掷两个骰子时掷出和为 3 的概率是多少？

概率意味着可能性。它说明事件即将发生的可能性。事件的概率只能存在于 0 和 1 之间，其中 0 表示事件不会发生，即不可能，1 表示肯定会发生，即确定性。

事件发生的概率越高或越小，事件发生或不发生的可能性就越大。

例如 –一枚无偏的硬币被抛一次。因此，结果的总数只能是 2，即“正面”或“反面”。两种结果的概率相等，即 50% 或 1/2。
因此，事件的概率是有利结果/结果总数。它用括号表示，即P(Event)。

P(Event) = N(Favorable Outcomes) / N (Total Outcomes)

Note: If the probability of occurring of an event A is 1/3 then the probability of not occurring of event A is 1-P(A) i.e. 1- (1/3) = 2/3

什么是样本空间？

事件的所有可能结果称为样本空间。

例子-

• A six faced dice is rolled once. So, total outcomes can be 6 and 
  Sample space will be [1, 2, 3, 4, 5, 6]
• An unbiased coin is tossed, So, total outcomes can be 2 and 
  Sample space will be [Head, Tail]
• If two dice are rolled together then total outcomes will be 36 and 
  Sample space will be 
  [ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)   
    (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
    (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)  
    (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) 
    (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) 
    (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) ]

活动类型

独立事件：如果两个事件（A 和 B）是独立的，那么它们的概率为

P(A 和 B) = P (A ∩ B) = P(A).P(B) 即 P(A) * P(B)

Example: If two coins are flipped, then the chance of both being tails is 1/2 * 1/2 = 1/4

互斥事件：

• 如果事件 A 和事件 B 不能同时发生，则称为互斥事件。
• 如果两个事件互斥，则两者发生的概率记为 P (A ∩ B) 并且
  P (A 和 B) = P (A ∩ B) = 0
• 如果两个事件互斥，则任一事件发生的概率记为 P (A ∪ B)
  P (A 或 B) = P (A ∪ B) = P (A) + P (B) - P (A ∩ B) = P (A) + P (B) - 0 = P (A) + P (二）

Example: The chance of rolling a 2 or 3 on a six-faced die is P (2 or 3) = P (2) + P (3) = 1/6 + 1/6 = 1/3

不互斥事件：如果事件不互斥，则

P (A or B) = P (A ∪ B) = P (A) + P (B) − P (A and B)

什么是条件概率？

对于某个事件 A 的概率，给出了某个其他事件 B 的发生。写成 P (A ∣ B)

P (A ∣ B) = P (A ∩ B) / P (B)

示例-在一袋 3 个黑球和 2 个黄球（共 5 个球）中，拿一个黑球的概率是 3/5，拿第二个球的概率是黑球或黄球取决于先前取出的球。因为，如果拿了一个黑球，那么再次捡到一个黑球的概率是 1/4，因为只剩下 2 个黑球和 2 个黄球，如果之前拿过一个黄球，那么再次捡到一个黑球的概率是黑球将是 3/4。

掷两个骰子时掷出和为 3 的概率是多少？

解决方案：

When two dice are rolled together then total outcomes are 36 and Sample space is 
[ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)    
  (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)    
  (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)    
  (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)    
  (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)    
  (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) ]

So, pairs with sum 3 are (1, 2) (2, 1)  i.e. total 2 pairs

Total outcomes = 36
Favorable outcomes = 2

Probability of getting the sum of 3 = Favorable outcomes / Total outcomes  = 2/36 = 1/18

So, P(sum of 3) = 1/18.

类似问题

问题1：两个骰子得到偶数的概率是多少？

解决方案：

When two dice are rolled together then total outcomes are 36 and Sample space is 
[ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)   
  (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)    
  (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)    
  (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)    
  (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)    
  (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) ]

So, pairs with both even numbers are (2, 2) (2, 4) (2, 6) (4, 2) (4, 4) (4, 6) (6, 2) (6, 4) (6, 6) i.e. total 9 pairs

Total outcomes = 36
Favorable outcomes = 9

Probability of getting both even numbers = Favorable outcomes / Total outcomes   = 9/36 = 1/4

So, P(E, E) = 1/4.

问题2：两个骰子都得到奇数的概率是多少？

解决方案：

When two dice are rolled together then total outcomes are 36 and Sample space is

[ (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)   
  (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)    
  (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)   
  (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)   
  (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)   
  (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) ]

So, pairs with both odds are (1,1) (1,3) (1,5) (3,1) (3,3) (3,5) (5,1) (5,3) (5,5)  i.e. total 9 pairs

Total outcomes = 36
Favorable outcomes = 9

Probability of getting both odds = Favorable outcomes / Total outcomes                                                       
                                                    = 9/36 = 1/4

So, P(O,O) = 1/4.

问题3：得到2之和的概率是多少？
解决方案：

When two dice are rolled together then total outcomes are 36 and Sample space is 
[ (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)   
  (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)    
  (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)    
  (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)    
  (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)    
 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) ]

So, pairs with sum 2 are (1,1) i.e. only 1 pair

Total outcomes = 36
Favorable outcomes = 1

Probability of getting the sum of 2 = Favorable outcomes / Total outcomes  = 1/ 36 

So, P(2) = 1/36.