum of a sufficiently large number of IID random variables converges to Normal distribution. So is the following relationship always true?

Let X be a separable Banach space. The p-Wasserstein space on X is defined as
$W_p(X)=\{\mathrm{\mu <!--\; \mu \; -->}\in <!--\; \in \; -->P(X):{\int <!--\; \int \; -->}_X||x|{|}^{p}d\mathrm{\mu <!--\; \mu \; -->}(x)<\mathrm{\infty <!--\; \infty \; -->}\},p\ge <!--\; \ge \; -->1$

Bayes' Theorem in Conditional Probability
We are testing for a disease D that we think is present, D+, with probability 0.4, and absent, D-, with probability 0.6. We believe that a test has sensitivity $P\{T+|D+\}=0.75$ and specificity $P\{T-<!--\; -\; -->|D-<!--\; -\; -->\}=0.8$.
Q1: What is our probability that the disease is present if we perform the test and it is positive, T+, and our probability the disease is absent if that test is negative, T-?
My ans:
We can apply Bayes' formula to calculate $P\{D+|T+\}$ and $P\{D-<!--\; -\; -->|T-<!--\; -\; -->\}$.
$P\{D+|T+\}=5/7$
$P\{D-<!--\; -\; -->|T-<!--\; -\; -->\}=24/29$
Q2: Suppose we perform three tests, conditionally independent given D. Given each possible number of positive test results, 0, 1, 2, or 3, what is our probability that the disease is present?
My ans:
Let k denote the number of positive test results.
We know that P{ [exactly] k successes in n trials | p } $=(nCk)x(p^k)\times <!--\; \times \; -->((1-<!--\; -\; -->p{)}^{n-k})$, hence we can calculate $P\{k=0|D+\}andP\{k=0|D-<!--\; -\; -->\}$.
We can then apply Bayes' formula to calculate $P\{D+|k=0\}$.
$P\{D+|k=0\}=0.692$
Using the same approach for $k=1:3$, we derive:
$P\{D+|k=1\}=0.5$
$P\{D+|k=2\}=0.308$
$P\{D+|k=3\}=0.165$
Would really appreciate it if someone can verify whether my reasoning and answers are correct.

A survey was carried out to find out if the occurrence of different kinds of natural disasters varies from on part of the town to the other. The town was divided into two as center town and town outskirts, and the natural disasters were divided into tornadoes, lighting, and fires.The survey showed these results:
Tornadoe Lightning Fires
Town Center 10 220 120
Town Outskirts 20 150 200
Do the above findings provide enough proof at 95 confidence level to indicate that the happening of natural disasters depends on the part of town?

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. $C(200,2)2^{101}(-<!--\; -\; -->3{)}^{99}$

Two events $E$ and $F$ are independent if any one of the following conditions are met:
(i) $P(E/F)=P(E)$
(ii) $P(F/E)=P(F)$
(iii) $P(EnF)=P(E)P(F)$
Is it correct to then assume that if $E$ and $F$ are independent having met condition (i) or (ii), then $E$ and $F$ are also mutually exclusive?

why doubling the number in a contingency table changes the p-value?
i am doing a facts trouble, which is checking out if the evaluation of someone is impartial of the individual's intercourse. i'm given a contingency desk, I calculated the expected fee for each access and calculated the chi-rectangular value then I got a p-fee.
Then the query requested me to do the same thing after doubling all entries in the contingency desk, I were given a p-price smaller than the only I got earlier than. Why does this take place? Can all and sundry supply me an cause of the distinction?

1. Is there evidence of diffetence in mean time spent exercising between males and females? In the sample, we have a mean of 12.4 hours a week for 20 males and a mean of 9.4 hours a week for 30 females.
2. Is there evidence of a negative correlation between blood pressure and heart rate? In a sample of 200 2. Is there patients, we found a sample correlation of -0.057.

Let X be a random variable with finite expectation, and suppose that $E[e^{-<!--\; -\; -->X}]\le <!--\; \le \; -->1$. Prove that $E[X]\ge <!--\; \ge \; -->P(X\ge <!--\; \ge \; -->2)$.<br<My first thought was to use Markov's inequality, but I can only use that if X is non-negative, and in any case, the resulting inequality doesn't seem very helpful. Even if I assume that $X\ge <!--\; \ge \; -->-<!--\; -\; -->c$ and apply it to $Y=X+c$, I'm left to show $P(X\ge <!--\; \ge \; -->2)\ge <!--\; \ge \; -->\frac{c}{c+1}$, which doesn't seem like it's always true.
I'm primarily unsure of how to use the $E[e^{-<!--\; -\; -->X}]$ condition; Jensen's inequality tells me that the expectation is non-negative but there's presumably more to it.