问题1.在板球比赛中，女球手在她打的30球中击中6次边界。找出她没有达到界限的可能性。
解决方案：
Data given in the question:
Total number of balls batswoman plays = 30
Numbers of boundary hit by batswoman = 6
To find number of time batswoman didn’t hit boundary , we will substract
⇒(Total number of balls batswoman plays) – (Numbers of boundary hit by batswoman)
⇒ 30 – 6 = 24
Probability of that she did’t hit a boundary = 24/30 = 4/5
问题2：随机选择1500个有2个孩子的家庭，并记录以下数据：
| Number of girls in a family | 2 | 1 | 0 |
| Number of families | 475 | 814 | 211 |
计算一个随机选择的家庭拥有的概率
(i)2个女孩(ii)1个女孩(iii)没有女孩
解决方案：
According to question
Total numbers of families given in the question 1500
(i) Numbers of families having 2 girls = 475
Probability of choosen 2 girls = Numbers of families having 2 girls / Total numbers of families
= 475/1500 = 19/60
Probability of choosen 2 girls is 19/60
(ii) Numbers of families having 1 girls = 814
Probability of choosen 1 girl = Numbers of families having 1 girl / Total numbers of families
= 814/1500 = 407/750
Probability of choosen 1 girl is 407/750
(iii) Numbers of families having 2 girls = 211
Probability of choosen 0 girl = Numbers of families having 0 girls/Total numbers of families
= 211/1500
Sum of the probability = (19/60)+(407/750)+(211/1500)
= (475+814+211)/1500
= 1500/1500 = 1
Yes, the sum of these probabilities is 1.
问题3。请参阅示例5，第14章的14.4节。找出班上一名学生8月出生的概率。
解决方案：
According to questions:
Total number of students in the class in the given question = 40
Numbers of students born in August = 6
The probability that a student of the class was born in August = (Total numbers of students in the class) /
(Numbers of students born in August)
= 6/40 = 3/20
问题4.将三枚硬币同时投掷200次，其结果发生频率如下：
| Outcome | 3heads | 2heads | 1 head | No heads |
| Frequency | 23 | 72 | 77 | 28 |
如果同时抛掷这三个硬币，请计算出现2个正面的概率。
解决方案：
Number of times 2 heads come up (in the given question) = 72
Total number of times the coins were tossed = 200
The probability of 2 heads coming up = (Number of times 2 heads come up) / (Total number of times the coins were tossed)
= 72/200 = 9/25
问题5.一个组织随机选择了2400个家庭，并对其进行了调查，以确定收入水平与家庭中车辆数量之间的关系。
下表中列出了收集的信息：
|
Monthly income (in ₹) |
0 | 1 | 2 | Above 2 |
| Less than 7000 | 10 | 160 | 25 | 0 |
| 7000-10000 | 0 | 305 | 27 | 2 |
| 7000-10000 | 1 | 535 | 29 | 1 |
| 13000-16000 | 2 | 469 | 59 | 25 |
| 16000 or more | 1 | 579 | 82 | 88 |
假设选择了一个家庭。找到所选择的家庭是
(i)每月赚取10000到13000卢比，恰好拥有2辆车。
(ii)每月赚取₹16000或更多，并且正好拥有1辆车。
(iii)每月收入少于₹7000，并且不拥有任何车辆。
(iv)每月赚取13000 – 16000卢比，拥有2辆以上的汽车。
(v)拥有不超过1辆车。
解决方案：
Total number of families = 2400 (According to question)
(i) Numbers of families earning ₹10000 –13000 per month and owning exactly 2 vehicles = 29
The probability that the family chosen is earning ₹10000 – 13000 per month and owning exactly 2 vehicles =
(Numbers of families earning ₹10000 –13000 per month and owning exactly 2 vehicles) / (Total number of families)
= 29/2400
(ii) Number of families earning ₹16000 or more per month and owning exactly 1 vehicle = 579
The probability that the family chosen is earning ₹16000 or more per month and owning exactly 1 vehicle =
(Number of families earning ₹16000 or more per month and owning exactly 1 vehicle) / (Total number of families)
=579/2400
(iii) Number of families earning less than ₹7000 per month and does not own any vehicle = 10
The probability that the family chosen is earning less than ₹7000 per month and does not own any vehicle =
(Number of families earning less than ₹7000 per month and does not own any vehicle)/(Total number of families)
= 10/2400 = 1/240
(iv) Number of families earning ₹13000-16000 per month and owning more than 2 vehicles = 25
The probability that the family chosen is earning ₹13000 – 16000 per month and owning more than 2 vehicles =
(Number of families earning ₹13000-16000 per month and owning more than 2 vehicles ) / (Total number of families)
= 25/2400 = 1/96
(v) Number of families owning not more than 1 vehicle = 10+160+0+305+1+535+2+469+1+579 = 2062
The probability that the family chosen owns not more than 1 vehicle = 2062/2400 = 1031/1200
问题6.请参阅表14.4.7，第14章。
(i)找出学生在数学测验中获得少于20％的概率。
(ii)找出学生获得60分或以上的概率。
解决方案：
Total number of students = 90
(given in question)
(i) Number of students who obtained less than 20% in the mathematics test = 7
The probability that a student obtained less than 20% in the mathematics test =
( Number of students who obtained less than 20% in the mathematics test)/(Total number of students)
= 7/90
(ii) Number of students who obtained marks 60 or above = 15+8 = 23
The probability that a student obtained marks 60 or above =
(Number of students who obtained marks 60 or above ) / (Total number of students)
= 23/90
问题7.为了了解学生对学科统计的看法，对200名学生进行了调查。
数据记录在下表中
| Opinion | Number of students |
| like | 135 |
| dislike | 65 |
找出学生随机选择的概率
(i)喜欢统计资料，(ii)不喜欢统计资料。
解决方案：
Total number of students = 135+65 = 200 (According to question)
(i) Number of students who like statistics = 135
The probability that a student likes statistics = (Number of students who like statistics) / (Total number of students)
= 135/200 = 27/40
(ii) Number of students who do not like statistics = 65
The probability that a student does not like statistics =
(Number of students who do not like statistics) / (Total number of students)
= 65/200 = 13/40
问题8。请参阅问题2 ，练习14.2。工程师生存的经验概率是多少：
(i)距她的工作地点不到7公里?
