Question 2.28: A mouse caught in a maze has to maneuver through three succe......
A mouse caught in a maze has to maneuver through three successive escape hatches in order to escape. If the hatches operate independently and the probabilities for the mouse to maneuver successfully through them are 0.6, 0.4, and 0.2, respectively, calculate the probabilities that the mouse: (i) will be able to escape, (ii) will not be able to escape.
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.