问题1.掷骰子。令E为事件“骰子显示4”，而F为事件“骰子显示偶数”。 E和F是否互斥?
解决方案：
Let us consider that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.
Hence, Space = (1, 2, 3, 4, 5, 6)
As per the conditions given in the question
E be the event “die shows 4”
E = (4)
F be the event “die shows even number”
F = (2, 4, 6)
E ∩ F = (4) ∩ (2, 4, 6) = 4
4 ≠ ∅ -(since there is a common element in E and F)
For that reason, we can conclude that E and F are not mutually exclusive event.
问题2.掷骰子。描述以下事件：
( i)答：小于7的数字
(ii)B：大于7的数字
(iii)C：3的倍数
(iv)D：小于4的数字
(v)E：大于4的偶数
(vi)F：一个不少于3的数字
还要找到A∪B，A∩B，B∪C，E∩F，D∩E，A – C，D – E，E∩FI，FI
解决方案：
Let us consider that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.
Hence, Space = (1, 2, 3, 4, 5, 6)
As per the conditions given in the question,
(i) A: a number less than 7:
Here, all the numbers in the die are less than 7,
Hence, A = (1, 2, 3, 4, 5, 6)
(ii) B: a number greater than 7:
There is no number present which is greater than 7 on the die
That’s why, B = (∅)
(iii) C: a multiple of 3:
Two numbers are present which are multiple of 3.
That’s why, C = (3, 6) -(As, 3×1 = 3, 3×2 = 6)
(iv) D: a number less than 4:
There are three numbers which are less than 4.
That’s why, D= (1, 2, 3)
(v) E: an even number greater than 4:
There is only one even number which is greater than 4
That’s why, E = (6)
(vi) F: a number not less than 3:
That means, we have to find the number(s) which is/are greater than 3
There are four numbers which are greater than 3.
That’s why, F = (3, 4, 5, 6)
According to the question, we also have to solve, A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, D – E, A – C, E ∩ F’, F’
Hence,
- A ∩ B = (1, 2, 3, 4, 5, 6) ∩ (∅) = (∅)
- B U C = (∅) ∪ (3, 6) = (3, 6)
- E ∩ F = (6) ∩ (3, 4, 5, 6) = (6)
- D ∩ E = (1, 2, 3) ∩ (6) = (∅)
- D – E = (1, 2, 3) – (6) = (1, 2, 3)
- A – C = (1, 2, 3, 4, 5, 6) – (3, 6) = (1, 2, 4, 5)
- F’ = S – F = (1, 2, 3, 4, 5, 6) – (3, 4, 5, 6) = (1, 2)
- E ∩ F’ = (6) ∩ (1, 2) = (∅)
问题3.实验涉及掷骰子并记录出现的数字。描述以下事件：
答：总和大于8
B：两个骰子上均发生2
C：总和至少为7且为3的倍数。
这些事件中的哪些对互斥?
解决方案：
Let us consider that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.
In the question, it is given that pair of dice is thrown, so sample space will be-
{(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)}
A: the sum is greater than 8:
According to the above sample space the maximum sum will be (6+6) = 12
Hence, the possible sum greater than 8 are: 9, 10, 11, and 12
Where A= {(3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}
B: 2 occurs in either die:
In this case, there are three possibilities:
(i) 2 may come in the first die: B1 = {(2, 1), (2, 3), (2, 4), (2, 5), (2, 6)}
(ii) 2 may come in the second die: B2 = {(1, 2), (3, 2), (4, 2), (5, 2), (6, 2)}
(iii) 2 may come in both the die simultaneously: B3 = {(2, 2)}
Hence, B = {B1 ∪ B2 ∪ B3} = {(2, 1), (2, 3), (2, 4), (2, 5), (2, 6),
(1, 2), (3, 2), (4, 2), (5, 2), (6, 2), (2, 2)}
C: The sum is at least 7 and multiple of 3:
According to this condition the sum can be 9 or 12 [As, 9 = (3×3) & 12 = (3×4)
both are multiple of 3 and greater than 7]
Hence, C= {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}
Now, we have to find pairs of these events are mutually exclusive or not.
A = {(3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}
B = {(2, 1), (2, 3), (2, 4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2), (2, 2)}
C = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}
(i) A∩ B = ∅
We find that, there is no common element in A and B
Therefore, A & B are mutually exclusive
(ii) B ∩ C = ∅
We find that, there is no common element between B and C
Therefore, B and C are mutually exclusive.
(iii) A ∩ C = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}
⇒ {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)} ≠ ∅
We find that, A and C has common elements.
Therefore, A and C are mutually exclusive.
问题4.将三枚硬币扔一次。设A表示事件“三个头显示”，B表示事件“两个头一个尾显示”，C表示事件“三个头显示尾”，D表示事件“第一个硬币上显示一个头”。哪些事件是
(i)互斥?
(ii)简单吗?
