ABy Admin
Apr 28'23
Exercise
ABy Admin
Apr 28'23
Solution: D
First note
[[math]]
\operatorname{P}[ A ∪ B ] = \operatorname{P}[ A] + \operatorname{P}[ B ] − \operatorname{P}[ A ∩ B ], \,
\operatorname{P}[ A ∪ B '] = \operatorname{P}[ A] + \operatorname{P}[ B '] − \operatorname{P}[ A ∩ B '].
[[/math]]
Then add these two equations to get
[[math]]
\begin{align*}
\operatorname{P}[ A ∪ B ] + \operatorname{P}[ A ∪ B '] &= 2 \operatorname{P}[ A] + ( \operatorname{P}[ B ] + \operatorname{P}[ B '] ) − ( \operatorname{P}[ A ∩ B ] + \operatorname{P}[ A ∩ B '] ) \\
0.7 + 0.9 &= 2 \operatorname{P}[ A] + 1 − P ⎡⎣( A ∩ B ) ∪ ( A ∩ B ') ⎤⎦
\\ 1.6 &= 2 \operatorname{P}[ A] + 1 − \operatorname{P}[ A] \\
\end{align*}
[[/math]]
which implies that [math]\operatorname{P}[A] = 0.6 [/math].