问题1.以下哪项不能是对以下结果的概率的有效分配：
样本空间S = {ω1，ω2，ω3，ω4，ω5，ω6，ω7}
| Assignment: | ω1 | ω2 | ω3 | ω4 | ω5 | ω6 | ω7 |
| (a) | 0.1 | 0.01 | 0.05 | 0.03 | 0.01 | 0.2 | 0.6 |
| (b) | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 |
| (c) | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.7 |
| (d) | -0.1 | 0.2 | 0.3 | 0.4 | -0.2 | 0.1 | 0.3 |
| (e) | 1/14 | 2/14 | 3/14 | 4/14 | 5/14 | 6/14 | 15/14 |
解决方案：
(a) To check whether a given assignment is valid or not we should check 2 conditions:
i) Given numbers should be positive.
ii) sum of probabilities is 1.
0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1
Therefore, the given assignment is valid.
(b) To check whether a given assignment is valid or not we should check 2 conditions:
i) Given numbers should be positive.
ii) sum of probabilities is 1.
= (1/7) + (1/7) + (1/7) + (1/7) + (1/7) + (1/7) + (1/7)
= 7/7
= 1
Therefore, the given assignment is valid.
(c) To check whether a given assignment is valid or not we should check 2 conditions:
i) Given numbers should be positive.
ii) sum of probabilities is 1.
= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7
= 2.8 > 1
Therefore, the 2nd condition is not satisfied
So, the given assignment is not valid.
(d) Since, the given numbers are negative, which doesn’t satisfy the 1st condition.
So, the assignment is not valid.
(e) To check whether a given assignment is valid or not we should check 2 conditions:
i) Given numbers should be positive.
ii) sum of probabilities is 1.
= (1/14) + (2/14) + (3/14) + (4/14) + (5/14) + (6/14) + (7/14)
= (28/14) ≥ 1
The second condition doesn’t hold true so the assignment is not valid.
问题2.抛硬币两次，至少出现一条尾巴的概率是多少?
解决方案：
The possible outcomes are Head(H) and Tail(T).
Here coin is tossed twice, then sample space is S = (TT, HH, TH, HT), n(S) = 4.
Let A be the event of getting at least one tail
n (A) = 3
P(Event) = Number of outcomes favorable to event/ Total number of possible outcomes
P(A) = n(A)/n(S)
= 3/4.
问题3.掷骰子，找出发生以下事件的概率：
(i)质数会出现，
(ii)出现大于或等于3的数字，
(iii)出现一个小于或等于1的数字，
(iv)将会出现6个以上的数字，
(v)小于6的数字将出现。
解决方案：
Here, S = {1, 2, 3, 4, 5, 6}
n(S) = 6
(i) A = {2, 3, 5}; n(A) = 3
P(A) = n(A)/n(S)
= 3/6
(ii) A = {3, 4, 5, 6}; n(A) = 4
P(A) = n(A)/n(S)
= 4/6
(iii) A = {1}; n (A) = 1
P(A) = n(A)/n(S)
= 1/6
(iv) A = {0}; n (A) = 0
P(A) = n(A)/n(S)
= 0/6 = 0
(v) A= {1, 2, 3, 4, 5}; n (A) = 5
P(A) = n(A)/n(S)
= 5/6
问题4.从52张纸牌中选择一张纸牌。
(a)样本空间中有多少个点?
(b)计算该牌是黑桃A的概率。
(c)计算该牌是(i)ace牌(ii)黑牌的概率
解决方案：
(a) Number of points in the sample space = 52
n(S) = 52
(b) Let us assume ‘A’ be the event of drawing an ace of spades.
A = 1
Then, n (A) = 1
P(A) = n(A)/n(S)
= 1/52
(c) Let us assume ‘A’ be the event of drawing an ace. There are four aces.
Then, n (A) = 4
P(A) = n(A)/n(S)
= 4/52
= 1/13
(d) Let us assume ‘A’ be the event of drawing a black card. There are 26 black cards.
