Sarah finds herself lost in a labyrinth and needs to find the exit. She comes to a room with three guards and two doors, one on the left wall and the other on the right. One of these doors will lead to the exit, while the other will lead to certain doom. Suppose that she is given the knowledge that one of the guards will always lie and give the wrong directions, while the other two will give the correct directions with probability 3/4. Suppose also that a guard will come up with an independent response, even under repeated questioning.
a) Sarah picks a guard at random and asks whether the right or left door leads to the exit. The guard says that the exit is through the door on the right wall. What is the probability that this is correct?
b) Sarah asks the same guard the same question. The answer is, again, that the exit is through the door on the right wall. What is the probability that this is the correct answer?
c) Sarah asks the same guard the same question for the third time, and receives the same response. What is the probability that this is correct?
d) Sarah asks the same guard the same question for the fourth time. The guard says that the exit is through the door on the left wall this time. What is the probability that the exit is truly through the door on the right wall?
A challenge probability questi
2012-02-03 4:56 pm
回答 (3)
2012-02-05 4:48 am
✔ 最佳答案
a)P(E)
= P(The guard not always lies) * P(The guard give the correct answer)
= 2/3 * 3/4
= 1/2
b)
P(E)
=
P(The guard is not always lies) * P(The guard give the correct answer twice)
---------------------------------------------------------------------------------------------------
P(The guard is always lies)
+ P(The guard is not always lies) * P(The guard give the correct answer twice)
+ P(The guard is not always lies) * P(The guard give the wrong answer twice)
=
2/3 * (3/4)²
-------------------------------------------- = 1/2
1/3 + 2/3 * (3/4)² + 2/3 * (1/4)²
c)
P(E)
=
P(The guard is not always lies) * P(The guard give the correct answer 3 times)
--------------------------------------------------------------------------------------------------------
P(The guard is always lies)
+ P(The guard is not always lies) * P(The guard give the correct answer 3 times)
+ P(The guard is not always lies) * P(The guard give the wrong answer 3 times)
=
2/3 * (3/4)³
----------------------------------------- = 9/20
1/3 + 2/3 * (3/4)³ + 2/3 * (1/4)³
d)
P(E)
=
P(The guard give the correct answer first 3 times and the 4th is wrong)
----------------------------------------------------------------------------------------------
P(The guard give the correct answer first 3 times and the 4th is wrong)
+ P(The guard give the wrong answer first 3 times and the 4th is correct)
=
(3/4)³ (1/4)
--------------------------------- = 9/10
(3/4)³ (1/4) + (1/4)³ (3/4)
2012-02-04 4:56 pm
Your answers are correct. How did you get 0.9 for part d?
2012-02-04 2:29 am
I think the answer should be :
(a) 0.5
(b) 0.5
(c) 0.45
(d) 0.9
2012-02-04 11:15:32 補充：
P(actual=R and answer=RRRL) =1/2 [1/3 (0)^3 (1)+2/3 (3/4)^3 (1/4) ]=9/256
P(actual=L and answer=RRRL)= /2 [1/3 (1)^3 (0)+2/3 (1/4)^3 (3/4) ]=1/256
P(actual=R)= P(actual=R and answer=RRRL)/ Probability(answer=RRRL)
=(9/256)/(9/256+1/256)=0.9
(a) 0.5
(b) 0.5
(c) 0.45
(d) 0.9
2012-02-04 11:15:32 補充：
P(actual=R and answer=RRRL) =1/2 [1/3 (0)^3 (1)+2/3 (3/4)^3 (1/4) ]=9/256
P(actual=L and answer=RRRL)= /2 [1/3 (1)^3 (0)+2/3 (1/4)^3 (3/4) ]=1/256
P(actual=R)= P(actual=R and answer=RRRL)/ Probability(answer=RRRL)
=(9/256)/(9/256+1/256)=0.9
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