the probability model
Increases the probability of 10 to 11
The probability of a normal non-safe buff was between 10% and 30%, depending on the equipment's quality, level, enchantment, and strengthening level. The probability of a safe buff was between 3% and 10%, but there was no separate probability of a 10 to 11 buff.
1 answer
2026-07-06 16:20
2 dice probability
For two dice, there were several possibilities:
1. ** Single number probability **:
- Each die had six sides, and the numbers were 1, 2, 3, 4, 5, and 6. When a die is rolled, the probability of each number appearing is 1/6. When two dice were thrown, for example, the probability of both dice rolling a 1 was 1/6. Therefore, the probability of both dice rolling a 1 (the combination of 1 and 1) was (1/6)×(1/6)=1/36. Similarly, any particular combination of numbers (such as 3 and 5) has a probability of 1/36.
2. ** Point sum probability **:
- The sum of the points was 2 (1 + 1), and there was only one possibility. The probability was (1/6)×(1/6)=1/36.
- The sum of the points is 3 (1+2 or 2 + 1). There are two possibilities, and the probability is 2/36.
- The sum of the points was 4 (1+3, 2+2, 3+1). There were 3 possibilities, and the probability was 3/36.
- The sum of the points is 5 (1+4, 2+3, 3+2, 4+1). There are 4 possibilities, and the probability is 4/36.
3. ** Dice size probability (1 - 6 is small, 7 - 12 is big)**:
- There were a total of 6x6 = 36 combinations of the two dice. Among them, there were 15 situations where the sum of points was 1 - 6 (small), and 21 situations where the sum of points was 7 - 12 (large). However, since it was impossible for two dice to have a point, the probability of a small point appearing was 5/11 (about 45.45%), and the probability of a big point appearing was 6/11 (about 54.55%).
4. ** The probability that the sum of the two dice numbers is odd (assuming the first die numbers are 1, 2, 3, 3, 5, 6, and the second die numbers are 1, 2, 4, 4, 5, 6)**:
- For the first die, the probability of an odd number appearing was 4/6 = 2/3, and the probability of an even number appearing was 1/3. For the second die, the probability of an odd number appearing was 2/6 = 1/3, and the probability of an even number appearing was 2/3. To get the odd sum, there are two situations: if the first die is odd and the second die is even, the probability is 2/3×2/3 = 4/9; if the first die is even and the second die is odd, the probability is 1/3×1/3 = 1/9. Therefore, the total probability was 4/9+1/9 = 5/9.
Hurry up and click on the link below to return to the super classic "Lord of the Mysteries"!
1 answer
2026-07-31 12:05
A probability problem related to the classical probability model chapter of the university is actually not difficult. I just learned it and don't know how to master it. Please help me analyze it.
下面是一道大学古典概型章节的概率问题:
设 $X$ 是一个服从参数为 $\mu$ 和 $\sigma^2$ 的二项分布的随机变量满足 $P(X=k)=\frac{\sigma^2}{k!}$其中 $k=12\ldots$.问在以下条件下$X$ 的概率密度函数为多少:
1 $\mu=0$$\sigma^2=1$;
2 $\mu=1$$\sigma^2=0$;
3 $\mu=\infty$$\sigma^2=\frac{1}{n}\sum_{i=1}^n i$ (其中 $n$ 是一个正整数).
求解上述三个条件中$X$ 发生概率最大的条件.
首先根据二项分布的密度函数性质当 $k=1$ 时$X$ 的分布函数为 $f_X(x)=P(X=1)=\frac{\sigma^2}{1!} = \frac{\sigma^2}{x!}$因此 $X$ 发生概率为 $\frac{1}{x!}$.
其次当 $\mu=1$ 且 $\sigma^2=0$ 时$X$ 的分布函数为 $f_X(x) = 1$因此 $X$ 发生概率为 0.
最后当 $\mu=\infty$ 且 $\sigma^2=\frac{1}{n}\sum_{i=1}^n i$ (其中 $n$ 是一个正整数)时$X$ 的分布函数为 $f_X(x) = \frac{1}{x\ln(n)}$因此 $X$ 发生概率为 $\frac{\ln(n)}{\frac{1}{n}\sum_{i=1}^n i}$.
根据古典概型的定义在条件 2 和条件 3 中$X$ 发生的概率可以分别计算为:
在条件 2 中$X$ 发生的概率为 $\frac{1}{x!}$;
在条件 3 中$X$ 发生的概率为 $\frac{\ln(n)}{\frac{1}{n}\sum_{i=1}^n i}$.
