问题1：一个盒子装有10个红色大理石，20个蓝色大理石和30个绿色大理石。从盒子中抽出5个弹珠，几率是
(i)一切都会是蓝色的吗? (ii)至少有一个是绿色的?
解决方案：
The total number of marbles in the box = 10 + 20 + 30 = 60 marbles
As 5 marbles are drawn from the box at a time, the number of ways drawing them = 60C5
i) all will be blue.
all the marbles would be blue when, out of 20 blue marbles, 5 are drawn.
So, the number of ways of drawing 5 blue marbles = 20C5
Hence, the probability of drawing all the blue marbles = 20C5 ÷ 60C5
ii) at least one will be green.
the number of ways in which the drawn marble is not green = number of ways drawing red marble + number of ways drawing blue marbles
i.e 10C5 + 20C5 = (10+20)C5 = 30C5
the probability that no marble is green = 30C5
So, the probability of at least getting one green marble = 1 – the probability of no green marble = 1 – 30C5
问题2：4张纸牌是从52张纸牌中抽调而来的。获得3颗钻石和1个锹的概率是多少?
解决方案：
the total number of ways in which 4 cards can be drawn from the deck = 52C4
In a deck of cards, there are 13 diamonds and 13 spades
number of ways of drawing 3 diamonds = 13C3
number of ways of drawing 1 spade = 13C1
total number of ways of drawing 3 diamonds and 1 spade = 13C3 x 13C1
Thus, the probability of obtaining 3 diamonds and 1 spade from a deck of cards = (13C3 x 13C1) ÷ 52C4
问题3：一个骰子有两个分别带有数字“ 1”的面，三个带有每个数字“ 2”的面和一个带有数字“ 3”的面。如果模具滚动一次，请确定
(一)P(2)
(ii)P(1或3)
(iii)P(不是3)
解决方案：
The total number of faces in a dice = 6
(i) P(2)
the number of faces with the number ‘2’ = 3
so, the probability of getting 2 = P(2) = 3/6 = 1/2
(ii) P(1 or 3)
the number of faces with number ‘1’ = 2
the number of faces with number ‘3’= 1
total number of faces with either number ‘1’ or number ‘3’ = 1 + 2 = 3
so, the probability of getting either number ‘1’ or ‘3’ = P(1 or 3) = 3/6 = 1/2
Alternative Solution of (ii)
P (1 or 3 ) = P(not 2 ) = 1 – P(2) = 1-1/2 = 1/2
(iii) P(not 3 )
the number of faces with the number ‘3’ = 1
probability of getting 3 = P(3)= 1/6
so, the probability of not getting 3 = P(not 3) = 1 – P(3)
= 1-1/6 = 5/6
问题4：在某彩票中，售出10,000张彩票，并颁发了十等奖。如果您购买不中奖的概率是多少
(a)一张票
(b)两张票
(c)10张票
解决方案：
the total number of lottery tickets sold = 10,000
number of prizes to be awarded = 10
(a) one ticket
probability of getting a prize = 10/10000 = 1/1000
probability of not getting a prize when one ticket is bought =
P(not getting a prize ) = 1 – P(getting a prize)
P(not getting a prize) = 1 – 1/1000
= 999/1000
(b) two tickets
If we buy two tickets, then
the number of tickets not awarded = 10000-10
= 9990
number of ways in which we can buy two tickets that are not awarded = 9990C2
number of ways in which we can buy any two tickets = 10000C2
Hence, the probability of buying two non-prize tickets = (9990C2) ÷ (10000C2)
(c) 10 tickets
If we buy two tickets, then
the number of tickets not awarded = 10000-10
= 9990
number of ways in which we can buy ten tickets that are not awarded= 9990C10
number of ways in which we can buy any ten tickets = 10000C10
Hence, the probability of buying ten non-prize tickets = (9990C10) ÷ (10000C10)
问题5：每100名学生中有40人和60人组成两个部分。如果您和您的朋友在这100名学生中，那么几率是
(a)你们都输入相同的部分?
(b)你们都输入了不同的部分?
解决方案：
Since, both I and my friend are among the 100 students,
so, the number of ways in which we can select 2 students among 100 students = 100C2
(a) you both enter the same section.
to be in the same section, we can either be in 40 students section or in 60 students section.
the total number of ways in which we both of us are in same section = 40C2 + 60C2
hence, the probability of both of us are in same section =
= ( 40C2 + 60C2 ) ÷ (100C2)
= (( 39 x 40 ) + (59 x 60) ) ÷ (99 x 100)
= 17/33
(b) you both enter the different sections.
probability of entering the different sections = 1 – P(getting in the same section)
P(entering in different sections) = 1 – 17/33 = 16/33
问题6：给三个人指定三个字母，并给每个人写一个信封，然后将这些字母随机插入信封中，以便每个信封中只包含一个字母。找出至少一个字母在其适当信封中的概率。
解决方案 ：
Let the letters be denoted by L1, L2, L3.
And the envelopes be denoted by E1,E2,E3.
The letters are inserted into the envelopes and each envelope contains exactly one letter. The combinations that can be made are:
E1L1 E2L2 E3L3
E1L2 E2L1 E3L3
E1L2 E2L3 E3L1
E1L1 E2L3 E3L2
E1L3 E2L1 E3L2
E1L3 E2L2 E3L1
So, the total number of combinations that can be made is 6.
the combinations in which at least one letter is in its proper envelope: E1L1 E2L2 E3L3, E1L3 E2L2 E3L1, E1L1 E2L3 E3L2, E1L2 E2L1 E3L3.
