Problem 5.A miner is trapped inside a mine and has access to 3 doors. The first door leads to a tunneland can take him to safety in 3 hours. The second door is a trap and brings him back to the mine againafter 5 hours through a tunnel. The third door also is a trap that brings him back to the mine after 7 hoursthrough a tunnel. The miner at all times is equally like to choose any of the doors. (In the darkness he canhardly distinguish the three). What is the expected length of time until he reaches safety

Expected time is 15 hours for him to get to safety.

Step-by-step explanation:

We define X as the time that this miner would get to safety.

We define Y as the door he chooses initially.

P(Y= 1) = P(Y=2)=P(Y=3) = 1/3

We have E[X|Y=1] = 3

E[X|Y] = 5 hours + E[X}

E[X|Y] = 7 hours + E[X]

Then we have

E[X] = 1/3(3 + 5 + E[X] + 7 + E[X])

We cross multiply

3*E[X] = (15 + 2E[x])

3E[X] - 2E[X] = 15

E[X] = 15

So the time it would take to get him to safety is 15 hours

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse

x + y = 90 5x – 10y = 90 90 – 10y = 5x 90 + 10y = 5x

step-by-step explanation:

last month maria hiked the 5-mile mountain trail a number of times and she hiked the 10-mile canal trail several times. let x represent the number of times she hiked the 5-mile trail, and let y represent the number of times she hiked the 10-mile trail. if she hiked a total of 90 miles, which equation can be used to find the number of times maria hiked each trail? x + y = 90 5x – 10y = 90 90 – 10y = 5x 90 + 10y = 5x