(iii)化合物?
解决方案：
When a coin is thrown, the possible outcomes are either Head(H) or Tail(T).
Now according to the question, three coins are tossed once so the possible sample space contains,
Space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTH}
Now,
A: ‘three heads’
A = (HHH)
B: “two heads and one tail”
B = (HHT, THH, HTH)
C: ‘three tails’
C = (TTT)
D: a head shows on the primary coin
D = (HHH, HHT, HTH, HTT)
(i) Mutually exclusive:
A ∩ B = (HHH) ∩ (HHT, THH, HTH) = ∅
Therefore, A and B are mutually exclusive.
A ∩ C = (HHH) ∩ (TTT) = ∅
Therefore, A and C are mutually exclusive.
A ∩ D = (HHH) ∩ (HHH, HHT, HTH, HTT) = (HHH)
A ∩ D ≠ ∅
So they aren’t mutually exclusive
B ∩ C = (HHT, HTH, THH) ∩ (TTT) = ∅
Since there’s no common element in B & C, in order that they are mutually exclusive.
B ∩ D = (HHT, THH, HTH) ∩ (HHH, HHT, HTH, HTT) = (HHT, HTH)
B ∩ D ≠ ∅
We find that, there are common elements in B & D,
So, they not mutually exclusive.
C ∩ D = (TTT) ∩ (HHH, HHT, HTH, HTT) = ∅
Since there’s no common element in C & D,
So they are mutually exclusive.
(ii) Simple event
If an occasion has just one sample point of a sample space,
it’s called an easy (or elementary) event.
A = (HHH)
C = (TTT)
Both A & C have just one element,
So, they are simple events.
(iii) Compound events
If an occasion has quite one sample point, it’s called a Compound event
B = (HHT, HTH, THH)
D = (HHH, HHT, HTH, HTT)
Both B & D have quite one element,
So, they’re compound events.
问题5.投掷三枚硬币。描述
(i)两个相互排斥的事件。
(ii)三个相互排斥和详尽的事件。
(iii)两个事件并不互斥。
(iv)两个互斥但并非穷举的事件。
(v)三个相互排斥但并不详尽的事件。
解决方案：
When a coin is thrown, the possible outcomes are either Head(H) or Tail(T).
Now according to the question, three coins are tossed once so the possible sample space contains,
Space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTH}
(i) Two events which are mutually exclusive.
Let us consider A be the event of getting only head
A = (HHH)
And also let us consider B be the event of getting only Tail
B = (TTT)
So, A ∩ B =∅
We find that, there is no common element in A & B so these two are mutually exclusive.
(ii) We find that, three events which are mutually exclusive and exhaustive
Now,
Let us consider P be the event of getting exactly two tails
P = (HTT, TTH, THT)
Let us consider Q be the event of getting at least two heads
Q = (HHT, HTH, THH, HHH)
Let us consider R be the event of getting only one tail
C= (TTT)
P ∩ Q = (HTT, TTH, THT) ∩ (HHT, HTH, THH, HHH) = ∅
We find that, there is no common element in P and Q,
Therefore, they are mutually exclusive
Q ∩ R = (HHT, HTH, THH, HHH) ∩ (TTT) = ∅
We find that, there is no common element in Q and R
Hence, they are mutually exclusive.
P ∩ R = (HTT, TTH, THT) ∩ (TTT)= ∅
We find that, there is no common element in P and R,
So they are mutually exclusive.
Hereby, P and Q, Q and R, and P and R are mutually exclusive
∴ P, Q, and R are mutually exclusive.
And also,
P ∪ Q ∪ R = (HTT, TTH, THT, HHT, HTH, THH, HHH, TTT) = Space
Hence, P, Q and R are exhaustive events.
(iii) Two events, which are not mutually exclusive
Let us consider ‘A’ be the event of getting at least two heads
A = (HHH, HHT, THH, HTH)
Let us consider ‘B’ be the event of getting only head
B = (HHH)
Now A ∩ B = (HHH, HHT, THH, HTH) ∩ (HHH) = (HHH)
A ∩ B ≠ ∅
We find that, there is a common element in A and B,
So, they are not mutually exclusive.
(iv) Here, two events which are mutually exclusive but not exhaustive
Let us consider ‘P’ be the event of getting only Head
P = (HHH)
Let us consider ‘Q’ be the event of getting only tail
Q = (TTT)
P ∩ Q = (HHH) ∩ (TTT) = ∅
We find that, there is no common element in P and Q,
Hence, these are mutually exclusive events.
But,
P ∪ Q = (HHH) ∪ (TTT)
= {HHH, TTT}
P ∪ Q ≠ Space
We find that, P ∪ Q ≠ Space, these are not exhaustive events.
