Suppose that a and B events are independent of each other, and the probability of only a and only B is 1 / 4, calculate the probability of a event and B event

Suppose that a and B events are independent of each other, and the probability of only a and only B is 1 / 4, calculate the probability of a event and B event

If event a and B are independent of each other, then:
P(AB)=P(A)×P(B)
P (a, b) = P (AB, b) = 1 / 4
Then:
P (a horizontal) P (b) = P (a) P (b horizontal) = 1 / 4
[1－P(A)]×P(B)=[1－P(B)]×P(A)=1/4
The results show that P (a) = 1 / 2; P (b) = 1 / 2

Suppose that the probability of two independent events a and B not occurring is one in nine, and the probability of a not occurring and B occurring is equal to that of a occurring and B not occurring, then find the probability of a?

Let a be x and B be y
There are,
X*Y=1/9；
（1-X）*Y=(1-Y)*X
By solving the equation, we can get x = y = 1 / 3
So the probability of a happening is 2 / 3
It should be like this

Suppose that events a and B are independent of each other, and the probability of occurrence of a and B is 0.6 and 0.9 respectively, then the probability of no occurrence of a and B is 0

If a and B don't happen, we can get 0.40.1 by subtracting 1. If a and B don't happen, we can multiply the probability that a doesn't happen by the probability that B doesn't happen to get 0.04

On the scaling effect of relativity
Today, the teacher talked about this problem. A 200m car has passed a 175m bridge at a relatively fast speed. Then people on the ground see that the car is shorter and smaller than 175, and the bridge on the car is obviously smaller than 175m
That's what the teacher said in class. Then I have a question. If there is no bridge here and the car's center of gravity is at the end (that is, it will fall only after passing through the bridge), will the car fall

The scale effect is deceptive. In Merkel's experiment, it is said that the distance between the two mirrors in the instrument shortens with the movement of light

450 oral arithmetic questions with answers

Please tell me more about the scale down and clock slow effect

You have to understand relativity
A moving ruler will be shorter;
A moving clock slows down

100 oral arithmetic questions for Grade 5 (primary school people's Education Press)

3.5+5.2= 6.4-3.6= 7.3-0.05= 5.68-5.6= 0.04+0.6=1.1-0.98= 9+1.3= 1-0.48= 7.25+2.05= 10-2.3= 3.6+0.4= 10-5.2= 2.63+0.8= 7.5-5= 6+2.4= 9-0.72= 4.3-3.1= 8.04+0.06= 0.2+4.9= 8.5-1.5= 3.6×0.4= 0.8÷2= 0.19...

The slow down problem of relativistic moving clock
I don't understand it all the time. Does a watch belong to a clock, or is it affected as long as it is a timer?
For example, I am wearing a watch and sitting on a C / 3 aircraft. There is a big clock on the ground (big enough to be seen at any time). The starting time is always 1 o'clock. When I see the clock on the ground reaching 2 o'clock, I quickly look at my watch. Will the watch time change? (A. also 2 o'clock B. less than 2 o'clock C. more than 2 o'clock)
`
Another question I want to ask is: if it's all at 1 o'clock at the beginning, I control the aircraft with an initial speed of C / 3 to slowly slow down and land towards the big clock on the ground, while constantly comparing the time of the big clock and the watch.
When it falls on the big clock, the speed just decreases to 0. If the big clock is 2 o'clock, what time does my watch indicate?
(A. also 2 points B. less than 2 points C. more than 2 points D random uncertainty)
Thank you~~

The first question is that the clock on the spaceship is more than 2 o'clock. When an object has speed relative to you, you will find that the time of the object moving at high speed slows down. The spaceship flies at the speed of C / 3, and people standing on the earth will see that your speed is C / 3, then your time will slow down. When the time on the earth is 2 o'clock, people on the earth will see that your spaceship

Grade 5 Volume 2 100 oral arithmetic questions
Come on, today! T）～

0.9×6= 0.12×6= 6.8÷4= 0.72÷12= 0.24×2= 1.2÷3= 14×0.5= 9.6÷6= 1.6×5= 0.48÷6= 12.5÷5= 0.12×5= 4.6÷23= 1.6÷0.8= 7.2÷0.6= 3.9÷3.9= 1.8×0.5= 6.3÷9= 3.7+4.8= 0÷6.81= 2.5÷10= 0.15×8= 4.2...

The slow down of the moving clock in relativity
If there is a train that emits a beam of light from the front to the rear, and the train and the light are moving towards each other, who is the longer time to measure on the train or on the ground?

The time measured from the car is long, because the speed of light is always the same, but on the ground, the rear of the car moves forward to meet the light for a certain distance, so it meets the light faster. On the car, the rear of the car does not move