简单来说,概率是在事件发生之前对事件发生的预测。我们在日常生活中的许多事情上都做预测,例如:
1)野餐前先预测天气。
2)预测选举结果。
3)预测谁将赢得比赛。
在所有这些情况下,我们都尝试通过考虑有利于该事件的所有条件来找到事件发生的可能性或机会。从上面的讨论中,概率可以用数学定义为:
概率是数学的一个分支,它告诉我们事件发生的机会是什么。事件的概率是介于0和1之间的数字,其中0表示事件不可能,而1表示确定性。因此, 0≤P(E)≤1,其中P(E)=事件发生的概率。
概率中使用的一些基本术语是:
1.实验:实验被称为一种事件,其中预期会有明确的结果。也称为样本空间。例如,样本空间S为S = {H,T},其中H表示头,T表示尾巴。
2.试验:试验被称为确定结果的单项事件。
3.结果:结果是实验的结果。 例如,板球比赛的获胜可能是结果。
4.随机实验:随机实验是指可能无法预先预测结果的实验。可以在许多条件下重复进行此操作。
5.不可能事件:当事件的概率为0时,该事件称为不可能事件。
6.确定事件:当事件的概率为1时,该事件称为确定事件。
概率公式
事件的实验概率或经验概率为
Probability(E) =
或者,
P(E) =
在这里,P(E)=事件发生的概率
N(E)=有利结果的总数
N(S)=所有可能结果的总数
现在让我们继续解决问题和更好地理解概率。
样本问题
问题1. Sumit正在和他的朋友打板球,要决定要去击球的人是通过掷硬币来决定的,取胜者将首先进行击球。假设Sumit和Mohit是两支分别选择了头和尾的球队的队长。找到Sumit先击球的机会。
解决方案:
We know, there are only two possible outcomes of the toss i.e. heads or tails.
Therefore, Sample space(s) = Total possible outcomes = {H, T}
Sumit needs heads to win the toss, therefore there is only one favorable outcome.
Probability of Sumit to win the toss = favorable outcome / total outcome
= 1/2 = 0.5
问题2。掷两个骰子。找出总分数是质数的概率吗?
解决方案:
Since, two dices are tossed therefore total no of combination = n(S) = (6 x 6) = 36 combinations.
Let us considered E be the event that the sum is a prime number.
All the favorable outcomes are (E) = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1),
(2, 3), (2, 5), (3, 2), (3, 4), (4, 1),
(4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}
Therefore, n(E) = 15
Probability of score to be prime number = n(E)/n(S) = 15/36 = 5/12
问题3.在彩票盒中,有10个奖品和25个空白。从抽奖箱中随机抽出一张纸条。获得奖品的几率是多少?
解决方案:
Given: Total number of prize = 10
Total number of blanks = 25
So, the total number of possible outcomes(i.e., n(S)) are = 10 + 25 = 35
Total number of prizes, n(E) = 10
According to the formula
P(E) = n(E)/n(S) = 1035 = 27
问题4.一个袋子包含8个蓝色的球和一些粉红色的球。如果画一个粉红色的球的概率是画一个蓝色的球的概率的一半,则找出袋子中粉红色的球的数量。
解决方案:
Let us considered the number of pink balls be n.
The number of blue balls = 8.
Therefore, the total number of balls present in the bag = n + 8.
Now, the probability of drawing a pink ball, i.e. P(X) = n/n + 8
the probability of drawing blue ball, i.e. P(B) = 8/n + 8
According to the question, the probability of drawing pink
ball is half of the probability of drawing the blue ball
So, P(X) = P(B)/2
n = 4.
So, number of pink balls present in the bag is 4.
问题5.将编号为1到20的卡片混合在一起,然后随机抽出一张卡片。开出的卡具有3或5的倍数的几率是多少?
解决方案:
Card are numbered from 1 to 20 therefore n(S) = {1, 2, 3, 4, …., 19, 20}.
Let us considered E be the event of getting a multiple of 3 or 5
So, n(E) = {3, 6, 9, 12, 15, 18, 5, 10, 20}.
According to the formula
P(E) = n(E)/n(S) = 9/20.
问题6:一张纸牌是从52张纸牌中抽出的,洗得很整齐。计算卡将出现的概率
(i)成为王牌,
(ii)不是王牌。
解决方案:
Well-shuffling ensures equally likely outcomes.
(i) There are 4 aces in a deck.
Let us considered E be the event the card drawn is ace.
The number of favorable outcomes to the event E = 4
The number of possible outcomes = 52
Therefore, P(E) = 4/52 = 1/13
(ii) Let us considered F be the event of ‘card is not an ace’
The number of favorable outcomes to F = 52 – 4 = 48
The number of possible outcomes = 52
Therefore, P(F) = 48/52 = 12/13
问题7.同时投掷一对骰子。求出总和大于7的概率。
解决方案:
Total number of combinations for a pair of dice is = n(S) = (6 x 6) = 36
Let us considered E be the event of getting a total more than 7
= {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4),
(5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Therefore, P(E) = n(E)/n(S)
= 15/36 = 5/12.
问题8.在一个由364名工人组成的公司中,有91人结婚。查找选择未婚工人的可能性。
解决方案:
Given,
Total workers (i.e. Sample space) = n(S) = 364
Total married workers = 91
Now, total workers who are unmarried = n(E) = 364 – 91 = 273
Method 1: So, the probability of unmarried worker P(NM) = n(E)/n(S) = 273/364 = 0.75
Method 2: P(M) + P(NM) = 1
Here, P(M) = 91/364 = 0.25
So, 0.25 + P(NM) = 1
P(NM) = 1 – 0.25 = 0.75
问题9.从一袋黄色和棕色的球中,捡到一个红色球的概率是x / 2。如果捡到一个棕色球的概率为2/3,则找到“ x”。
解决方案:
Given, in the bag only yellow and brown balls.
P(picking a yellow ball) + P(picking a brown ball) = 1
x/2 + 2/3 = 1
3x + 4 = 6
3x = 2
Or, x = 2/3
问题10.将两枚硬币同时投掷360次。 “ 2尾巴”出现的次数是“无尾巴”出现的次数的三倍,“ 1尾巴”出现的次数是“无尾巴”出现的次数的两倍。找到获得“两条尾巴”的可能性。
解决方案:
Total number of outcomes = 360
Let us considered the number of times ‘No Tail’ appeared be z
Then, number of times ‘2 Tails’ appeared = 3z
Number of times ‘1 Tail’ appeared = 2z
Now, z + 2z + 3z = 360
6z = 360
z = 60
Hence, the probability of getting ‘two tails’ = (3 x 60)/360 = 1 /2