Then, n (A) = 26
P(A) = n(A)/n(S)
= 26/52
= 1/2
问题5.扔了一个在一个面上刻有1个，在另一个面上刻有6个，并且有一个公平的骰子的公平硬币。求出出现的数字总和是(i)3(ii)12的概率
解决方案：
1,2,3,4,5,6 are the possible outcomes
sample space S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 12
(i) A = {(1, 2)}; n (A) = 1
P(A) = n(A)/n(S)
= 1/12
(ii) A = {(6, 6)}; n (A) = 1
P(A) = n(A)/n(S)
= 1/12
问题6.市议会有四男六女。如果选择一名理事会成员
对于一个随机的委员会来说，这是一个女人的可能性有多大?
解决方案：
Total members in the council = 4 + 6 = 10; n (S) = 10
Number of women are 6
n (A) = 6
P(A) = n(A)/n(S)
= 6/10
问题7.将一枚公平的硬币扔四次，一个人每头赢得1卢比，而每出现一条尾巴则损失1.50卢比。从样本空间中，计算出四次抛硬币后您可以拥有多少不同数量的货币，以及每一个拥有这些货币的概率。
解决方案：
Here, Head(H) and Tail(T) are the possible outcomes.
S = (HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT,
HTTH, THTH, TTHH, TTTH, TTHT, THTT, HTTT, TTTT)
(i) For 4 heads = 1 + 1 + 1 + 1 = Rs 4
So, he wins Rs 4
(ii) For 3 heads and 1 tail = 1 + 1 + 1 – 1.50
= 3 – 1.50
= Rs 1.50
So, he will be winning Rs 1.50
(iii) For 2 heads and 2 tails = 1 + 1 – 1.50 – 1.50
= 2 – 3
= – Rs 1
So, he will be losing Rs 1
(iv) For 1 head and 3 tails = 1 – 1.50 – 1.50 – 1.50
= 1 – 4.50
= – Rs 3.50
So, he will be losing Rs. 3.50
(v) For 4 tails = – 1.50 – 1.50 – 1.50 – 1.50
= – Rs 6
So, he will be losing Rs. 6
Now the sample space is
S = {4, 1.50, 1.50, 1.50, 1.50, – 1, – 1, – 1,
– 1, – 1, – 1, – 3.50, – 3.50, – 3.50, – 3.50, – 6}
Then, n (S) = 16
P (winning Rs 4) = 1/16
P (winning Rs 1.50) = 4/16
= 1/4
P (winning Rs 1) = 6/16
= 3/8
P (winning Rs 3.50) = 4/16
= 1/4
P (winning Rs 6) = 1/16
= 3/8
问题8.将三枚硬币扔一次。找到获得的可能性
(i)3头(ii)2头(iii)至少2头
(iv)最多2个头(v)没有头(vi)3个头
(vii)正好有两条尾巴(viii)最多没有两条尾巴(ix)
解决方案：
Here, Head(H) and Tail(T) are the possible outcomes.
