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

How to condition this problem involving Urns
There are two urns one with 6 blue objects an 4 yellow objects, and the other with 3 blue objects and 4 yellow objects. We choose an object at random.
(1) If an object we select is yellow, what is the probability it came from the first urn?
(2) Assume we put the yellow object from (1) back in the urn it was drawn from and then draw an object from the same urn. This object is also yellow. What is the probability we chose the first urn?
I have already determined that
$$\mathbb{P}(\{\text{We drew from the first urn}\}|\{\text{The object was yellow}\})=\frac{1}{2}.$$
$$\mathbb{P}(\{\text{We drew from the first urn}\}|\{\text{The object was yellow}\})$$
We compute this using Bayes' theorem.
$$\frac{\mathbb{P}(\{\text{The object was yellow}\}|\{\text{We drew from the first urn}\})\mathbb{P}(\{\text{We drew from the first urn}\})}{\mathbb{P}\{\text{The object was yellow}\}}$$
Implicit in the question are the following probabilities:
$$\mathbb{P}(\{\text{The object was yellow}\})=\frac{8}{17}$$
$$\mathbb{P}(\{\text{The object was yellow}\}|\{\text{We drew from the first urn}\})=\frac{4}{10}$$
$$\mathbb{P}(\{\text{We drew from the first urn}\})=\frac{10}{17}$$
Combining these results together, we have that
$$\mathbb{P}(\{\text{We drew from the first urn}\}|\{\text{The object was yellow}\})=\frac{\frac{4}{10}\frac{10}{17}}{\frac{8}{17}}=\frac{1}{2}$$
Part (b) looks like it is asking to compute
$\mathbb{P}(\{\text{We drew the second object the first urn}\}|\{\text{the set of outcomes such that the first and second draws are from the same urn and both objects are yellow}\})$

$\mathbb{P}({\mathrm{\tau <!--\; \tau \; -->}}_a\le <!--\; \le \; -->{\mathrm{\tau <!--\; \tau \; -->}}_{-<!--\; -\; -->a}\le <!--\; \le \; -->t)=\mathbb{P}({\mathrm{\tau <!--\; \tau \; -->}}_{3a}\le <!--\; \le \; -->t)$ for a Brownian motion?
Let ${\mathrm{\tau <!--\; \tau \; -->}}_a$ denote the first passage time to $a>0$ for a Brownian motion W(t). Then is the claimed identity true?
$$\mathbb{P}({\mathrm{\tau <!--\; \tau \; -->}}_a\le <!--\; \le \; -->{\mathrm{\tau <!--\; \tau \; -->}}_{-<!--\; -\; -->a}\le <!--\; \le \; -->t)=\mathbb{P}({\mathrm{\tau <!--\; \tau \; -->}}_{3a}\le <!--\; \le \; -->t)$$
I had a heuristic (diagram) idea through reflection principle, but I am not sure if it is correct. The form of reflection principle that I am using is,
$$\mathbb{P}({\mathrm{\tau <!--\; \tau \; -->}}_a\le <!--\; \le \; -->t,W(t)\le <!--\; \le \; -->b)=\mathbb{P}(W(t)\ge <!--\; \ge \; -->2a-<!--\; -\; -->b)$$
where $b\le <!--\; \le \; -->a$. I am not able to rigorously prove or disprove the identity above since I am not able to deal with the joint distribution of the first passage time with the reflection principle.

The probability that a one-car accident is due to faulty brakes is 0.04, the probability that a one-car accident is correctly attributed to faulty brakes is 0.82, and the probability that a one-car accident is incorrectly attributed to faulty brakes is 0.03. What is the probability that (a) a one-car accident will be attributed to faulty brakes; (b) a one-car accident attributed to faulty brakes was actually due to faulty brakes?
Is my work below correct?
My Approach:
Event A- Car accident is due to faulty break
Event B- It gets correctly attributed to faulty break
Event D- It gets incorrectly attributed to faulty break
Event C- It gets attributed to faulty breaks ; then
$P(A)=0.04$
$P(B)=0.82$
$P(D)=0.03$
$P(A^{\prime })=0.96$
(a) I use total probability rule i.e.
$P(C)=P(A)P(B|A)+P(A^{\prime })P(D|A^{\prime })$
$P(c)=(0.04)(0.82)+(0.96)(0.03)$
$P(C)=0.0328+0.0288$
$P(C)=0.0616$
(b) For this I used Bayes theorem;
$$\begin{gathered} P(A|C)=P(AnC)/P(C)=P(A)P(C|A)/P(C) \\ =(0.04)(0.82)/0.0616=0.0328/0.0616 \end{gathered}$$
$P(A|C)=0.532467532$

