stats question, i dont know how to do this question: A student, Tom, sits a multi-choice exam in a subject he knows nothing about. There are twenty questions, and each question has four answers, only one of which is correct. Suppose another student, Sarah, tried a slightly different guessing strategy, by avoiding picking the same answer twice in a row (e.g. if she answers (b) in one question then she randomly picks one of (a), (c) or (d) in the next). If the answers were placed by the examiners in random order for each question, does her plan improve her chance of passing?
@dan815
her plan shouldnt improve her chances
how can you tell? can you please explain
Lets label the questions as follows $$ \Large { a ~ b ~ c ~ d \\ a ~ b ~ c ~ d \\ a ~ b ~ c ~ d \\ a ~ b ~ c ~ d \\ ... } $$
each question has 4 possible choices, a,b,c or d
so that would be 1-p(a)
oh, what if the test submitter picked a,a,a,a,a,... for all the questions. sure it is unlikely, but according to randomness its possible
Lets say the answers for the questions are 1. a 2.b 3.c 4. d 5. d 6.a .. 19. c 20.c The person who does not allow repeating answers is impossible to get them all right Because it is random, it is like flipping a coin, each selection of a,b,c,d are independent of what you picked for the last question. They are independent choices.
Each time the test maker picks a letter a,b,c,d , it is like flipping a 4 sided die. You could get `a,a,a,a,`... each time, because they are random . each event is memoryless , has no memory of the past selection
If the test submitter used a random number generator 1 through 4 , for each question , its possible that he could consecutive repetitions. it is just like if you flip a coin, you could have heads followed by heads followed by heads, etc. A coin flip would be a good way to decide which letter to pick on a 2 multiple choice question exam. In any case, her strategy will lessen the chance of getting all of them right.
so the probability would be the same. regardless.
it would be less
oh.. i see
lets say theres a 4 question test, with 4 multiple choices. here are the possible ways to set the answers. a,a,a,a a,a,a,b a,a,b,a a,b,a,a b,a,a,a ... too many to count, but you see that she will never guess all of these consecutive cases
the key here is to realize that if the test maker is randomly picking numbers one through 4, its possible he can pick two of the same numbers consecutively. each experiment of picking a random number is separate, and each number is equally likely assume the numbers correspond to letters 1 = a, 2 = b, 3 = c, 4= d
yup i get it! thanks so much!!
please can you help me with one more question.
Sawn logs are loaded onto a boat. The weight of each log is uniformly distributed between 200kg and 400kg. Suppose that the log weights are independent of each other, and that 100 logs are loaded onto a boat. (iv) Without doing any further calculations, for what weight W is there a 50% probability that T is at least W? Explain briefly.
what are parts i, ii) , iii)
(i) What type of distribution would you use to describe the total weight T of these 100 logs? Briefly justify your answer. (ii) What is the mean and standard deviation of T? (iii) What is the probability that T is at least 31,000kg?
for v1) i think its using the z-score formula, but i can't find the standard deviation
so range would be 200? 400-200
so standard deviation would be 50?
right
i usd norminv( 0.5,300,50) in excel and i got 300
for a uniform. distributed variable on [a,b]
isn't it normal distribution
a uniform distribution looks flat, like this
|dw:1431333536470:dw|
thats the distribution, and we need the area to add up to 1. therefore the probability density curve is f(x) = 1/ (400 - 200) , for 200<= x <= 400
oh i see.
|dw:1431333720333:dw|
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