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In general, if you flip a fair coin $n$ times, where $n$ is odd, then the probability that more than half the outcomes are heads [$(n+1)/2$ or more heads] is 0.5. … And the probability that less than half the outcomes are tails [$(n-1)/2$ or fewer tails] is also 0.5, in fact this is the same event. …

The fact is that all 52 cards are treated symmetrically, hence all must have the same probability of being selected. … Given your setup, you could display the probability using the law of total probability: P(choose ace spades) = P(choose ace spades | ace of spades is in my half of deck) x P(ace of spades is in my half …

w huber points out that a quirk of the Gregorian Calendar causes the starting day of a leap year to be not quite uniformly distributed, so the true probability of 53 Sundays is 1% or so greater than 2/ …

No, but you can conclude that the probability of any shared events is zero. Disjoint means that $A_i \cap A_j=\emptyset$ for any $i\ne j$. … Any shared elements must have probability zero. Same goes for all higher-order intersections as well. In other words, you can say, with probability 1, that none of the sets can occur together. …

You either win (with probability $0.3 \times 0.3$), or lose (with probability $0.7 \times 0.7$) or get back to duece again (with probability $0.3 \times 0.7 + 0.7 \times 0.3$). … for the required probability. …

If indeed you simply want to estimate the probability by simulation, as your question states, then much simpler and faster code would be: > n <- 1e6 > X <- rnorm(n, 0, 1) > Y <- rnorm(n, 1, 5) > mean … Of course, you could compute the desired probability to full machine accuracy in R without using simulations by: pnorm(1, sd=sqrt(26), lower.tail=FALSE). …

While that notation is technically possible, because it is possible to keep the probability function $P()$ separate from the event $P$, it nevertheless is needlessly confusing. … If you do want to use $P$ to represent an event, then it would be wise to prepare for that by choosing a different notation other than $P()$ for the probability function. …

The process of thinning out the number of eggs laid to the number of eggs hatched is called binomial thinning, and the distribution of the number of eggs that hatch is known to be Poisson with mean $p …

Let's continue your example with two decks of cards, supposing that one card is chosen randomly from each deck. You have proposed that $A$ is the event that the ace of spades is chosen from the first …

In other words, we "thin out" the random processes by keeping each of the original events with probability $p$: $X_1|Y_1 \sim {\rm Binomial}(Y_1,p)$ $X_2|Y_2 \sim {\rm Binomial}(Y_2,p)$ $X_3|Y_3 \sim …

What is the probability that 14 or fewer of the B's balls are white (also received an offer from A)? … > phyper(14, m=50, n=50, k=20, lower.tail=FALSE) [1] 0.01141749 The probability is only 0.011, meaning that the probability of having so much overlap between A's and B's offers would have been unlikely …

The definition of a "uniform distribution" is that the density function is constant for all $x,y$ within the support region. So one must have $$f_{X,Y}(x,y) = \frac{1}{A}$$ where $A$ is the area of ei …

For self-study questions like this the CV policy is to give hints rather than complete answers. Here the question focuses on something that is special to finite populations, and it seems to me to be …

From the log-likelihoods you give in your post, it is obvious that the largest probability will be exactly 1 and most of the rest will be exactly 0. … This shows that the second largest normalized probability will be about $10^{-139}$. …

Unfortunately, the Wikipedia article on "F-test of equality of variances" is incorrect. When the variances are unequal, the distribution of $F$ is neither $F$ nor non-central $F$, it is simply scaled …