Your question is not clear, but I will attempt to interpret it as best I can. When you first learn about probability, you are taught to list out the possible outcomes. If all outcomes are equally probable, then the probability is easy to calculate. Probability distributions are functions which provide probabilities of events or outcomes. A probability distribution may be discrete or continuous. The range of both must cover all possible outcomes. In the discrete distribution, the sum of probabilities must add to 1 and in the continuous distribtion, the area under the curve must sum to 1. In both the discrete and continuous distributions, a range (or domain) can be described without a listing of all possible outcomes. For example, the domain of the normal distribution (a continuous distribution is minus infinity to positive infinity. The domain for the Poisson distribution (a discrete distribution) is 0 to infinity. You will learn in math that certain series can have infinite number of terms, yet have finite results. Thus, a probability distribution can have an infinite number of events and sum to 1. For a continuous distribution, the probability of an event are stated as a range, for example, the probability of a phone call is between 4 to 10 minutes is 10% or probability of a phone call greater than 10 minutes is 60%, rather than as a single event.
The mean absolute deviation for a set of data is a measure of the spread of data. It is calculated as follows:Find the mean (average) value for the set of data. Call it M.For each observation, O, calculate the deviation, which is O - M.The absolute deviation is the absolute value of the deviation. If O - M is positive (or 0), the absolute value is the same. If not, it is M - O. The absolute value of O - M is written as |O - M|.Calculate the average of all the absolute deviations.One reason for using the absolute value is that the sum of the deviations will always be 0 and so will provide no useful information. The mean absolute deviation will be small for compact data sets and large for more spread out data.
Presuming a standard 52 card deck, you have 5 cards in that deck which satisfy your requirement. Therefore the probability of getting one of those cards is 5/52.
The odds of of an odd number on the first role are 3/6 since the odds are 1,3, and 5 and there are 6 numbers. The odds of an even are the same. Since they are independent we can multiply the two probabilities and the answer is .5x.5=.25 If this is confusing, consider the sample space where I denote O for an odd number and E for an even number. In two rolls you can have: OO, OE,EO, OO So there are 4 possible outcomes. We are interested only in OE which is 1 of the 4 outcomes so the probability of this happening is 1 in 4 or .25.
NO
0% and 100% or o and 1
Henrik O. Madsen has written: 'On stochastic load combination' -- subject(s): Probabilities, Structural design
You cannot. The sum of any two odd numbers is even (O+O=E), the sum of any two even numbers is even (E+E=E) and the sum of an odd and an even (or even and odd) is odd (O+E=E+O=O). O+O+O+O+O = O+O+O+O+O = E + E +O = E +O = O Addition is commutative, that is, the order in which you add the numbers does not matter. Some trial and error will demonstrate commutativity, though a proof is more complicated.
The sample space is {p, r, o, b, a, i, l, t, e, s}
The correct pronunciation is "possum" the "o" is slient.
Sum is pronounced as "soom". Quod is pronounced as "kwod", with a short 'o' (note to Americans: not the common American short 'o' that verges on 'ah'; more like the 'au' in "taught").
Yes, the word "some" has a short "o" sound, as in "sum."
o.875
sum is zero. +1 for H, +5 for N, -2 for each O. So sum is zero
54 and 48
No. The O is pronounced as a short U vowel sound (sum). The E is silent.
17