Why is a zero probability event not necessarily impossible

Why is a zero probability event not necessarily impossible


For continuous random variables, such as a drop of water from the basin, the probability of a drop of water being taken is 1 / N, n tends to infinity, so the probability is zero
Probability theory says that the probability of an impossible event is zero, but a zero probability event can happen. For example, if you take a person in the universe, you can get your probability. This is an example of a zero probability event that can happen!
Random variables are divided into continuous and discrete, and their distribution descriptions are different
For continuous random variables, the probability density value of a specific point is a bounded constant, which can be arbitrary (including 0 and 1). But because the point has no length, the probability density integral of the point is 0 (because the probability density value of the point is bounded), that is, the probability of the event corresponding to the point is 0, but the event is still possible, Similarly, the probability density of a point is 1, but the probability density integral of the point is still 0, so the event with probability 1 does not necessarily occur. In short, for continuous random variables, it is meaningless to discuss the probability of a single point (all are 0), The probability that the random variable falls into an interval
For a discrete random variable, if its event domain is finite, it can be considered that the event with a probability of 0 will not happen, and the event with a probability of 1 will happen. But if the event is infinite, it needs to be analyzed concretely
Since all events with zero probability are possible, events with a probability approaching zero are indeed possible. However, when we deal with problems, we usually count events with a probability approaching zero as events with zero probability, which is not absolute



If x2 + (m-2) x + 9 is a complete square, then the value of M is______ .


∵ x2 + (m-2) x + 9 is a complete square form, ∵ x2 + (m-2) x + 9 = (x ± 3) 2, and (x ± 3) 2 ∵ x2 ± 6x + 9, ∵ m-2 = ± 6, ∵ M = 8 or M = - 4



If x2 + 2mx + 9 is a complete square, then M=______ .


∵ x2 + 2mx + 9 is a complete square form, ∵ x2 + 2mx + 9 = (x ± 3) 2 = x2 ± 6x + 9, ∵ 2m = ± 6, M = ± 3



Please write a mixed operation problem of one to rational number and solve it. The problem should meet the following conditions at the same time
1. It must contain the power of addition, subtraction, multiplication and division 2. The divisor must be negative 3. The calculation result is equal to 2012


Square of 2 × [(- 1000) / (- 2) + 3]
=Square of 2 × [500 + 3]
=4*503
=2012



How to deduce the formula of the instantaneous velocity of the midpoint of the displacement of the uniform linear motion
The specific formula is VN = [xn + X (n + 1)] / 2T (generally used in the experimental problem of speed measurement by dot timer). This formula is hard to come up with~
What I want is not the formula of V = root ((V0 ^ 2 + VT ^ 2) / 2), but the formula of displacement above!


V(n-1)=Xn/T
V(n+1)=X(n+1)/T
Vn=[V(n-1)+V(n+1)]/2
=[Xn+X(n+1)]/2T



1. If the number of a increases by 25% and is equal to the number of B, what percentage of the original number of a is that of B?
2. Form a formula with five 3's, and the results are 0,1,2,3 respectively


1.1÷(1+25%)=80%
2.3+3-3+3-3=3
3-3÷3-3÷3=1
3-3+3-3÷3=2
(3-3)×3×3×3=0



Derivation of the relationship between the instantaneous velocity in the middle of a section of displacement and the initial and final velocity of this section of displacement
How to get V (s / 2) ^ 2-v0 ^ 2 = 2A * s / 2 from vt ^ 2-v0 ^ 2 = 2As in the derivation process of V = root sign (VT ^ 2 + V0 ^ 2) / 2


Uniform speed change
In VT ^ 2-V ^ 2 = 2As
In V ^ 2-VO ^ 2 = 2As (total displacement 2S)
therefore
Vt ^ 2-V middle ^ 2 = V middle ^ 2-VO ^ 2
In 2V ^ 2 = VT ^ 2 + VO ^ 2
V middle = √ [(VT ^ 2 + VO ^ 2) / 2]



Fifth grade next semester mathematics formulaic calculation question, has how many writes how many


1.3/7 × 49/9 - 4/3 2.8/9 × 15/36 + 1/27 3.12× 5/6 – 2/9 ×3 4.8× 5/4 + 1/4 5.6÷ 3/8 – 3/8 ÷6 6.4/7 × 5/9 + 3/7 × 5/9 7.5/2 -( 3/2 + 4/5 ) 8.7/8 + ( 1/8 + 1/9 ) 9.9 × 5/6 + 5/6 10.3/4 ×...



The formula of median velocity of uniform acceleration motion is deduced
How to deduce the median velocity = root ((V0 ^ 2 + VT ^ 2) / 2)? Note: it is not the median velocity


Vt ^ 2-v0 ^ 2 = 2As VT ^ 2-vm ^ 2 = 2A * s / 2 = as VM ^ 2-v0 ^ 2 = 2A * s / 2 = as
Vt ^ 2-vm ^ 2 = VM ^ 2-v0 ^ 2, so VM = root ((V0 ^ 2 + VT ^ 2) / 2)



Sixth grade mathematics application problem (how to do? Urgent, fast!)
There are 60 tons of cement in a construction site, 1 / 4 of which is used in the first day, and 9 / 2 in the second day. How many tons of cement are still on the construction site?


60 * (1-1 / 4) - 9 / 2 = 81 / 2 = 40.5 tons