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Probability of drawing two balls from a bag

You can solve it both by using the conditional probability and independence. The probability of selecting the first ball as red is obviously $$P(r_1) = \frac{4}{9}.$$ Then for each of this selection ...
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3 votes

Probability of drawing two balls from a bag

You are using the general multiplication rule, so you are actually computing $P(A)*P(B|A),$ as the two events are not independent, as you have suspected. That is why you reduced the denominator for ...
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4 votes
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$P[A > B, A > C]$, where $A$, $B$, and $C$ are i.i.d. random variables

This is the simplest example I could think of. If $A$, $B$, and $C$ are independent coin tosses with heads 1 and tails 0 then $$P(A>B, A>C)=1/8$$ since this only happens if A=1 and B and C are ...
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2 votes

$P[A > B, A > C]$, where $A$, $B$, and $C$ are i.i.d. random variables

Does the independence of A, B, and C imply the independence of the events A>B and A>C? No. The event of $A>B$ is evidence that $A$ is large, thus making it more likely that $A>C$ too. ...
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1 vote
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Odds of selecting all other outcomes besides desired outcome in a series (more BAG O’ MARBLES probability)

The comment of lulu is conclusive. The (somewhat inelegant) way of reaching the same result, through direct multiplication of the probability of the respective events is $$\frac{9}{10} \times \frac{8}...
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0 votes

Solve integral of $\Phi(a+bx)^k(1-\Phi(a+bx))^{n-k}\phi(x)$

Using R and the default precision (which can be tweaked) here is a numerical approach to find the probability mass function given $a$, $b$, and $n$. ...
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1 vote

Odds of selecting all other outcomes besides desired outcome in a series (more BAG O’ MARBLES probability)

If all 10 balls are distinct, then there are $10! = 10 \times 9 \times \ldots \times 1 = 3,628,800$ different orders the balls can come out of the bag, and the probability of drawing them in any ...
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1 vote
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Can this probability question be solved using $3$D cube volume?

Kindly, consider the cube above. It's a $6\times6\times6$ cube. $\qquad\qquad(661)(662)\dots(666)$ $\qquad\qquad\dots$ $\quad\quad(631)(632)\dots(636)$ $\quad(621)(622)\dots(626)$ $(611)(612)\dots(616)...
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3 votes
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Simplifying a double summation to find alpha

Hint: In your first sum $$\sum_{m=0}^{\infty}\sum_{n=0}^{\infty} \frac{2m}{m!n!} (p)^{m+n} $$ the sum over $m$ can start at $m=1$ instead of $m=0$. Now define a new index of summation, say $M:=m-1$ ...
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4 votes
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Game theory statistics problem

Hint: Since the two players have equal skill, their collective 101 attempts can be lumped together into a single pool. You're then asking the probability that a particular distinguished one of those ...
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0 votes

Can this probability question be solved using $3$D cube volume?

Yes. You're basically listing all possible $6^3$ outcomes and finding which ones result in "success". You can use the geometric structure to make the problem more intuitive, which you are &...
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1 vote
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Change of variables and Tonelli applied to a negative function

If the functions $f$ and $g$ are such that for each $x\in\mathbb R$, $f(x,Y)$, $f(X,Y)$, $g(X)$ are all integrable but possibly take negative values, then either the Fubini-Tonelli theorem or writing $...
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0 votes

Prove that prove that $E[X^k] \le E[X]^k$ for any even integer $k\ge 1$

As stated in the comments, the statement is not true. Your proof is wrong since you are falsely assuming that $X$ is uncorrelated with itself, which is impossible. To see why, suppose that there ...
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2 votes
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Prove that prove that $E[X^k] \le E[X]^k$ for any even integer $k\ge 1$

The supposed inequality $E[X^k]\le (E[X])^k$ is not always true: Let $X$ be a random variable that takes the value $0$ with pribability $\frac{9}{10}$ and the value $10$ with probability $10$. Then ...
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On an exercise concerning continuous random variables

$\newcommand{\d}{\,\mathrm{d}}$A series of strong hints. When asked to compute the distribution function in $(1)$, they mean find: $$F_X=\int_{-\infty}^xf_X(t)\d t$$Which is known as the cumulative ...
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1 vote

Start with $1,000. Suppose you can flip a coin any number of times. Heads you quadruple your money. Tails you lose it all. How many times do you flip?

No simple answer is known to this question, which is called the St. Petersburg paradox. For detailed discussion of the issues, see the Wikipedia article or the SEP article.
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Start with $1,000. Suppose you can flip a coin any number of times. Heads you quadruple your money. Tails you lose it all. How many times do you flip?

