Probability Examples and Explanations

Find each of the following probabilities when n independent Bernoulli trials are carried out with probability of success p.
a) the probability of no successes
b) the probability of at least one success
c) the probability of at most one success
d) the probability of at least two successes
e) the probability of no failures
f) the probability of at least one failure
g) the probability of at most one failure
h) the probability of at least two failures

What is the probability an event occurs, given the event occurs exactly N times in the next M trials?
Assume the probably an event occurs are equal, but what if the probabilities were unequal for each trial?

Approximating the result of probability
10 dices are rolled simultaneously. What is the probability of getting the sum of numbers appearing on top of all the dice as 35?
If we calculate manually, I think it's pretty tedious with no calculators allowed. So, I thought of approximating the answer. We know that rolling 10 dice, has total number of ways as $6^{10}$ which is ~1e7 which makes it ideal for using a normal distribution. So, mean of 10 throws is 35 and SD of 10 throws is ~5.4 and so $Z={\displaystyle \frac{(X-<!--\; -\; -->35)}{5.4}}$ and so we want $P(\le <!--\; \le \; -->35)-<!--\; -\; -->P(\le <!--\; \le \; -->34)$ as my answer, $P(\le <!--\; \le \; -->35)=1/2$ and the other one can be obtained from the Normal graph and hence, we have the answer.
But to my surprise, the interviewer didn't seem convinced! Is there something wrong with my method? Please point out.

People tend to attend a party on time with a probability of 0.7. For your birthday party, 45 people are invited. What is a probability that at least 30 arrive on time?

Domain of joint continuous random variables - Geometrically
Given the joint P.D.F of X and Y:
$f_{X,Y}(x,y)=6x^{2}y(x,y)\in <!--\; \in \; -->(0,1)\times <!--\; \times \; -->(0,1)$
$f_{X,Y}(x,y)=0$ elsewhere
The following double integral should be the probability of $P(Y>2X)$:
$P(Y>2X)={\int <!--\; \int \; -->}_0^1{\int <!--\; \int \; -->}_0^{y/2}6x^{2}y\mathrm{d}x\mathrm{d}y$
However what I don't understand is why X takes values from 0 to y/2.
I tried to sketch the corresponding area - upper left triangle of $1\times <!--\; \times \; -->1$ square above $Y=2X$ - and my best guess is that for each value of Y then $X=\frac{Y}{2}$
Finally how would one calculate the $P(Y<2X)$ area where X and Y move?
Right now what I struggle with is finding the corresponding area of $R^2$ regardless of the problem/question

How to show that $\int <!--\; \int \; -->\Vert <!--\; \Vert \; -->y-<!--\; -\; -->y_0\Vert <!--\; \Vert \; -->\mathrm{\nu <!--\; \nu \; -->}(dy)<\mathrm{\infty <!--\; \infty \; -->}?$
Let X be a uniformly random variable on [0,1]. Since Y is supported on [cos(2),1], then we can define a probability measure on R so that $Y\sim <!--\; \sim \; -->\mathrm{\nu <!--\; \nu \; -->}$
How to show that for any $y_0\in <!--\; \in \; -->R$ we have
$$\int <!--\; \int \; -->\Vert <!--\; \Vert \; -->y-<!--\; -\; -->y_0\Vert <!--\; \Vert \; -->\mathrm{\nu <!--\; \nu \; -->}(dy)<\mathrm{\infty <!--\; \infty \; -->}?$$

What is the formula to find the permutation of r objects given that there are n total object and out of them n1 and n2 are indistinguishable?
The question is I've n the total number of objects, out of them I've to find the permutation of r objects given that $r<n$. It's also given that there m and t are indistinguishable objects in n such that $m+t=n$. What will be the formula to find the permutation in such a case?
As an example, we've 8 marbles out of which 5 are red and 3 and blue, and the number of objects we have to choose is 2. The permutation we know would be equal to 4. But I don't don't the formula to evaluate this value of 4.

Complete the following probability table:
$\begin{array}{|cccc|}\hline \text{Prior Probability}& \text{Conditional Probability}& \text{Join Probability}& \text{Posterior Probability}\\ P(B)0.70& P(A|B)0.10& P(A\cap <!--\; \cap \; -->B)0.07& P(B|A)\\ P(B^C)0.30& P(A|B^C)0.65& P(A\cap <!--\; \cap \; -->B^C)& P(B^C|A)\\ \text{Total}1& & P(A)& \text{Total}0.9999\\ \hline\end{array}$

Probability: What is probability of a student solving at least x questions from n randomly chosen questions?
Out of a pool of 50 questions, 7 random questions are selected for the exam. If a student prepares a total of 16 questions perfectly, what is the probability that he can solve at least one of the 7 random questions. My solution is :
- Number of possible question combinations:C(50,7)
- Combinations where no prepared question occurs:C(34,7)
P(no questions solved) : C(34,7)/C(50,7)
P(atleast one question): $1-<!--\; -\; -->(C(34,7)/C(50,7))$
But I am not sure if this approach is correct. In addition, I have to also prepare a script with monte carlo method to verify it empirically. Which I am not sure how to proceed with.

Conditional expectation of X given (X, Y, Z)
Suppose there are three random variables X, Y, Z (may be dependent on each other).
Then, $E[X|X,Y,Z]=X$, regardless of their inter-relationship?

