let's say there's two events, a and b both a and b are uniformly distributed and have a range of [100,400] a and b are independent i know that the probability that a=A is 1/300 and the probability that b=B is 1/300   but let's say also that c = b - a   what's the probability distribution of c? like... what's the probability that c=C? how do i figure this out?
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Let $X$ and $Y$ both be distributed between $[1,2]$, what is the distribution of $Z=X-Y$?
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I recently read a math problem and, having not yet taken anything beyond calculus 1, was curious about how to solve it correctly.   Problem:  Calculate the probability of two people meeting at the same location between 1 and 2 p.m. Assume both people show and person 1 will wait 15 minutes for person 2.     Doesn't this probability increase as the time frame shrinks?  For example, Isn't there a much smaller chance the pair will meet if person 1 arrives at 1:00 vs arrives at 1:45 (the probability of meeting becomes 1).  Is this somewhat easily answered using basic statistics?  Just curious if my thought process is accurate or way wrong.
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I have two normal distributions defined by their averages and standard deviations.  Sample 1: Mean=5.28; SD=0.91  Sample 2: Mean=8.45; SD=1.36  You can see how they look like in the next image:    How can I get the probability to obtain an individual from the overlapping area (green)? Is the probability the same as the area?
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Two players A and B shoot a target alternately, until somebody hits it and wins the game. Player A takes the first shot. Each time they take a shot, their probabilities of hitting the target are $p$ and $q$ respectively. Which is the probability $\mathbb{P}$ that A wins?   Is the following solution correct ?  A wins if :  he wins with the $1$st shot or  they both fail to hit it during the first $2$ shots, and A hits it with the $3$rd shot or  they both fail to hit it during the first $4$ shots, and A hits it with the $5$rd shot or  ...  So,  $$ \mathbb{P} = p + (1-p)(1-q)p + (1-p)^2(1-q)^{2}p + ... =  p\sum_{n=0}^{\infty}[(1-p)(1-q)]^n = \cfrac{p}{p+q-pq} $$  Thank you in advance
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I'm self-studying a textbook () (not for course credit) and wanted to post my solution to a problem related to conditional probability.   Given:  $$ \begin{align} &amp;P((A \cup B)^C) = 0.6 \\ &amp;P(A \cap B) = 0.1 \\ &amp;E = A\ xor\ B\ \text{(A or B but not both)} \end{align} $$  Find $P(E\ |\ A \cup B)$  Below is my solution. Is this correct or else where did I go wrong?   $$ \begin{align} P(E\ |\ A \cup B) &amp;= \frac{P(E\ \cap\ (A \cup B))}{P(A \cup B)} \tag 1 \\  \\ P(E\ \cap (A \cup B)) &amp;= P((E \cap A)\ \cup\ (E \cap B)) \tag 2 \\ \\ &amp;= P((A \cup B) \cup (A \cup B)) \\ \\ &amp;= P(A \cup B) \\ \\ \therefore \\ \\ P(E\ |\ A \cup B) &amp;= \frac{P(A \cup B)}{P(A \cup B)} = 1 \tag 3 \\ \\ \end{align} $$  Noting that $E \cap B$ = [(A xor B) or B] = $A \cup B$ and similarly $E \cap A$ = [(A xor B) or A] = $A \cup B$.
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Lets say I have three independent sources and each of them make predictions for the weather tomorrow. The first one says that the probability of rain tomorrow is 0, then the second one says that the probability is 1, and finally the last one says that the probability is 50%. I would like to know the total probability given that information.  If apply the multiplication theorem for independent events I get 0, which doesn't seem correct. Why is not possible to multiply all three if all sources are independent? Is there some Bayesian way to update the prior as I get new information?  Note: This is not homework, is something that I was thinking about.
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Given three Uniform random variables between 0 and 1, $(x,y,z)$, that are i.i.d., what is the probability $x+y+z &lt; 1$?
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Given two random integers $a$, $b$ calculate probability of $a^2$ + $b^2$ is divisible by $10$.  I've tried to simulate this process and got a result about $0.18$ (maybe incorrect), but have no idea why.