The number of such combinations is 4.
Hence, the probability of getting at least one letter in its proper envelope = 4/6 = 2/3
问题7：A和B是两个事件，因此P(A)= 0.54，P(B)= 0.69和P(A B)= 0.35。找 ：
(i)P(A∪B)
(ii)P(A´∩B´)
(iii)P(A∩B´)
(iv)P(B∩A´ )
解决方案：
It is given that, P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35.
(i) P(A ∪ B)
we know that, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
=> P(A ∪ B) = 0.54 + 0.69 – 0.35 = 0.88
(ii) P(A´ ∩ B´)
we know that, by De Morgan’s Law, (A´ ∩ B´) = (A ∪ B)´
=> P(A´ ∩ B´) = P(A ∪ B)´
= P(A´ ∩ B´) = 1 – P(A ∪ B)
= P(A´ ∩ B´) = 1- 0.88 = 0.12
(iii) P(A ∩ B´)
we know that, P(A ∩ B´) = P(A) – P(A ∩ B)
=> P(A ∩ B´) = 0.54 – 0.35 = 0.19
(iv) P(B ∩ A´)
we know that, n(B ∩ A´) = n(B) – n(A ∩ B)
(n(B ∩ A´) ÷ n(S) ) = (n(B) ÷ n(S)) – (n(A ∩ B) ÷ n(S))
=> P(B ∩ A´) = P(B) – P(A ∩ B)
=> P(B ∩ A´) = 0.69 – 0.35 = 0.34
问题8.从公司的雇员中选出5人代表他们参加公司的管理委员会。五个人的详细情况如下：
| S.No | Name | Sex | Age in years |
| 1. | Harish | M | 30 |
| 2. | Rohan | M | 33 |
| 3. | Sheetal | F | 46 |
| 4. | Alis | F | 28 |
| 5. | Salim | M | 41 |
从该组中随机选择一个人作为发言人。发言人是男性还是35岁以上的概率是多少?
解决方案：
Total number of employees in the company = 5
Number of males in employees = 3
number of persons above 35 years of age = 2
number of persons who are male and above 35 years of age = 1
the probability of selecting a male = 3/5 = P(E)
the probability of selecting a person above 35 years of age = 2/5 = P(F)
P(E∩F) = 1/5
So, the probability of selecting the spokesperson who will be either a male or over 35 years = P(E∪F)
=> P(E∪F) P(E) + P(F) – P(E∩F)
=> P(E∪F) = 3/5 + 2/5 – 1/4 = 4/5
问题9.如果从数字0、1、3、5和7中随机形成大于5,000的4位数字，那么在以下情况下形成被5整除的数字的概率是多少，
(i)这些数字是重复的吗? (ii)不允许重复数字吗?
解决方案：
(i) the digits are repeated.
Since we have to make 4-digit numbers greater than 5,000, so the left-most digit could either be 5 or 7.
the remaining 3 digits can be filled by any of the digits 0,1,3,5, and 7 as repetition is allowed. So, the total number formed greater than 5,000 are
= 2 x 5 x 5 x 5 = 250
A number is divisible by 5 if its one’s place is either filled by 0 or 5. the numbers that can be formed which are divisible by 5 are
= 2 x 5 x 5 x 2 = 100
So, the probability of forming numbers that are divisible by 5 =
=> 100/250 = 2/5
(ii) the repetition of digits is not allowed.
to make 4-digit numbers greater than 5,000 the left-most place, i.e, thousands place can be filled by either 5 or 7.
the remaining 3 places can be filled by the other 4 digits. So, the total numbers formed greater than 5,000 when repetition is not allowed are
= 2 x 4 x 3 x 2 = 48
A number is divisible by 5 if its one’s place is either filled by 5 or 0.
when the thousand’s place is taken by 5 and one’s place can only be filled with 0 to be divisible by 5. and the rest of the places can be filled by the remaining 3 digits.
the numbers that can be formed starting with 5 and ending at 0 are = 1 x 3 x 2x 1 = 6
and when the thousand’s place is taken by 7 and one’s place can be filled with 0 or 5 to be divisible by 5, the rest of the places can be filled by the remaining 2 digits.
the numbers that can be formed starting with 7 and ending at 5 or 0 are = 1 x 3 x 2 x 2 = 12
the total numbers formed greater than 5,000 and divisible by 5 when repetition of digits are not allowed = 12 + 6 = 18
So, the probability of forming numbers that are divisible by 5 when repetition is not allowed =
=> 18/48 = 3/8
问题10.手提箱的数字锁有4个轮子，每个轮子都标有10位数字，即从0到9。该锁以四位数字的序列打开，没有重复。一个人以正确的顺序打开行李箱的概率是多少?
解决方案：
Here, the number lock has 4 wheels each labeled with ten digits i.e from 0 to 9.
the number of ways of selecting 4 digits out of 10 digits = 10C4
as there are 4 wheels, each combination made can be arranged in 4! ways.
number of 4-digits combinations that can be made when no digit is repeated = 10C4 x 4! = 5040
As there is only one sequence which can open the suitcase.
Thus, the required probability is 1/5040.