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
$\mathbb{P}(\{\text{We drew the second object the first urn}\}|\{\text{the set of outcomes such that the first and second draws are from the same urn and both objects are yellow}\})$

Theorem 16.13. If f is nonnegative, then
$${\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}f(T\mathrm{\omega <!--\; \omega \; -->})\mathrm{\mu <!--\; \mu \; -->}(d\mathrm{\omega <!--\; \omega \; -->})={\int <!--\; \int \; -->}_{{\mathrm{\Omega <!--\; \Omega \; -->}}^{\mathrm{\prime <!--\; \prime \; -->}}}f\left({\mathrm{\omega <!--\; \omega \; -->}}^{\mathrm{\prime <!--\; \prime \; -->}}\right)\mathrm{\mu <!--\; \mu \; -->}{T}^{-<!--\; -\; -->1}\left(d{\mathrm{\omega <!--\; \omega \; -->}}^{\mathrm{\prime <!--\; \prime \; -->}}\right).$$
A function f (not necessarily nonnegative) is integrable with respect to $\mathrm{\mu <!--\; \mu \; -->}{T}^{-<!--\; -\; -->1}$ if and only if fT is integrable with respect to $\mathrm{\mu <!--\; \mu \; -->}$, in which case (16.17) and
$${\int <!--\; \int \; -->}_{T^{-<!--\; -\; -->1}{A}^{\mathrm{\prime <!--\; \prime \; -->}}}f(T\mathrm{\omega <!--\; \omega \; -->})\mathrm{\mu <!--\; \mu \; -->}(d\mathrm{\omega <!--\; \omega \; -->})={\int <!--\; \int \; -->}_{A^{\mathrm{\prime <!--\; \prime \; -->}}}f\left({\mathrm{\omega <!--\; \omega \; -->}}^{\mathrm{\prime <!--\; \prime \; -->}}\right)\mathrm{\mu <!--\; \mu \; -->}{T}^{-<!--\; -\; -->1}\left(d{\mathrm{\omega <!--\; \omega \; -->}}^{\mathrm{\prime <!--\; \prime \; -->}}\right)$$
hold. For nonnegative f, (16.18) always holds.
(2.47) Theorem. Suppose $\mathrm{\Omega <!--\; \Omega \; -->}$ is an open set in ${\mathbf{R}}^n$ and $G:\mathrm{\Omega <!--\; \Omega \; -->}\to <!--\; \to \; -->{\mathbf{R}}^n$ is a $C^1$ diffeomorphism. (a) If f is a Lebesgue measurable function on $G(\mathrm{\Omega <!--\; \Omega \; -->})$, then $f\circ <!--\; \circ \; -->G$ is Lebesgue measurable on $\mathrm{\Omega <!--\; \Omega \; -->}$. If $f\ge <!--\; \ge \; -->0$ or $f\in <!--\; \in \; -->L^1(G(\mathrm{\Omega <!--\; \Omega \; -->}),m)$, then
$${\int <!--\; \int \; -->}_{G(\mathrm{\Omega <!--\; \Omega \; -->})}f(x)dx={\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}f\circ <!--\; \circ \; -->G(x)|\det <!--\; \; -->D_{x}G|dx$$
(b) If $E\subset <!--\; \subset \; -->\mathrm{\Omega <!--\; \Omega \; -->}$ and $E\in <!--\; \in \; -->{\mathcal{L}}^n$, then $G(E)\in <!--\; \in \; -->{\mathcal{L}}^n$ and $m(G(E))=$ ${\int <!--\; \int \; -->}_E|\det <!--\; \; -->D_{x}G|dx$.

Fair die roll 8 times, number of of perfect squares
You have a fair die you roll 8 times. Let X be the number of times you roll a perfect square. What is $E(X^2)$?
I tried doing summation from 1 to 8 of $(n/8)\cdot <!--\; \cdot \; -->2.5)+(8-<!--\; -\; -->n)/8\cdot <!--\; \cdot \; -->(3.5)$ and divide by 8 but that was wrong. G

$\mathbb{P}({\mathrm{\tau <!--\; \tau \; -->}}_a\le <!--\; \le \; -->{\mathrm{\tau <!--\; \tau \; -->}}_{-<!--\; -\; -->a}\le <!--\; \le \; -->t)=\mathbb{P}({\mathrm{\tau <!--\; \tau \; -->}}_{3a}\le <!--\; \le \; -->t)$ 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;
$$\begin{gathered} P(A|C)=P(AnC)/P(C)=P(A)P(C|A)/P(C) \\ =(0.04)(0.82)/0.0616=0.0328/0.0616 \end{gathered}$$
$P(A|C)=0.532467532$

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), $P(Y\ge <!--\; \ge \; -->61)=P(Y\ge <!--\; \ge \; -->60.5)=P(z\ge <!--\; \ge \; -->2.1)$

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:
$$\begin{gathered} A=\left[\begin{array}{cc}3& -<!--\; -\; -->2\\ 2& 1\end{array}\right] \\ x{=}^{(d)}N(O,I_{2\times <!--\; \times \; -->2})=\left[\begin{array}{c}x\\ y\end{array}\right] \\ Ax=\left[\begin{array}{c}3X-<!--\; -\; -->2Y\\ 2X+Y\end{array}\right] \end{gathered}$$
$$\begin{gathered} 3X-<!--\; -\; -->2Y{=}^{(d)}N(0,3^2+2^2)=N(0,13) \\ 2X+Y{=}^{(d)}N(0,4^2+1)=N(0,5) \end{gathered}$$

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

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?

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.)