S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}; n(S) = 8
(i) Possibility of getting 3 heads is 1; n(A) = 1
P(A) = n(A)/n(S)
= 1/8
(ii) Possibility of getting 2 heads is 3; n(A) = 3
P(A) = n(A)/n(S)
= 3/8
(iii) Possibility of getting at least 2 heads is 4; n(A) = 4
P(A) = n(A)/n(S)
= 4/8
(iv) Possibility of getting at most 2 heads is 7; n(A) = 7
P(A) = n(A)/n(S)
= 7/8
(v) Possibility of getting no head is 1; n(A) = 1
P(A) = n(A)/n(S)
= 1/8
(vi) Possibility of getting 3 tails is 1; n(A) = 1
P(A) = n(A)/n(S)
= 1/8
(vii) Possibility of getting 2 tails is 3; n(A) = 3
P(A) = n(A)/n(S)
= 3/8
(viii) Possibility of getting no tail is 1; n(A) = 1
P(A) = n(A)/n(S)
= 1/8
(ix) Possibility of getting at most 2 tails is 7; n(A) = 7
P(A) = n(A)/n(S)
= 7/8
问题9.如果2/11是事件的概率，那么事件“非A”的概率是多少。
解决方案：
2/11 is the probability of an event A
P (A) = 2/11
P (not A) = 1 – P (A)
= 1 – (2/11)
= (11 – 2)/11
= 9/11
问题10.从“暗杀”一词中随机选择一个字母。找出字母是的概率
(i)元音(ii)辅音
解决方案：
Total letters in the given word = 13
Number of vowels in the given word = 6
Number of consonants in the given word = 7
Then, the sample space n(S) = 13
(i) a vowel
sample space n(S) = 6
P(A) = n(A)/n(S)
= 6/13
(ii) a consonant
n(A) = 7
P(A) = n(A)/n(S)
= 7/13
问题11：在彩票中，一个人从1到20中随机选择六个自然数，如果这六个数与彩票委员会已经确定的六个数相匹配，则中奖。在游戏中赢得奖金的概率是多少? [数字的提示顺序并不重要。]
解决方案：
Total numbers of numbers in the draw = 20
Numbers to be selected = 6
So, n(S) = 20C6
Now, let us assume the X be the event that chosen six numbers to
match with the six numbers already fixed by the lottery committee
n(A) = 6C6 = 1
So, the probability of winning the prize is
P(A) = n(A)/n(S) = 6C6/20C6
= 6×5×4×3×2×1×14! / 20×19×18×17×16×15×14!
= 1/38760
问题12.检查是否一致定义了以下概率P(A)和P(B)
(i)P(A)= 0.5，P(B)= 0.7，P(A∩B)= 0.6
(ii)P(A)= 0.5，P(B)= 0.4，P(A∪B)= 0.8
解决方案：
(i) P(A ∩ B) > P(A)
Here, the given probabilities are not consistently defined.
(ii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.8 = 0.5 + 0.4 – P(A ∩ B)
P(A ∩ B) = 0.9 – 0.8
= 0.1
Therefore, P(A ∩ B) < P(A) and P(A ∩ B) < P(B)
So, the given probabilities are consistently defined.
问题13.填写下表中的空白：
| P(A) | P(B) | P(A∩B) | P(A∪B) | |
| (i) | 1/3 | 1/5 | 1/15 | ……. |
| (ii) | 0.35 | ……. | 0.25 | 0.6 |
| (iii) | 0.5 | 0.35 | ……. | 0.7 |
解决方案：
(i) P(A) = 1/3, P(B) = 1/5, P(A ∩ B) = 1/15, P(A ∪ B) = ?
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= (1/3) + (1/5) – (1/15)
= ((5 + 3)/15) – (1/15)
= 7/15
(ii) P(A) = 0.35, P(B) = ?, P(A ∩ B) = 0.25, P(A ∪ B) = 0.6
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.6 = 0.35 + P(B) – 0.25
P(B) = 0.6 + 0.25 – 0.35
= 0.5
(iii) P(A) = 0.5, P(B) = 0.35, P(A ∪ B) = 0.7, P(A ∩ B) = ?
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.7 = 0.5 + 0.35 – P(A ∩ B)
P(A ∩ B) = 0.85 – 0.7
= 0.15
问题14：给定P(A)= 5/3，P(B)= 1/5。如果A和B是互斥事件，则找到P(A或B)。
解决方案：
P(A) = 5/3 and P(B) = 1/5
P(A ∪ B) or P(A or B) = P(A) + P(B)
= (3/5) + (1/5)
= 4/5
问题15：如果E和F是P(E)=¼，P(F)= 1/2和P(E和F)= 1/8的事件，则求
(i)P(E或F)，(ii)P(不是E也不是F)
解决方案：
(i) P(E∪F) = P(E) + P(F) – P(E ∩ F)
= 1/4 + 1/2 – (1/8)
= 5/8
(ii) P(E’ ∩ F’) = P((E U F)’) = 1 – P(E U F )
= 1 – (5/8)
= (8 – 5)/8
= 3/8
问题16：事件E和F使得P(不是E还是F)= 0.25，说明E和F是否互斥。
解决方案：
P(E’ U F’) = 0.25
P((E ∩ F)’) = 0.25
1 – P(E ∩ F) = 0.25
P(E ∩ F) = 0.75
P(E ∩ F) is not equal to 0
So, E and F are not mutually exclusive events.