Normal n-distribution and Variance-covariance matrix
Let x be a standard normal vector of length n.<br.Let $A\in <!--\; \in \; -->{\mathbb{R}}^{m\times <!--\; \times \; -->n}$ be a matrix.
Then we say that Ax has normal distributed components. Or we could say that Ax is a vector whose components are normally distributed with length m.
Example:
$$\begin{gathered} A=\left[\begin{array}{cc}3& -<!--\; -\; -->2\\ 2& 1\end{array}\right] \\ x{=}^{(d)}N(O,I_{2\times <!--\; \times \; -->2})=\left[\begin{array}{c}x\\ y\end{array}\right] \\ Ax=\left[\begin{array}{c}3X-<!--\; -\; -->2Y\\ 2X+Y\end{array}\right] \end{gathered}$$
$$\begin{gathered} 3X-<!--\; -\; -->2Y{=}^{(d)}N(0,3^2+2^2)=N(0,13) \\ 2X+Y{=}^{(d)}N(0,4^2+1)=N(0,5) \end{gathered}$$

A city water supply system involved three pumps, the failure of any one of which crashes the system. The probabilities of failure for each pump in a given year are .025, .034, .02, respectively. Assuming the pumps operate independently of each other, what is the probability that the system does crash during the year?

Find $\mathrm{\alpha <!--\; \alpha \; -->}$ for this dice rolling game.
There is the point A(10) on the number line. Let's following the die rolling game rules like below with its pips are 1 to 6 as commonly we've known.
- The 1st rule) A point moves as much as +2[positive direction with magnitude 2] for the even pips like 2,4 and 6
- The 2nd rule) A point moves as much as −1[negative direction with magnitude 1] for the odd pips like 1,3 and 5.
After doing this trial 100 times, "A" lied on the right direction compared with 90. We get $P(Z\ge <!--\; \ge \; -->\mathrm{\alpha <!--\; \alpha \; -->})$ considering the continuity correction. Find the real number $\mathrm{\alpha <!--\; \alpha \; -->}$. (Z is the random variable following normal distribution.)
From here, my solution starts. Let the $X_i$ be the movement of a point for ith trial (Here Each i, $1\le <!--\; \le \; -->i\le <!--\; \le \; -->100$)
$$\begin{array}{clcr}{X}_i& 2& -<!--\; -\; -->1& \\ p(X_i)& 1/2& 1/2& 1\end{array}$$
From the table I got $E(X_i)=\frac{1}{2}$ and $V(X_i)=\frac{9}{4}$
So only have to consider is $X(=X_1+X_2+...+X_{100})>80$, Plus $E(X)=50$ and $V(X)={15}^2$
Therefore $P(X\ge <!--\; \ge \; -->80.5)=P(Z\ge <!--\; \ge \; -->\frac{80.5-<!--\; -\; -->50}{15})$ from continuity correction for $P(X>80)$. (In other words, $\mathrm{\alpha <!--\; \alpha \; -->}=\frac{61}{30}(=\frac{30.5}{15})$ )
The answer sheet claimed the $\mathrm{\alpha <!--\; \alpha \; -->}=2.1$. I can't find any errors in my solution. Please let me know which points did I wrong.
p.s.) The answer sheet solution. (The reason why the author claimed 2.1).Say Y be the number of the even pips doing the 100 times trials, Then only just find $P(10+2Y-<!--\; -\; -->(100-<!--\; -\; -->Y)>90)$
In other words $P(Y\ge <!--\; \ge \; -->61)$. Since the Y~B(100,1/2), $P(Y\ge <!--\; \ge \; -->61)=P(Y\ge <!--\; \ge \; -->60.5)=P(z\ge <!--\; \ge \; -->2.1)$

Decide whether this statement is true or false: Let $(\mathrm{\Omega <!--\; \Omega \; -->},\mathbb{F},\mathbb{P})$ be a probability space, if for two events $A,B\in <!--\; \in \; -->\mathbb{F}$ $\mathbb{P}(A\cup <!--\; \cup \; -->B)=\mathbb{P}(A)+\mathbb{P}(B)$ holds, then $A\cap <!--\; \cap \; -->B=\mathrm{\varnothing <!--\; \varnothing \; -->}$.

Suppose the discrete random variable of interest, x, is the number of rolls that result in a composite when rolling a die three times (4 and 6 are possible composite numbers when rolling a die).
a) How many outcomes are there in the tree diagram?
b) Calculate the probabilyty for each of the outcomes in the tree diagram.