因此当 $\mu=0$$\sigma^2=1$ 时$X$ 发生概率最大的条件为 $\mu=1$$\sigma^2=0$即条件 3.
需要注意的是上述解析仅适用于二项分布的情况如果涉及到其他的概率分布需要根据具体情况进行解析.
1 answer
2024-09-16 12:08
Destiny, 2000 words probability
I'm not sure what exactly you mean by the '2,000-word probability of fate' you mentioned. If you can provide more context or detailed information, I will try my best to help you.
While waiting for the anime, you can also click on the link below to read the classic original work of The King's Avatar!
1 answer
2024-10-21 19:43
Hegemony reward probability
"We can come to the following conclusion: the probability of an orange card appearing in the Hegemony Card Pack is 5.6%. However, the exact number of orange card draws was not certain, because different card packs had different guarantee mechanisms. Some card packs guaranteed an orange card after a certain number of draws, while others did not have a minimum number of draws. According to the information provided, the probability of obtaining other Hegemony rewards cannot be known.
1 answer
2024-12-25 08:09
10,000 words signing probability
The signing rate of Zongheng was less than 0.8%. Some authors had received more than 10,000 words of short messages, but this did not accurately indicate the probability of signing a contract for 10,000 words. Although there were cases of successful signing, the overall signing rate was relatively low. There were also cases where the signing threshold was lowered but the quality was not strictly controlled. There were many novels signed on the website but the general quality was not high. Therefore, it was difficult to determine the specific signing probability at 10,000 words.
1 answer
2026-06-23 19:44
Mythical equipment probability
The probability of obtaining mythical equipment in different games was different:
- In the games that had daily dungeons (which would be opened when the character reached level 35, and each person would have three chances a day, and the number of chances would increase at a specific time) and team dungeons (which would be opened when the character reached level 50), the specific probability of obtaining Mythical equipment was not clearly stated. It only meant that there was a certain chance of obtaining Mythical equipment after completing the stages in the dungeon. Mythical equipment had a high drop rate in team dungeons, but the class was random.
- In the DNF:
- Nether Tower Jar (From the Nether Tower Instance Dungeon, Level 100. You can exchange coins for jars after clearing each level, including weapon jars, armor jars, jewelry jars, etc.) The probability of obtaining Mythical equipment is about 1%.
- Path of the Strong (Source: Path of the Strong Instance Dungeon. Exchanged with Trial Proof materials). Mythical equipment is a random event.
- [Cirlock Jar (Originated from Cirlock's dungeon, exchanged with 500 Eternal Aria). Can randomly unlock all Level 100 Epic (excluding work clothes) and Level 100 Mythical equipment. Chance is low.]
- In the Epic Road event, the Mythical drop buff would increase the Mythical drop rate as the number of passes increased. Each character would be calculated separately. After the current character dropped Mythical equipment, the Mythical drop rate would return to normal.
- In the Legend of Wonderland mobile game:
- There was a chance of obtaining Level 20 Mythical Equipment after clearing the Lonely Desert Abyss difficulty dungeon, and there was a chance of obtaining Level 40 Mythical Equipment after clearing the Yellow Springs difficulty dungeon, but the specific probability value was not given.
- Killing mini-monsters can obtain the God's Bless Secret Box, which can drop mythical equipment. There is no mention of the probability.
- Daily Instance Dungeons can be entered at level 35, 3 chances per day (3 additional chances for activating special privileges). After clearing the dungeon, there is a chance to obtain Mythical equipment, but the specific probability value is not given.
- The Tower of Pride Shop could use Pride Silver Coins obtained from clearing the tower to exchange for Mythical equipment. There was no mention of the probability.
1 answer
2026-07-09 03:33
Hearthstone painting probability
Under normal circumstances, an average of 181 normal packs would give out a mutant orange. However, on December 7, 2022, after Hearthstone's patch 25.0 was released, players found that the probability of the game's standard packs giving out mutant orange cards was extremely high. Therefore, Blizzard temporarily closed the opening function of the standard packs to investigate. The special drawing card can be opened through the package.(Use in-game gold coins or real money to buy card packs to activate), participate in activities (You can get a special drawing card bag or get it through the event reward route), participate in the competition (Official Stairway or Single Player Challenge), Combination (There is a certain chance of getting a special drawing version from the normal version), joining the guild (through the guild treasure chest and missions), pre-ordering the super collection (such as the "Lich King's March" super collection can get two special drawing cards), completing the Death Knight Prologue (can unlock the Death Knight profession and get 32 core series cards including 6 special drawing cards for free), etc., but these methods did not mention the specific probability of getting special drawing.
