Question

A prisoner is trapped in a cell containing 3 doors. The first door leads to a tunnelthat returns him to his cell after 2 days’ travel. The second leads to a tunnel thatreturns him to his cell after 4 day’s travel. The third door leads to freedom after 1day of travel. If it is assumed that the prisoner will always select doors 1,2,and 3with respective probabilities 0.5,0.3, and 0.2, what is the expected number of daysuntil the prisoner reaches freedom?

Answer 1

Answer:

2 days

Step-by-step explanation:

Expected number of days until prisoner reaches freedom=E(x)=?

E(x)=x*p(x)

Where x is the number of days and p(x) is the probability associated with them.

X    1    2    3

P(x) 0.5 0.3 0.2

E(x)=1*0.5+2*0.3+3*0.2

E(x)=0.5+0.6+0.6

E(x)=1.7.

Thus, the expected number of days until prisoner reaches freedom are 2 days.