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Quiz Quandary (Posted on 2003-12-26) Difficulty: 3 of 5
A teacher said that she had observed that how well a student does on a particular quiz depends on how well or poorly he or she did on the last quiz. Then she gave the following statistics:

If you did well on a quiz, there is an 80% chance you will do well on the next quiz, a 15% chance you will do so-so, and a 5% chance you will do poorly.

If you did so-so on a quiz, there is a 20% chance you will do well on the next quiz, a 60% chance you will do so-so, and a 20% chance you will do poorly.

If you did poorly on a quiz, there is a 3% chance you will do well on the next quiz, a 15% chance you will do so-so, and an 82% chance you will do poorly. The teacher then asked the following question (which she said we'd be able to answer once we had successfully completed the class):

If you did well on the first quiz, what is the probability that you will do well on the fifth quiz in the class?

See The Solution Submitted by DJ    
Rating: 3.7778 (9 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Markov chains | Comment 11 of 14 |
(In reply to re: Markov chains by Charlie)

Additional comments on the markov chain point of view:

If we introduce a "state space" with three states:
S1: you did well on the previous quiz
S2: you did so-so on the previous quiz
S3: you did poorly on the previous quiz

then the matrix A=
0.8 0.15 0.05
0.2 0.6 0.2
0.03 0.15 0.82

will be the transition matrix. That is, element A(1,1) (read: A(row,col)) is the probability that you'll remain in S1 if you started in S1, element A(1,2) the probability that you'll go to S2 if you started in S1, and so on (this matrix is usually denoted with "P" in standard textbooks on Markov chains, at least the ones I've read).

Since we know for certain that we start out in S1, the initial state probability vector will be I=
1
0
0

In other words, it is certain that we did well on the last quiz (quiz number 1). The state probability vector after the second quiz will have propagated to A*I=
4/5
1/5
3/100

In the same way the state probability vector after quiz five will be A^4*I (as Charlie pointed out in the previous post)=
10436837/20000000
1676387/5000000
2128893/12500000

I.e. the first element of the above vector is the probability of doing well on the fifth quiz (which is equivalent to ending up in S1 after the fifth quiz). An approximation of this fraction is:

0.52184

  Posted by Finkle on 2003-12-27 19:21:26

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