当两个骰子同时掷出时,得到一个和小于 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)
示例:如果两个硬币被翻转,那么两个硬币都是反面的机会是 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 (B)
示例:在六面骰子上掷出 2 或 3 的机会是 P(2 或 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。
当两个骰子同时掷出时,得到一个和小于 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) ]
P(Sum < 9 ) = P(1) + P(2) + P(3)…… + P(8)
So, pairs with sum less than 9 are (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2,3) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (4,1) (4,2) (4,3) (4,4) (5,1) (5,2) (5,3) (6,1) (6,2) i.e. total 26 pairs
Total outcomes = 36
Favorable outcomes = 26
Probability of getting the sum less than 9 = Favorable outcomes / Total outcomes
= 26 / 36 = 13/18
So, P(sum < 9) = 13/18
类似问题
问题 1:当掷两个骰子时,它们上出现的数字之和小于 5 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) ]
P(Sum < 5 ) = P(1) + P(2) + P(3) + P(4)
So, pairs with sum less than 5 are (1, 1) (1, 2) (1, 3) (2,1) (2,2) (3,1) i.e. total 6 pairs
Total outcomes = 36
Favorable outcomes = 6
Probability of getting the sum less than 5 = Favorable outcomes / Total outcomes
= 6 / 36 = 1/6
So, P(sum < 5) = 1/6
问题 2:当两个骰子被掷出时,它们上出现的数字之和大于 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) ]
Maximum sum possible is 12.
P(Sum > 9 ) = P(10) + P(11) + P(12)
So, pairs with sum more than 9 are (4,6) (5,5) (5,6) (6,4) (6,5) (6,6) i.e. total 6 pairs
Total outcomes = 36
Favorable outcomes = 6
Probability of getting the sum more than 9 = Favorable outcomes / Total outcomes = 6/36 = 1/6
So, P(sum>9) = 1/6
问题 3:当掷两个骰子时,它们上出现的数字之和小于 4 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) ]
Minimum sum possible is 2.
P(Sum < 4 ) = P(2) + P(3)
So, pairs with sum less than 4 are (1,1) (1,2) (2,1) i.e. total 3 pairs
Total outcomes = 36
Favorable outcomes = 3
Probability of getting the sum less than 4 = Favorable outcomes / Total outcomes = 3 / 36 = 1/12
So, P(sum < 4) = 1/12