Ever thought of an exam that has passing probability of 0.9 or greater? It indeed sounds cool to attend one! But not so cool, when it boils down to design one. There was a similar problem of designing a question paper in my MSc(Engg.) SERC test at IISc.
“The probability of answering a question correctly is ½. A candidate is required to attempt 2 questions correctly in order to pass the exam. What are the minimum number of questions required in the exam so that the probability of passing is at least 0.9?”
Though the question looks very straight forward but when I started solving it using tree diagram, I started facing lot of issues and it took lot of time. Indeed, it is not a straight forward question but reverse version of a usually asked question. Suppose, had the question asked – “What is the probability of passing when there are, say 5 questions?” This one is pretty straight forward and lame. Since the total search space or the possible number of response combinations for a 5 question exam is 25=32. Now out of these we are interested in responses that have at least 2 correct answers. So total number of such responses would be –
= Number of responses with 2 correct answers +
Number of responses with 3 correct answers +
Number of responses with 4 correct answers +
Number of responses with 5 correct answers
= 5C2 + 5C3 + 5C4 + 5C5
= 10 + 10 + 5 +1
= 26
Thus Probability of passing is 26/32 = 0.81 (approx.)
Aah, so close to our required probability of 0.9! Now instead of calculating so many combinations, a direct option would be –
Probability of passing = Probability of at least 2 correct answers
= 1 – Probability of at most 1 correct answer
= 1 – (5C0 + 5C1)/32
= 1 – (1+ 5)/32
= 26/32 = 0.81
Here 5C0 indicates number of responses with 0 correct answers or all 5 wrong answers and there is only one possibility – all wrong answers. And 5C1 indicates the other part, where only one answer is correct and hence 5 possible responses i.e. any of the 5 questions is answered correctly.
But our task is exactly the reverse. We need to find out the minimum number of questions required such that the passing probability is at least 0.9. Though I couldn’t find the straight forward solution but I could attack it from reverse direction and keep on calculating the probabilities for each number until we find the required probability. Hence probability of passing when the number of questions, N = 6,
= 1 – (6C0 + 6C1)/26
=57/64 = .89
Similarly for N = 7,
= 1 – (7C0 + 7C1)/27
= 120/128 = 0.93
That’s it. We have the required probability and hence the answer, equal to 7. The exam needs to have at least 7 questions so that the passing probability is at least 0.9. Wow, how good it would be if everyone designed such exam!
Though, this is not the way it should be solved. But I found only this! So let me delve more into it until I find one, if any, or I faint!
No comments:
Post a Comment