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I have no idea how to do this question. If you have a loaded dice such that the probability to obtain an outcome of 1 is 2p/3, the probability of obtaining 2, 3, 4 or 5 is p each, and the probability of obtaining 6 is 3p/2. Find p.   Any help would be much appreciated!
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I have seen , and  and a few YouTube videos, and I'm still stuck.   I understand how the probability integral transform gives rise to the result that the CDF of the p-values will have a uniform distribution.  What I don't understand is why that implies that the p-values themselves have a uniform distribution.  That is, I understand this much:  Suppose X ~ Unif(a, b). Then the CDF of X is:  $$P(X \le x) =  \begin{cases} 0\ \ {\rm{if}}\ \ x \le a \\ (x-a)/(b-a)\ \ {\rm{if}} \ \ a \le x &lt; b \\ 1\ \ {\rm{if}}\ x \ge b \end{cases}$$  So if X ~ Unif(0, 1), then $$P(X \le x) = x$$ (just substituting a=0 and b=1).  Now suppose $$Y = F(X)$$, and we want to know the probability distribution of Y. That is, we want to know the probability distribution of the CDF of X.  We know that the CDF of a distribution is a unique identifier of a distribution, so if you see, for example, $P(X \le x) = x$ then you know X ~ Unif(0, 1).   We also know that CDFs are right-continuous, and they go from 0 to 1. So it's reasonable to pick a value, f, that lies between 0 and 1 and try to find the probability that the CDF, Y, takes a value less than or equal to f:  $$\begin{align*} P(Y \le f) &amp;= P(F(X) \le f) \\ &amp;= P(X \le F^{-1}(f)) \ {\rm{assuming\ F\ is\ invertible}} \\ &amp;= F(F^{-1}(f)) \\ &amp;= f \end{align*}$$  So since $P(Y \le f) = f, Y = F(X)$ must follow a uniform distribution.  This implies that for any continuous random variable (that satisfies some properties that I'm not sure of), the CDF of that continuous random variable will have a Unif(0, 1) distribution.  It does NOT imply that the random variable itself has a Unif(0, 1) distribution. That is, it does not mean that X has a Unif(0, 1) distribution, only that F(X) has a Unif(0, 1) distribution.  So if a test statistic has a continuous distribution, then the CDF of that test statistic has a Unif(0, 1) distribution. Why does this mean that the p-values have a uniform distribution?  Wait...are p-values the CDF of a test statistic?   Clearly I'm tying myself in knots here. Any help would be appreciated.  EDIT (responding to a comment):   Here's my line of thinking since sleeping on it.  If we have $P(X \le x) = x$, then X ~ Unif(0,1).   Since $P(F(X) \le f) = f$, that means $F(X)$ ~ Unif(0,1), right?  But why does this lead us to think that p-values are uniformly distributed if the null hypothesis is true?   Suppose for instance we have:  $$H_0: \mu \ge 0$$ $$H_a: \mu &lt; 0$$,  and $\sigma$ is known. Let $ts$ be the test statistic, which has a non-standard normal distribution. After standardization, let the z-score associated with the test statistic be $z_{ts}$.  Then we would reject $H_0$ if $P(Z &lt; z_{ts}) &lt; 0.05$. That is, we would reject $H_0$ if the p-value is less than 0.05.   The form $P(Z &lt; z_{ts})$ is the same kind of form as a CDF, right? If the test statistic is continuous then this is the same as $P(Z \le z_{ts})$.   Now let $F(Z) = P(Z \le z_{ts})$.   Is this really a CDF? If so, then what?  What about when we have other alternative hypotheses (like $H_a: \mu &gt; 0$ or $H_a: \mu \ne 0$)?
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There are two Random Variables X and Y. Both of them are Normally distributed with 0 mean and variance a and b respectively. X and Y are independent of each other.  What is the probability of   What should be the easiest way to do this?
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Two squares are chosen at random on a chessboard. What is the probability that they have a side in common? I have got the total no of events by using 64 C 2. But I am unable to find the numerator(no. of favorable events).