(ii)距她的工作地点超过或等于7公里?
(iii)在她工作地点的½公里范围内?
解决方案：
The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
5 3 10 20 25 11 13 7 12 31 19 10 12 17 18 11 3 2
17 16 2 7 9 7 8 3 5 12 15 18 3 12 14 2 9 6
15 15 7 6 12
Total numbers of engineers = 40
(According to question)
(i) Number of engineers living less than 7 km from their place of work = 9
The probability that an engineer lives less than 7 km from her place of work =
(Number of engineers living less than 7 km from their place of work) / (Total numbers of engineers )
= 9/40
(ii) Number of engineers living more than or equal to 7 km from their place of work = 40-9 = 31
Probability that an engineer lives more than or equal to 7 km from her place of work =
(Number of engineers living more than or equal to 7 km from their place of work ) / (Total numbers of engineers)
= 31/40
(iii) Number of engineers living within ½ km from their place of work = 0
The probability that an engineer lives within ½ km from her place of work =
(Number of engineers living within ½ km from their place of work) / (Total numbers of engineers)
=0/40 = 0
问题9：活动：注意两轮车，三轮车和四轮车在一定时间间隔内越过学校大门的频率。找出您观察到的所有车辆中有一辆是两轮车的概率。
解决方案：
The question is an activity to be performed by the students.
问题10：活动：要求班上所有学生写一个3位数的数字。从房间随机选择任何学生。她/他的数字被3整除的概率是多少?请记住，如果数字的总和可被3整除，则该数字可被3整除。
解决方案：
The question is an activity to be performed by the students.
问题11. 11袋小麦粉，每个标记为5公斤，实际上包含以下重量的面粉(以kg为单位)：
4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00
找出随机选择的这些袋子中的任何一个包含超过5千克面粉的可能性。
解决方案：
Data given in the question
Total number of bags present = 11
Number of bags containing more than 5 kg of flour = 7
The probability that any of the bags chosen at random contains more than 5 kg of flour =
(Number of bags containing more than 5 kg of flour) / (Total number of bags present)
= 7/11
问题12。在第5题练习14.2中，要求您准备一个频率分布表，该表关于某城市30天中空气中二氧化硫的浓度(百万分之几)。使用此表，在这些天中的任何一天中，找到介于0.12-0.16之间的二氧化硫浓度的概率。
30天获得的数据如下：
0.03 0.08 0.08 0.09 0.04 0.17 0.16 0.05 0.02 0.06 0.18 0.20 0.11 0.08 0.12 0.13 0.22 0.07 0.08 0.01 0.10 0.06 0.09 0.18 0.11 0.07 0.05 0.07 0.01 0.04
解决方案：
Total number of days in which the data was recorded = 30 days (According to the question)
Numbers of days in which sulphur dioxide was present in between the interval 0.12-0.16 = 2
The probability of the concentration of sulphur dioxide in the interval 0.12-0.16 on any of these days =
(Numbers of days in which sulphur dioxide was present in between the interval 0.12-0.16) /
(Total number of days in which the data was recorded )
= 2/30 = 1/15
问题13。在第1题练习14.2中，要求您准备一张关于班上30名学生的血型的频率分布表。使用此表确定随机选择的该班学生的血型为AB的概率。
记录了30名VIII级学生的血型：
A，B，O，O，AB，O，A，O，B，A，O，B，A，O，O，A，AB，O，A，A，A，O，O，AB，B，A， O，B，A，B，O.
解决方案：
Total numbers of students = 30 (according to questions)
Number of students having blood group AB = 3
The probability that a student of this class, selected at random, has blood group
AB = (Number of students having blood group AB) / (Total numbers of students)
= 3/30 = 1/10