The expectation of how much money you get after $n$ turns is $$\$\frac1{2^n}\times 4^n,000=\$2^n,000$$ so there is no point at which you stop playing the game at an advantage over continuing.
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1 vote

Facing problem in a question of probability arrangements

For at least three English book together: If you want to see another approach , you can use complement rule such that $$1 -\text{there is not any three adjacent Eng. book}$$ So , if there is not any ...
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1 vote

Facing problem in a question of probability arrangements

Taking the question to mean P(at least $3$ English books are together) Since probability has been asked for, we can simply use combinations Also, here it is simpler to use the complement, weeding ...
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0 votes

What is the optimal strategy for the following bidding game?

Long comment : (I think there's a minor mistake below , numerical simulation roughly agrees with me but seems to tilt towards the left and maximized at $a=75$ with expected gain $39.4803$ .) For (3) . ...
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1 vote
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Conditional Probability and Random testing.

If all 30 bulbs are from the same batch, then you can absolutely take that into account. Your statistic is $X = $ the number of bulbs tested before finding a faulty bulb, and if the probability of ...
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Facing problem in a question of probability arrangements

Addendum added to (also) answer the alternative question of exactly 3 English books together. From the posted analysis, I am inferring (perhaps wrongly) that the question is asking for the ...
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0 votes

Finding the moment generating function of the product of two standard normal distributions

You can use conditioning and representation for a simpler, cleaner approach. You have by conditional expectations that $$ E\left(e^{tUV}\right) = E\left(E\left(e^{tUV}| U\right)\right) = E\left(e^{\...
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0 votes

Determining CDF of a function with variable inside absolute value

Because if $y \geq 1$, $F_X(-y) = 0$, so it might be confusing to answer both intervals in the same manner. Indeed, you could take $F_X(y) - F_X(-y)$ in the whole interval [0,2], noting that in the ...
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  • 320
0 votes

Stein Kernel for nonzero mean

It is just a name to the function. Let $U = X-\theta$, you will have $$E[Ug(U)] = E[T_Ug'(U)]$$ for every $g$. Now for a function $f$, let $g(u) = f(u+\theta)$ so $f(x) = g(x-\theta)$ and $f'(x) = g'(...
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1 vote

Markov chain: Find the probability he won yesterday’s game.

$\newcommand{\prob}{\mathrm{P}}\newcommand{\R}{{\rm I\!R}}$Essentially you have a Markov chain, where you know $$ \prob[x_{t+1} = j | x_t = i] = p_{ij}, $$ but you are looking for $\prob[x_t = i | x_{...
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0 votes

Probability of a white ball chosen from a bag if there are two bags.

I used a tree diagram. You draw all the options and their respective probabilities, then you can multiply them to get the probability for the event at the bottom of the tree. Here, getting White from ...
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1 vote

Probability of a white ball chosen from a bag if there are two bags.

It's easier not to use Bayes thereom to calculate $P(W|A)$. You are being confused by the notation of $|A$. But that just says if your universe of options is pulling marbles from a bag that is $A$ ...
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Probability of a white ball chosen from a bag if there are two bags.

Firstly, for probability of drawing white given A, don't use P(W\A), the correct symbol is $P(W|A)$ And P(W|A) is simply the ratio of white balls to total balls in $A$, so there is nothing to prove ...
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4 votes

Probability that one of the tickets is number 7 and none number 18?

You need to be very careful here. The simplest way would be to compute it in the typical hypergeometric way, viz $$\dfrac{\binom11\binom{18}2}{\binom{20}3} =\; \frac{51}{380}$$ If you wish to go by ...
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1 vote
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Conditional Probability involving two decks of cards and coin tosses

Let's define the events: $A:=$ "picking a card of the same color as the first" $B:=$ "the result of flipping the coin is heads". In question (1), we're looking for $\mathbf P\{A\}...
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0 votes

Exercise on the random variables associated with specific number of tosses

Hint $M$ follow a negative binomial law with parameter $(n,p)=(2,\frac{1}{5})$. Finally, $$\mathbb P\{N=1,M=2\}=\mathbb P\{M=2\mid N=1\}\mathbb P\{N=1\}.$$ $$\mathbb P\{M=N+1\}=\sum_{k\geq 1}\mathbb ...
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1 vote
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An exercise of probability

Let $R_mG_n$ represent the event of choosing $m$ red and $n$ green apples. We now have For the remaining two apples to be differently coloured, we must have chosen $2$ apples of one colour and $1$ of ...
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2 votes
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Dice and Bernoulli random variables

As per protocol, I am not supposed to give answers unless you show your effort, so I'll give you hints HINTS To get 5H, Alice must have got $X=0$ with probability $(1-p)$ and got $5H$ in $6$ tosses, ...
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0 votes

Probability that a person finds their seat in an auditorium.

