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Leona [35]
3 years ago
12

A prisoner is trapped in a cell containing 3 doors. The first door leads to a tunnel that returns him to his cell after 2 days’

travel. The second leads to a tunnel that returns him to his cell after 4 days’ travel. The third door leads to freedom after 1 day of travel. If it is assumed that the prisoner will always select doors 1, 2, and 3 with respective probabilities .5, .3, and .2, what is the expected number of days until the prisoner reaches freedom?
Mathematics
1 answer:
BigorU [14]3 years ago
5 0

Answer:

12 days

Step-by-step explanation:

E[X]=E[E[X|Y]]

E[E|Y=1]P{Y=1}+E[E|Y=2]P{Y=2}+E[E|Y=3]P{Y=3}

(2+E[X])\frac{1}{2} +(4+E[X])\frac{3}{10}+1(\frac{2}{10} )

Solving for E(x) we get 12

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4[5-8(4c-3)]=12(1-13c)-8
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\boxed{c = -4}

First, we will use the distributive property (order of operations) to simplify the equation:

(4)(5) + (4)(−8(4c−3)) = (12)(1) + (12)(−13c)+ (−8) =

20 + (−128c) + 96 = 12 + (−156c)+ (−8)

Now, we will combine like terms to simplify the equation:

(−128c) + (20 + 96) = (−156c) + (12+ (−8) =

−128c + 116 = −156c + 4

Now we will add 156 to both sides, using inverse operations.

−128c + 116 + 156c = −156c + 4 + 156c =

28c + 116 = 4

Now we will subtract 116 from both sides:

28c + 116 − 116 = 4 − 116 =

28c = −112

Lastly, we will divide both sides by 28:

28c/28 = c

-112/28 = -4

The equation now looks like:

c = -4

Therefore, your answer is -4.

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