Tossing 2 Coins, Fair coin and a two headed coin
There are two coins. One is two headed coin, the other one is a fair coin. A coin is selected randomly and flipped.
a) what is the probability of selecting a fair coin?
b) what is the probability of head from the flipped coin?
c) what is the probability of selecting a fair coin and having head on the top?
So A and B are relatively simple, because A is 1/2 and for B its 3/4 but I am concerned with C
Shouldn't be the chance of having a fair coin + head is basically 1/4? Correct me if I am wrong but
If I want to pick a fair coin, the probability is 0.5, lets say we picked it, and now we are going to throw it, isn't also chance between head and tail is 0.5? Which effectively means that getting a head from a fair coin
the intersection between them is $0.5\times <!--\; \times \; -->0.5$ which equals 0.25?

Prove that $\underset{k=0}{\overset{m}{\sum <!--\; \sum \; -->}}\frac{(\genfrac{}{}{0ex}{}{m}{k})}{(\genfrac{}{}{0ex}{}{n}{k})}=\frac{n+1}{n+1-<!--\; -\; -->m}$

Probability of a number randomly chosen out of given numbers
Three numbers are chosen at random without replacement from {1, 2, 3, ..., 10}. What is the probability that the minimum of the chosen number is 3, or their maximum is 7?
Approach 1:
Let A and B denote the following events
A:minimum of the chosen number is 3
B: maximum of the chosen number is 7
We have,
$P(A)=P(\text{choosing 3 and two other numbers from 4 to 10})=$ $\frac{(\genfrac{}{}{0ex}{}{7}{2})}{(\genfrac{}{}{0ex}{}{10}{3})}=\frac{(7\ast <!--\; \ast \; -->6)}{2}\ast <!--\; \ast \; -->\frac{3\ast <!--\; \ast \; -->2}{10\ast <!--\; \ast \; -->9\ast <!--\; \ast \; -->8}=\frac{7}{40}$
$P(B)=P(\text{choosing 7 and two other numbers from 1 to 6})=$ $\frac{(\genfrac{}{}{0ex}{}{6}{2})}{(\genfrac{}{}{0ex}{}{10}{3})}=\frac{6\ast <!--\; \ast \; -->5}{2}\ast <!--\; \ast \; -->\frac{3\ast <!--\; \ast \; -->2}{10\ast <!--\; \ast \; -->9\ast <!--\; \ast \; -->8}=\frac{1}{8}$
$P(A\cap <!--\; \cap \; -->B)=P(\text{Choosing 3 and 7 and one other number from 4 to 6})$ = $=\frac{3}{(\genfrac{}{}{0ex}{}{10}{3})}=\frac{3\ast <!--\; \ast \; -->3\ast <!--\; \ast \; -->2}{10\ast <!--\; \ast \; -->9\ast <!--\; \ast \; -->8}=\frac{1}{40}$
Now $P(A\cup <!--\; \cup \; -->B)=P(A)+P(B)-<!--\; -\; -->P(A\cap <!--\; \cap \; -->B)$
$=\frac{7}{40}+\frac{1}{8}-<!--\; -\; -->\frac{1}{40}=\frac{11}{40}$
Approach 2:
Chance of a number being drawn out of the given 10 numbers is equiprobable. Hence
P(any one number drawn )= $\frac{1}{10}$
P(A)=$(\genfrac{}{}{0ex}{}{7}{2})\frac{1}{10}\ast <!--\; \ast \; -->\frac{7}{9}\ast <!--\; \ast \; -->\frac{6}{8}=21\ast <!--\; \ast \; -->\frac{42}{720}=\frac{14\ast <!--\; \ast \; -->7}{80}=\frac{98}{80}$ which is wrong as it is above 1.
Here, $\frac{1}{10}$ is the probability of getting 3 and $\frac{7}{9}\ast <!--\; \ast \; -->\frac{6}{8}$ is the probability of getting two number above 3 which can be chosen in $(\genfrac{}{}{0ex}{}{7}{2})$ ways.
Why I cannot find the P(A) like this ? Please tell me what is wrong in approach 2

How many phone numbers are possible that have the 507 area code? In other words, how many phone numbers of the form (507) _ _ _-_ _ _ _? (Assume the first digit after 507 is not zero.)

Let g be a probability density function, what is the ratio $\frac{g(x)}{g(t)}$ for $x>t$?