Consider the problem from the perspective of mathematical induction. WLOG, we assume that the assigned seats are $123\ldots (n-1)n$. Let $P_n$ denote the answer (probability) w.r.t $n$. When $n=2$, as ...
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3 votes

Coin tosses with unknown success probability

To be more explicit, what you are describing can be viewed in conditional probability with an uninformative uniform prior on $p$. That is, first draw $p \sim U[0,1]$, then draw $k \mid p \sim \...
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0 votes

Conditional probability (Bayes theorem)

We want to find $\mathrm{P}(X = 2| Y < 4)$. By the definition of conditional probability, this is equal to $$\frac{\mathrm{P}(X = 2 \land Y < 4)}{\mathrm{P}(Y < 4)}$$ Equivalently, using the ...
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Multivariate Confidence Distribution Definition

Yes, in the multivariate case we have the coverage probability for intervals of the form $I_{X}:=(-\infty,X)$: $$C(\psi,Y) = P\left(\bigcap_{i=1}^{|X|}X_i\leq \psi_i\right)$$
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Probability of 10 heads

In this situation, we normally assume independence, in which case the number of heads has a binomial distribution: $$N_{\text{Heads}}\sim \text{Binomial}(0.5,20)$$ In this case, we want $P(N_{\text{...
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2 votes

Tossing two coins with unknown success probability

The probability is not hard to compute. I will be using the same argument as here. Simulate, all independently, $$ \begin{align} X_1, X_2, \dots, X_{N+1} &\sim \text{Unif}[0,1), \quad p_1:=X_1,\\ ...
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4 votes

Expected Value Notation Question

It means $E[(X-E(X))^2]$. It is obtained by expanding the square of the sum and we use the formula $(a+b)^2=a^2+b^2+2ab.$ \begin{align} E[(X-b)^2] &= E([(X-EX)+(EX-b)]^2)\\ &= E[(X-EX)^2] + (...
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3 votes

Why is the probably four all receive the wrong dish not $\frac{1}{2}$?

The reason becomes clear when you explicitly enumerate these outcomes. Let $(z_1, z_2, z_3, z_4)$ where $z_i \in \{A, B, C, D\}$ for $i = 1, 2, 3, 4$ be an ordered quadruplet that denotes the dish ...
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0 votes

Probability and coin tosses

If I toss a coin 8 times and want to know the odds of heads coming up 3 times then I use $^8C_3$ for my total # of possible valid results over $2^8$. I use $^nC_r$ because its unordered. More ...
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0 votes

Expected value of the number of distinct results of die rolls in $N$ trials

I just wanted to give another way to answer the question. Instead of partitioning by whether the number $i$, $1 \leq i \leq 6$, has appeared, one can consider $X_i$ to be the indicator random variable ...
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12 votes
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Coin tosses with unknown success probability

Of course there is an intuition. To see it let's ask the question differently. For that I will be considering $X_0, X_1,\ldots,X_{N}$ as $N+1$ independent uniform distributed random variables on $(0,1)...
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4 votes

Coin tosses with unknown success probability

Ok. I've calculated (using Wolfram) that it is exactly $\frac{1}{N+1}$, i.e. every $n$ is equiprobable. That result is correct, but we don't need no stinkin' Wolfram Alpha to do this. $$P(\text{# of ...
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0 votes

Tossing two coins with unknown success probability

In general, assuming uniform $B(1,1)$ priors on $p_1$ and $p_2$, you have $$f_X(x) = \text{PDF of }B(1 + n_1, 1 + (N-n_1)),$$ $$f_Y(y) = \text{PDF of }B(1 + n_2, 1 + (N-n_2)).$$ So the PDF of $Z = X - ...
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2 votes
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Computing the expectation and using the derivation trick

The author slightly overcomplicated this by separating out the first term of the sum. By the power rule and chain rule, $\frac{d}{dp}(1-p)^n=-n(1-p)^{n-1}$. Also, if $S=\sum_{k=1}^\infty r^k$, then $...
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2 votes

Lower bound on tail when $\mathbb E[X] = \infty$

I may be misunderstanding your question, but if all you know is that $E[X]=\infty$, then the only lower bound on $P(X>\tau)$ is zero. Pick large $\tau>0$ and small $\varepsilon>0$ and ...
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Mean re-arrivals for interarrival process with arbitrary IID distribution

Never mind - some additional searching and the triggering of a few fading neurons yielded the insight that it's a renewal process. If anyone comes across this post, see e.g. "Introduction to ...
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