Question 2.28: A mouse caught in a maze has to maneuver through three succe......

Denote by $~H_{1},\,H_{2},~$ and $~H_3~$ the events that the mouse successfully maneuvers through the three hatches, and by $E$ the event that the mouse is able to escape. We have that $~H_{1},\,H_{2},~$ and $~H_3~$ are independent, $P(H_{1})=0.6,P(H_{2})=0.4,$ and $P(H_{3})=0.2,$ and $E=H_{1}\cap H_{2}\cap H_{3}.$ Then: (i) $P(E)=$ $P(H_{1}\cap H_{2}\cap H_{3})\,=\,P(H_{1})P(H_{2})P(H_{3})\,=\,0.6\times0.4\times0.2\,=\,0.048,$ and (ii) $P(E^{c})=1-P(E)=1-0.048=0.952.$