since the experiment has exactly the three possible mutually exclusive outcomes A, B and C, they must be exhaustive events.
\(\Rightarrow S=A\cup B\cup C\)
Therefore, by axioms of probability
\(P(A)\ge 0,P(B)\ge P(C)\ge 0\) and
\(P(A\cup B\cup C)=P(A)+P(B)+P(C)=P(S)=1\)
Even though P(A) + P(B) + P(C) = 0.421 + 0.527 + 0.042 = 0.990 < 1
therefore, the assignment is not permissible

since the experiment has exactly the three possible mutually exclusive outcomes A, B and C, they must be exhaustive events.
\(\Rightarrow S=A\cup B\cup C\)
Therefore, by axioms of probability
\(P(A)\ge 0,P(B)\ge P(C)\ge 0\) and
\(P(A\cup B\cup C)=P(A)+P(B)+P(C)=P(S)=1\)
The assignment is permissible because
\(P(A)=\frac { 1 }{ \sqrt { 3 } } \ge 0,\quad P(B)-1-\frac { 1 }{ \sqrt { 3 } } \ge 0,\quad P(C)-0\ge 0\)
\(P(S)=P(A)+P(B)+P(C)=\frac { 1 }{ \sqrt { 3 } } +1-\frac { 1 }{ \sqrt { 3 } } +0-1\)

Since the experiment has exactly the three possible mutually exclusive outcomes A, B and C, they must be exhaustive events.
\(\Rightarrow S=A\cup B\cup C\)
Therefore, by axioms of probability
\(P(A)\ge 0,P(B)\ge P(C)\ge 0\) and
\(P(A\cup B\cup C)=P(A)+P(B)+P(C)=P(S)=1\)
Since 0.2 P (C) = −0.2 is negative, the assignment is not permissible

Since the experiment has exactly the three possible mutually exclusive outcomes A, B and C, they must be exhaustive events.
\(\Rightarrow S=A\cup B\cup C\)
Therefore, by axioms of probability
\(P(A)\ge 0,P(B)\ge P(C)\ge 0\) and
\(P(A\cup B\cup C)=P(A)+P(B)+P(C)=P(S)=1\)
Given that \(P(A)=\frac { 2 }{ 5 } \ge ,\quad P(B)=\frac { 1 }{ 5 } \ge 0,\quad P(C)=\frac { 3 }{ 5 } \ge 0\)
But \(P(S)=P(A)+P(B)+P(C)=\frac { 2 }{ 5 } +\frac { 1 }{ 5 } +\frac { 6 }{ 5 } >1\)
Therefore the assignment is not permissible

Since the experiment has exactly the three possible mutually exclusive outcomes A, B and C, they must be exhaustive events.
\(\Rightarrow S=A\cup B\cup C\)
Therefore, by axioms of probability
\(P(A)\ge 0,P(B)\ge P(C)\ge 0\) and
\(P(A\cup B\cup C)=P(A)+P(B)+P(C)=P(S)=1\)
Given that \(P(A)=\frac { 4 }{ 7 } \ge 0,\quad P(B)=\frac { 1 }{ 7 } \ge 0,\quad P(C)=\frac { 2 }{ 7 } \ge 0\)
Also \(P(S)=P(A)+P(B)+P(C)=\frac { 4 }{ 7 } +\frac { 1 }{ 7 } +\frac { 2 }{ 7 } =1\)
Therefore the assignment of probability is permissible

If A and B are mutually exclusive, then
\(P(A \cap B)=0\)
But if A and B are independent, then
\(P(A \cap B)=P(A) \cdot P(B)\)
So if A and B are non empty, then they are not mutually exclusive and independent simultaneously.

Given \(P(A)=0.35\)
\(P(A \text { or } B)=0.85 \text { and } P(A \text { and } B)=0.15\)
\((i) P(A \cup B)=P(A)+P(B)-P(A \cap B)\)
\(0.85 =0.35+P(B)-0.15 \)
\(P(B) =0.85-0.20\)
\(=0.65\)
\((ii) P(\bar{B})=1-P(B)=1-0.65=0.35\)
\((iii) P( only\ A)=P(A)-P(A \cap B)\)
\(=0.35-0.15=0.20\)

\((i) P(\bar{A})=1-P(A)=1-\frac{3}{8}=\frac{5}{8}\)
\((ii) P(A \cup B)=p(A)+P(B)=\frac{3}{8}+\frac{1}{8}=\frac{4}{8}=\frac{1}{2}\)
\((iii) P(\bar{A} \cap B)=P(B)-P(A \cap B) =P(B)=\frac{1}{8} \quad(\because P(A \cap B)=0)\)
\((iv) P(\bar{A} \cup \bar{B})=1-P(A \cap B)=1\)

The sample space S = {1, 2, 3, 4, 5, 6} \(\times\) {1, 2, 3, 4, 5, 6}
S = {(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)}
Number of possible outcomes = 6² =36 = n(S)
Let A be the event of getting sum 7,B be the event of getting the sum 9 and C be the event of getting sum 12. Then
A = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) ⇒ n(A) = 6
B = {(3,6), (4,5), (5,4), (6,3)}  ⇒ n(B) = 4
C = {(6,6)}  ⇒ n(C) = 1
(i) P (getting sum 7) = P(A)
=\(\frac { n(A) }{ n(S) } =\frac { 6 }{ 36 } =\frac { 1 }{ 6 } \)
(ii) P (getting sum 7 or 9) = P(A or B) = P(A∪B)
= P(A) + P(B)
Since A and B are mutually exclusive that is, A ก B - Ø)
\(\frac { n(A) }{ n(S) } +\frac { n(B) }{ n(S) } =\frac { 6 }{ 36 } +\frac { 4 }{ 36 } =\frac { 5 }{ 18 } \)
(iii) P (getting sum 7 or 12) = P(A or C) = P⋃C)
= P(A) + P(C) Since A and C are mutually exclusive)
\(\frac { n(A) }{ n(S) } +\frac { n(C) }{ n(S) } =\frac { 6 }{ 36 } +\frac { 1 }{ 36 } =\frac { 7 }{ 36 } \)

Given P(A) = 0.5 P(B) = 0.3
A and B are mutually cxclusive events \(P(A \cap B)=0\)
\((i) P(A \cup B)=P(A)+P(B)=0.5+0.3=0.8\)
\((ii) P(A \cap \bar{B})=P(A)-P(A \cap B)=0.5-0=0.5\)
\(\text { (iii) } P{\bar{A} \cap B}=P(B)-P(A \cap B)=0.3-0=0.3\)