a)
Sample space:
\[ \Omega = \{HHH, TTT, HHT, TTH, HTT, THH, HTH, THT\} \]
b)
\[ S_1 = \{HTT, TTH, THT\} \]
\[ S_2 = \{HTT, TTH, THT, TTT\} \]
\[ S_3 = \{HHH, HHT, HTH, HTT\} \]
a) Event A and Event B of the same sample space \(\Omega\) are disjoint, so:
\[ P(A \cup B) = P(A) + P(B) = 0.8+0.4 = 1.2 \]
As both A and B are on the sample space \(\Omega\), their union must belongs to \(\Omega\), and the probability of that union must be \(≤\) 1. However,
\[ P(A \cup B) = 1.2 > 1 \]
=> It is impossible to have P(B) = 0.8
b)
The complement of event B is the set of all outcomes in sample space \(\Omega\) that are not in B.
Event A and Event B of \(\Omega\) are disjoint, so all outcomes in A are not in B but must be in \(\Omega\). Therefore, A must be the subset of B’s complement (\(A \subseteq B^c\)). Then:
\[ A \cap B^c = A => P(A \cap B^c) = P(A) = 0.4 \]
a) \(R^c\)
b) \(R \cup F\)
c) \(R^c \cup F^c\)
a)
\[ P(R^c) = 1 - P(R) = 1 - 0.1 = 0.9 \]
b)
\[ P(R \cup F) = P(R) + P(F) - P(R \cap F) = 0.1 + 0.07 - 0.03 = 0.14 \]
c) Using De Morgan’s Law:
\[ P(R^c \cup F^c) = P((R \cap F)^c) = 1 - P(R \cap F) = 1 - 0.03 = 0.97 \]
a)
\[ |\Omega| = 6 \times 6 = 36 \]
b)
\[ E^c = \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\} \]
Because \(E^c\) has 6 outcomes, E has 36 - 6 = 30 outcomes
Probability of E is:
\[ P(E) = \frac{|E|}{|\Omega|} = \frac{30}{36}=\frac{5}{6} \]
c) Probability of each (x,x) from 1 -> 5:
\[ P(x,x) = \frac{1}{10} \times \frac{1}{10} = \frac{1}{100} \]
Probability of (6,6):
\[ P(6,6) = P(6) \times P(6) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]
Then
\[ => P(E) = 1 - P(E^c) = 1 - P(6,6) - 5\times P(x,x) \]
\[ = 1 - \frac{1}{4} - 5\times \frac{1}{100} \]
\[ = \frac{7}{10} \]