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Let $\pi_1,...,\pi_n$ be a random permutation of numbers $1,...,n$. If you are told that $\pi_k &gt; \pi_1,...,\pi_k &gt; \pi_{k-1}$, what is the probability that $\pi_k = n$?  What I've tried:  Let $A_j$ be the event that $\pi_k = j$. Let $H$ = be event that $\pi_k &gt; \pi_1,...,\pi_k &gt; \pi_{k-1}$.  Then $P(A_n|H) = \frac{P(H|A_n)P(A_n)}{\sum_{j = 1}^n P(H|A_j)P(A_j)} = \frac{\frac{1}{n}}{\sum_{j=1}^n P(H|A_j)\frac{1}{n}} = \frac{1}{\sum_{j=1}^n P(H|A_j)} = \frac{1}{\sum_{j=1}^n P(j &gt; \pi_1,...,j &gt; \pi_{k-1})}$.  How to calculate that last sum?
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Take two problems:  Andrew is 35, and the probability he will be alive in 10 years is .72. Ellen is 35, for her, .92. Assuming these are independent, what is the probability they both will be alive in 10 years? Answer: .66, Method: Use the multiplication rule .72*.92 = .66  Suppose your street has two traffic lights. The chance that the first light is red is .40, and the second light .30. The chance of them being red at the same time is .10. What is the the probability that neither light is red? Answer: .40, Method: Subtract .1 from .4 and .3, and then subtract all 3 from 1.0 = .4   Why doesn't the multiplication rule work for #2? In other words, why can't I multiply .40 and .30 to get .12? And then furthermore, if .12 is the chance they will both be red at the same time shouldn't (1-.12)=.88 be the chance that neither is on?
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I'm having trouble with the following problem     There are 4 different colored balls: Red, Green, Black, Blue what is the probability that I will get 2 of the same colored balls by drawing 4 balls with replacement.   I attempted to solve this problem by splitting the question into two parts. First I calculated the probability of drawing 4 balls where I have 2 of one color and 2 of another color. Call this event $A$.  $$P(A) = \frac{4 \times 1 \times 3 \times 1 \times \binom{4}{2}}{ 4^4}= \frac{72}{4^4}$$  Next I calculated the probability where 2 balls are the same color and the other 2 are different. Call this event $B$ $$P(B)= \frac{4\times1\times3\times2\times4P2}{4^4}$$ I understand that if I sum these two probabilities I should get the answer but for some reason the event B was not calculated correctly. Any feedback and help is appreciated.
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I found on this website  proof that the probability of two people in a room having the same birthday equates to 50% when when there are 23 people in the room  The only thing is, the explanation is too complicated for the average Middle School student to comprehend. So, my question is, is there any simplifies explanation of how this works? I tried Wikipedia but the explanations were even more complicated.  Thanks!
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Say I have two independent random variables $X \sim N(u_1, \sigma_1)$ and $Y \sim N(u_2, \sigma_2)$. I want to get the conditional distribution of X given whether X is bigger than Y or not.  $P(X|X&lt;Y)$ = ... and  $P(X|X&gt;Y)$ = ...  I am thinking solving this in this way:  \begin{align} P(X|X&lt;Y) &amp;= \frac{P(Y&gt;X|X)P(X)}{P(Y&gt;X)} \\ &amp;= \frac{(1-\Phi(\frac{x-\mu_2}{\sigma_2}))N(\mu_1,\sigma_1)}{\Phi(\frac{\mu_2-\mu_1}{\sqrt{\sigma_2^2+\sigma_1^2}})}\\ P(X|X&gt;Y) &amp;= \frac{P(Y&lt;X|X)P(X)}{P(Y&lt;X)} \\ &amp;= \frac{\Phi(\frac{x-\mu_2}{\sigma_2})N(\mu_1,\sigma_1)}{1-\Phi(\frac{\mu_2-\mu_1}{\sqrt{\sigma_2^2+\sigma_1^2}})} \end{align}  My questions are:   (1) Whether above solution is correct  (2) How to get the mean and sd for $P(X|X&lt;Y)$ and $P(X|X&gt;Y)$ if they are still normal?