Probability

The answer is 1 in 1000.
Rolling a 4 on a ten sided die is a 1 in 10 shot.
Rolling threes 4's is 1/10 x 1/10 x 1/10 which equals 1/1000.

1/2 x 1/2 x 1/2 = 1/8 or 0.125 also 12.5%

P(A given B')=[P(A)-P(AnB)]/[1-P(B)].

In words: Probability of A given B compliment is equal to the Probability of A minus the Probability of A intersect B, divided by 1 minus the probability of B.

p(a) = 1/3, p(b) = 1/2, p(a and b) = p(a)*p(b) = 1/6

From a probability perspective fair means equal probability.

2:7

edit:

That makes the assumption that:

-birth isn't induced for various reasons - people tend to try for induced births more on weekends so that it doesn't interfere with the work week

-the ratio of people born on any given day of the week is 1:1:1:1:1:1:1

Let A be the amount of cards in a normal deck that are not considered to be red. Let B be the total amount of cards in that same deck. Answer: A / B

2

10 Green marbles, 13 Blue marbles.

The probability is 0. Consider the event of tossing a coin . The possible events are occurrence of head and tail. they are mutually exclusive events. Hence the probability of getting both the head and tail in a single trial is 0.

The same units as the mean itself. If the units of the mean, are, for example miles; then the error units are miles.

Suppose a normal random variable has a mean of 72 inches and a standard deviation of 2 inches. Suppose the random variable X measures the height of adult males in a certain city. One may therefore conclude that approximately 84% of the men in this population are shorter than?

Assuming you refer to a normal deck of playing cards, either 0.481 (with jokers) or 0.5, one-half (without).

there are 3 types of conditional probability:

1. the indicate: if antecedent happens, then evidence manifests itself

example -> when tossing a coin, if it lands on tails, then you win the game

2. the subjunctive: if antecedent would happen, then evidence would manifest itself

example -> when a coin was tossed, if it would have landed on tails, i would have won the game

it is recommended for optimal bayesian inferences that your a priori distribution is indicative. If not, you could be dealing with inproper, uninformative, or hyper priors, which make decision-making and posterior determination more complex, if even possible.

Posterior distributions could very well be subjunctive. Suppose i have won the game, i could have tossed tails, but i could also have tossed heads.

50/50

It can be thought of as a generalization of the Chi-square distribution. See the link to a related WikiAnswer question below.

3/8

In tossing a coin, there are two outcomes; head or tail. Since there are two outcomes, probability of each outcome is: 0.5. Assuming a fair coin, probability of a head or tail is: 0.5. So, probability a team will win a toss is: 0.5.

Standard error (which is the standard deviation of the distribution of sample means), defined as σ/√n, n being the sample size, decreases as the sample size n increases.

And vice-versa, as the sample size gets smaller, standard error goes up. The law of large numbers applies here, the larger the sample is, the better it will reflect that particular population.

There are 2 ways to do this problem. 1. Go to a Binomial Distribution Table where n = 4 (4 children) and P=0.5(50% probability of a girl). Probability of at least 1 girl = 1 - probability of no girls. From Binomial Distribution Table n = 0 probability is .0625. So, 1 - 0.0625 = .9375 = probability of at least 1 girl. 2. The other way is to list all the possible ways to have 4 children and count the number of ways at least 1 girl exists divided by the total number of ways to have 4 children. There are 42 ways to have 4 children, all 16 listed below: bbbb bbbg bbgb bgbb gbbb bbgg bggb ggbb gbbg gbgb bgbg bggg gggb ggbg gbgg gggg Since 15 of the 16 have at least 1 girl, the Probability of at least 1 girl = 15/16 = 0.9375, the same answer as above.

There is 2 outcomes for flipping the coin, and 6 outcomes for rolling the cube. The total outcomes for both are 2*6 = 12.

We calculate the numerator of the desired probability:

There are 6 ways any number can show up on the first die

AND

There is only 1 way that same number can show up on the second die

So since "AND" means "MULTIPLY", there are 6x1 or 6 ways

that both dice can come up the same. So the numerator of

the probability is 6.

We calculate the denominator of the desired probability:

There are 6 ways any number can show up on the first die

AND

There are 6 ways any number can show up on the second die

So since "AND" means "MULTIPLY", there are 6x6 or 36 ways

the two dice can all come up any number. So the denominator

of the probability is 36.

yes it can be defined more commonly as a ratio between the number of and numbr of possible outcomes

There are 108 cards in a Phase 10 deck.

probability of tails on a coin is 1/2 probablility of 6 on a die is 1/6