当两个骰子同时掷出时,得到和为 9 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-
- 一个六面骰子掷一次。因此,总结果可以是 6 和
样本空间将是 [1, 2, 3, 4, 5, 6] - 抛一枚无偏的硬币,因此,总结果可以是 2 和
样本空间将是 [Head, Tail] - 如果两个骰子一起滚动,则总结果将为 36 和
样本空间将是
[ (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 and B) = P (A ∩ B) = P(A).P(B) i.e. 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,则两者发生的概率表示为
P (A ∩ B) 和 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 (B)
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。
当两个骰子同时掷时,两面之和为 9 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 9 are (3, 6) (4, 5) (5, 4) (6, 3) i.e. total 4 pairs
Total outcomes = 36
Favorable outcomes = 4
Probability of getting the sum of 9 = Favorable outcomes / Total outcomes
= 4 / 36 = 1/9
So, P(sum of 9) = 1/9.
类似问题
问题 1:两个骰子的和为 10 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 10 are (4,6) (5,5) (6,4) i.e. only 3 pairs
Total outcomes = 36
Favorable outcomes = 3
Probability of getting sum 10 = Favorable outcomes / Total outcomes
= 3 / 36 = 1/12
So, P(10) = 1/12.
问题2:得到11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 11 are (5,6) (6,5) i.e. total 2 pairs
Total outcomes = 36
Favorable outcomes = 2
Probability of getting the sum of 11 = Favorable outcomes / Total outcomes = 2/36 = 1/18
So, P(sum of 11) = 1/18.
问题 3:得到 12 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 12 are (6,6) i.e. only 1 pair
Total outcomes = 36
Favorable outcomes = 1
Probability of getting the sum of 12 = Favorable outcomes / Total outcomes = 1 / 36
So, P(12) = 1/36