Is the likelihood function $L(\mathrm{\theta <!--\; \theta \; -->}|X)$ equal to or proportional to $p(X|\mathrm{\theta <!--\; \theta \; -->})$?
Sometimes I find that the likelihood function is written as $L(\mathrm{\theta <!--\; \theta \; -->}|X)=p(X|\mathrm{\theta <!--\; \theta \; -->})$ while other times $L(\mathrm{\theta <!--\; \theta \; -->}|X)\propto <!--\; \propto \; -->p(X|\mathrm{\theta <!--\; \theta \; -->})$.
which is correct?
If $L(\mathrm{\theta <!--\; \theta \; -->}|X)\propto <!--\; \propto \; -->p(X|\mathrm{\theta <!--\; \theta \; -->})$, then what is the constant?

Let $\mathrm{\Omega <!--\; \Omega \; -->}={\mathbb{N}}_0^2,\mathcal{A}=Pot(\mathrm{\Omega <!--\; \Omega \; -->})$ and P the product of two poisson distributions with paremeter ${\mathrm{\lambda <!--\; \lambda \; -->}}_1,{\mathrm{\lambda <!--\; \lambda \; -->}}_{>}0$, i.e
$$\mathbb{P}(\{(n_1,n_2)\})=\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}_1^{n_1}{\mathrm{\lambda <!--\; \lambda \; -->}}_2^{n_2}}{n_1!n_2!}{e}^{-<!--\; -\; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_1-<!--\; -\; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_2}$$
Define then $X:\mathrm{\Omega <!--\; \Omega \; -->}\to <!--\; \to \; -->{\mathbb{N}}_0,(n_1,n_2)\mapsto <!--\; \mapsto \; -->n_1+n_2$. Prove that X is poisson distributed with parameter ${\mathrm{\lambda <!--\; \lambda \; -->}}_1+{\mathrm{\lambda <!--\; \lambda \; -->}}_2$

What is the key word that indictes that the Addition Rule for Probabilities will be used?

Problem:
In a fictional stats class, 40% of students are female, and the rest are male. Of the female students, 30% are less than 20 years old and 90% are less than 30 years old. Of the male students, half are less than 20 years old and 70% are less than 30 years old.
(a) Make a contingency table to describe these two variables
(b) Find the probability that a randomly selected studet is 30 years or older
(c) If a student is 20 years or older, what is the probability that the student is female?
(d) If a student is less than 30 years old, what is the probability that the student is 20 years or older?
My Thoughts:
(b) P(<30 years) = 1 - 0.78 = 0.22
(c) What I first did was find P(S2 given 'not A1'), but the answer doesn't make sense because the denominator ended up being smaller than the nominator.
(d) Do I solve this problem by doing 'not 20 years'?

How do you use the Counting Principle to find the probability of choosing the 5 winning lottery numbers when the numbers are chosen at random from 0 to 9?

Use the Standard Normal Distribution table to find the indicated area under the standard normal curve.
Q1: Between z = 0 and z = 2.24
$$\begin{gathered} A1:0=0.5000 \\ 2.24=0.9875 \\ 0.9875–0.5000=0.4875 \end{gathered}$$
$$\begin{gathered} Q2:\text{ To the left of }z=1.09 \\ A2:0.8621 \end{gathered}$$
Q3: Between z = -1.15 and z = -0.56
$$\begin{gathered} A3:-<!--\; -\; -->1.15=0.1251 \\ -<!--\; -\; -->0.56=0.2877 \\ 0.1251-<!--\; -\; -->0.2877=-<!--\; -\; -->0.1626 \end{gathered}$$
Q4: To the right of z = -1.93
$A4:1–0.0268=0.9732$
Section 5.2: Normal Distributions: Find Probabilities
Q5: The diameters of a wooden dowel produced by a new machine arenormally distributed with a mean of 0.55 inches and a standarddeviation of 0.01 inches. What percent of the dowels will have adiameter greater than 0.57?
$$\begin{gathered} A5:z=x-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}/\mathrm{\sigma <!--\; \sigma \; -->} \\ =0.57–0.55/0.01=2 \\ =P(x>0.57) \\ =P(z>2) \\ =1–P(z<2) \\ =1–0.9772 \\ =0.0228 \end{gathered}$$
Q6: A loan officer rates applicants for credit. Ratings arenormally distributed. The mean is 240 and the standard deviation is50. Find the probability that an applicant will have a ratinggreater than 260.
$$\begin{gathered} A6:z=x-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}/\mathrm{\sigma <!--\; \sigma \; -->}=260–240/50 \\ =0.4 \\ =P(x>260) \\ =P(z>0.4) \\ =1–P(z<0.4) \\ =1–0.6554 \\ =0.9772 \end{gathered}$$