问题17. A和B是这样的事件，P(A)= 0.42，P(B)= 0.48，P(A和B)= 0.16。确定(i)P(不是A)，
(ii)P(非B)和(iii)P(A或B)
解决方案：
(i) P(not A) = 1 – P(A)
= 1 – 0.42
= 0.58
(ii) P(not B) = 1 – P(B)
= 1 – 0.48
= 0.52
(iii) P(A not B) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.42 + 0.48 – 0.16
= 0.74
问题18：在学校的XI班中，有40％的学生学习数学，有30％的学生学习生物学。 10％的学生同时学习数学和生物学。如果从班级中随机选择一名学生，请找出他将学习数学或生物学的可能性。
解决方案：
P(A) = 40/100 = 2/5
P(B) = 30/100 = 3/10
P(A ∩ B) = 10/100 = 1/10
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 2/5 + 3/10 – 1/10
P(A ∪ B) = 3/5
问题19.在根据两次考试进行分级的入学考试中，随机选择的学生通过第一次考试的概率为0.8，通过第二次考试的概率为0.7。通过其中至少一项的概率为0.95。两者同时通过的概率是多少?
解决方案：
P(A ∪ B) = 0.95, P(A) = 0.8, P(B) = 0.7
P(A ∪ B) = P(A) + P(B) – P(A∩B)
0.95 = 0.8 + 0.7 – P(A∩B)
P(A ∩ B) = 1.5 – 0.95
= 0.55
问题20：学生通过英语和北印度语的期末考试的机率是0.5，而两者都不通过的机率是0.1。如果通过英语考试的概率为0.75，那么通过印地语考试的概率是多少?
解决方案：
Given that, P(A) = 0.75, P(A ∩ B) =0.5, P(A’ ∩ B’) = 0.1
P(A’ ∩ B’) = 1 – P(A ∪ B)
Then, P(A ∪ B) = 1 – P(A’ ∩ B’)
= 1 – 0.1
= 0.9
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.9 = 0.75 + P(B) – 0.5
P(B) = 0.9 + 0.5 – 0.75
= 0.65
问题21.在60名学生中，有30名选择了NCC，32名选择了NSS，有24名选择了NCC和NSS。如果随机选择这些学生之一，则求出
(i)学生选择了NCC或NSS。
(ii)学生未选择NCC或NSS。
(iii)学生选择了NSS，但未选择NCC。
解决方案：
The total number of students in class = 60
n(S) = 60
Assume NCC be ‘A’ and NSS be ‘B’
n(A) = 30, n(B) = 32 , n(A∩B) = 24
P(A) = n(A)/n(S)
= 30/60
= 1/2
P(B) = n(B)/n(S)
= 32/60
= 8/15
P(A ∩ B) = n(A ∩ B)/n(S)
= 24/60
= 2/5
(i) The student opted for NCC or NSS.
P (A or B) = P(A) + P(B) – P(A and B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= (1/2)+ (8/15) – (2/5)
= 19/30
(ii) P(student opted neither NCC nor NSS)
P(not A and not B) = P(A’ ∩ B’)
We know that, P(A’ ∩ B’) = 1 – P(A ∪ B)
= 1 – (19/30)
= 11/30
(iii) P(student opted NSS but not NCC)
n(B – A) = n(B) – n (A ∩ B)
32 – 24 = 8
= (8/60) = 2/15