(v) We find that, three events which are mutually exclusive but not exhaustive
Let us consider ‘X’ be the event of getting only head
X = (HHH)
Let us consider ‘Y’ be the event of getting only tail
Y = (TTT)
Let us consider ‘Z’ be the event of getting exactly two heads
Z = (HHT, THH, HTH)
Now,
X ∩ Y = (HHH) ∩ (TTT) = ∅
X ∩ Z = (HHH) ∩ (HHT, THH, HTH) = ∅
Y ∩ Z = (TTT) ∩ (HHT, THH, HTH) = ∅
Therefore, they are mutually exclusive
Also
X ∪ Y ∪ Z = (HHH TTT, HHT, THH, HTH)
X ∪ Y ∪ Z ≠ Space
So, X, Y and Z are not exhaustive.
Thus, it is proved that X, Y and X are mutually exclusive but not exhaustive.
问题6.掷两个骰子。事件A，B和C如下：
答：第一次死亡时得到偶数。
B：第一次死亡时得到一个奇数。
C：获得≤5的骰子上数字的总和。
描述事件
(i)A’
(ii)不是B
(iii)A或B
(iv)A和B
(v)A但不C
(vi)B或C
(vii)B和C
(viii)A∩B’∩C’
解决方案：
Let us consider that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.
In the question, it is given that pair of dice is thrown, so sample space will be-
Space = {(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)}
There are several conditions in the question and according to those conditions,
we have to solve proper solutions.
A: getting an even number on the first die:
So, we have to make the sample space in which the first die is even-
Hence, A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3),
(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
B: getting an odd number on the first die:
So, we have to make the sample space in which the first die is odd-
Hence, B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3),
(3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5),(5, 6)}
C: getting the sum of the numbers on the dice ≤ 5:
According to the above sample space the minimum sum will be (1+1)=2
So, the sum can be 2, 3, 4 & 5
Hence, C = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
(i) A’ = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3),
(3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} = B
(ii) B’ = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3),
(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} =A
(iii) A or B = A ∪ B = {(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)} = Space
(iv) A and B = A ∩ B = ∅
(v) A but not C = A – C = {(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5),
(4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
(vi) B or C =B ∪ C = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3),
(3, 4), (3, 5), (3, 6), (4, 1), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
(vii) B and C = B ∩ C = {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}
(viii) A ∩ B’ ∩ C’ = A ∩ A ∩ C’ = A ∩ C’
A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3),
(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
C = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
C’ = {(1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (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)}
A ∩ B’ ∩ C’ = A ∩ C’ = {(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5),
(4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
问题7。请参阅上面的问题6，陈述是或否：(给出回答的原因)
(i)A和B是互斥的
(ii)A和B是互斥且详尽的
(iii)A = B’
(iv)A和C是互斥的
(v)A和BI是互斥的。
(vi)A’，B’，C是互斥和穷举的。
解决方案：
Let us consider that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.
In the question, it is given that pair of dice is thrown, so sample space will be-
Space = {(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)}
There are several conditions in the question and according to those
conditions, we have to solve proper solutions.
A: getting an even number on the first die:
So, we have to make the sample space in which the first die is even-
Hence, A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3),
(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
B: getting an odd number on the first die:
So, we have to make the sample space in which the first die is odd-
Hence, B= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3),
(3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
C: getting the sum of the numbers on the dice ≤ 5:
According to the above sample space the minimum sum will be (1+1) = 2
So, the sum can be 2, 3, 4 & 5
Hence, C = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
(i) A and B are mutually exclusive:
A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3),
(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3),
(3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
There are no common elements between A & B
That’s why, (A ∩ B) = ∅
So, A & B are mutually exclusive.
Hence, the given statement is true.
(ii) A and B are mutually exclusive and exhaustive:
A ∪ B = {(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)} = Space
⇒ A ∪ B =Space
Hence, A and B are mutually exhaustive.
We already know from (i) that A and B are mutually exclusive.
Thus, A and B are mutually exclusive and exhaustive.
Hence, the given statement is true.
(iii) A = B’:
B’ = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3),
(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} = A
Hence, the given statement is true.
(iv) A and C are mutually exclusive:
A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3),
(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
C = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
A ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)}
As, A ∩ C ≠ ∅ , A and C are not mutually exclusive.
Hence, the given statement is false.
(v) A and B’ are mutually exclusive:
B’ = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3),
(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} = A
A ∩ B’ = A ∩ A = ∅
That’s why, A and B’ are not mutually exclusive.
Hence, the given statement is false.
(vi) A’, B’, C are mutually exclusive and exhaustive:
A’ = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2),
(3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
B’ = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3),
(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
C = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
A’ ∩ B’ = ∅
Hence, there is no common element between A’ and B’
So A’ and B’ are mutually exclusive.
B’ ∩ C ={{(2, 1), (2, 2), (2, 3), (4, 1)}
As, B’ ∩ C ≠ ∅, B and C are not mutually exclusive.
Thus, it is proved that A’, B’ and C are not mutually exclusive and exhaustive.
Hence, the given statement is false.