"The Legend of the Three Dragon Scales in the Milky Way Continent" is equally exciting. Everyone is welcome to click and read it!
1 answer
2026-03-15 19:34
The probability of dice rolling
Rolling dice was a classic example of probability.
For a single standard die, because it had six sides, each marked with the six numbers 1, 2, 3, 4, 5, and 6, the chances of each number appearing were equal. Therefore, when throwing a single die, the probability of any number appearing in the result was 1/6.
When there were multiple dice involved, the probability calculation would be more complicated. Take two dice as an example, if you consider the size of the dice,(1 - 6 is small, 7 - 12 is big), because there are a total of 6×6 = 36 combinations of the number of points on the two dice. Among them, there are 15 cases where the sum is 1 - 6 (small), and there are 21 cases where the sum is 7 - 12 (big). However, since it is impossible for the two dice to appear, the actual probability of small appearing is 5/11 (about 45.45%), and the probability of big appearing is 6/11 (about 54.55%).
If there were three dice, each die would have six situations, and there would be a total of 6x6x6 = 216 combinations. For example, if one calculated the probability of rolling a big die, there would be three kinds of big situations (namely, 4, 5, and 6 points) for each die. There were a total of 3×3×3 = 27 kinds. Therefore, the probability of a big situation appearing was 27 × 216 = 1/8. Similarly, the probability of a small situation appearing was also 1/8. The probability of a leopard appearing (three dice with the same number of points, such as 111, 222, etc.) was 6 × 216 = 1/36.
In addition, in theory, if the structure of the die was uneven, for example, the dots were hollowed out, and the six small round pits on the six dots were lighter than the one on the opposite side, then the probability of the six appearing would be greater than the one. However, in the ideal standard die situation, the probability of each number was 1/6. At the same time, there were many other situations in the probability calculation of dice rolling, such as the probability calculation of the sum of two dice numbers being odd, which could be solved by different probability methods (such as the step-by-step probability calculation method in junior high school or the combination knowledge method in high school).
Hurry up and click on the link below to return to the super classic "Lord of the Mysteries"!
1 answer
2026-07-31 02:15
dice-rolling probability
The condition probability refers to the probability of event A happening under the condition that event B has already happened. It is recorded as P(A| B)。The following is an example analysis of the probability of dice rolling.
Suppose we have two dice. Event A is that the sum of the two dice is 8, and Event B is that the first die rolls a 3.
First, calculate the probability of event A without any conditions. There were a total of 6×6 = 36 possibilities for the outcome of the two dice. There are five cases where the sum of points is 8:(2,6),(3,5),(4,4),(5,3), and (6,2), so P(A)=5/36.
Then, the probability of event B is calculated. The probability of the first die rolling a 3 is 1/6, which is P(B)=1/6.
The only situation where both events A and B occur at the same time is (3,5), so P(A Empyrean B)=1/36.
According to the formula P(A| B)=P(A Empyrean B)/P(B). Under the condition that the first die rolls a 3, the probability of the sum of the two dice points being 8 is P(A| B)=(1/36)/(1/6)=1/6。
For example, Event C was the difference between the two dice being 1, and Event D was the second die rolling a 4. Similarly, P(C) was first found. There were 10 cases where the difference between the points was 1:(1,2),(2,3),(3,4),(4,5),(5,6),(2,1),(3,2),(4,3),(5,4), and (6,5), so P(C)=10/36. The probability of event D is P(D)=1/6. There are two cases where events C and D occur at the same time: (3,4) and (5,4), so P(C Empyrean D)=2/36. Under the condition that the second die rolls a 4, the probability of the difference between the two dice being 1 is P(C| D)=(2/36)/(1/6)=1/3。
In short, when calculating the conditioned probability of rolling the dice, the key was to clarify the relationship between the events, separately calculate the probability of the event occurring alone and the probability of them occurring at the same time, and then use the conditioned probability formula to calculate.
Hurry up and click on the link below to return to the super classic "Lord of the Mysteries"!
1 answer
2026-08-01 02:05
Popular Answers
Popular Searches
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
s
t
u
v
w
x
y
z