Probability

To obtain a much better, simpler, and more practical understanding of the data distribution.

No because the term mutually exclusive implies the the trials that could result in these events are sequenced in time.

Pr(1 or 5) = 1/3

There are two sides to the coin, so the probability of getting heads or tails on one flip of the coin is 1/2 or 50%.

100% of all male offspring will be colorblind.

0% of all femal offspring will be colorblind.

Probability is the measure of how likely an event is. ... The probability of event A is the number of ways event A can occur divided by the total number of possible.

Odd Isøy died in 1983.

A man has an 81 percent chance to get married if they live the United States before the age of 40. A woman has an 86 percent chance of getting married.

It depends. If you're plugging in something into a wall outlet and you get zapped, your chances of surviving are probably 99.9%. But if you're electrocuted by an electrical conduit box in your house/building and you survive, God REALLY loves you!! Plain & simple, don't mess with electricity.

Probability is more of a mathematical statisic whereas chance is more of luck or fortunance.

I'm still working through this one. Yes, given N IID r.v., the maximum statistic is described by the beta distribution. Actually it is F(x) that is beta distributed. Remember that the argument of the beta distribution goes from 0 to 1. In fact, the beta distribution is the sampling distribution that describes any order statistic (k = 1 to n) taken from any distribution. Maximum and minimum become special cases. In all cases, the sample cdf of the maximum is G(x) = F(x)n. For the uniform, F(x) =(x-A)/(B-A) for a Uniform(A,B) continuous distribution. Now, if I'm using the beta cdf, B(x), for order statistics (k=1 is the minimum, k=n is the maximum), in the form, B(x|alpha, beta), then alpha = k and beta = n-k+1 and x = F(x) so the sampling distribution is: G(x) = B(F(x)|k,n-k+1) for any rank order statistic X(k), and for the maximum, B(F(x),n,1). The inverse of G(x) is interesting, G-1(p)=F-1(B-1(p))=x, where p is a probability of being equal or less than x. OK, this might not be so clear, and I try to post a bit more on this later. Your second question is difficult. Your sample is not taken independently. Each selection bears some relation to the prior selection. I don't see that even a large sample would be representative of the population, but I need to think on this some more. I thought a partial answer now is better than none at all.

In a book of 52 playing cards there are thirteen types of number cards, cards starting with Ace, 2 , 3 and ending as Jack, Queen and King . And king is number thirteen there are four kings each uniquely symbolized with Spade, Club, Diamond and Heart. Spade and Club are black colored while Heart and Diamond are Red colored. So apparently spade is black colored sign card which are thirteen starting from Ace, 2, 3 and ending as jack, queen and King at last.

The sample space contains 36 possible outcomes (1,1 & 1,2 & 1,3.....etc up to 6,6). Since 6,6 is the only way to get the sum of 12, the probability of 12 is 1/36.

It is 17/18.

No: the opposite.

With a fair die it is 5/6.

There are 52 cards in a deck (1-10, plus the Jack, Queen, King, and Ace).

There are four suits in a deck (hearts, spades, clubs, diamonds), and thus, there are 4 possible aces in the deck.

4/52 = 1/13

If you put the one card back in before trying it the second time, then the odds are slightly worse; otherwise, the odds are the same - it does not matter if you pull two cards at once or one card twice, subject to the noted limitation.

0.81 apex:)

It is 15/16.

The SD is the (positive) square root of the variance.

1/9

The probability is 1 as all values on a die are less than 7.

1 in 23328

The number of total outcomes on 3 tosses for a coin is 2 3, or 8. Since only 1 outcome is H, H, H, the probability of heads on three consecutive tosses of a coin is 1/8.