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In an examination of 100 questions,"p" is probability of getting "n" correct answers.for every correct answer there is 1 mark and for wrong answer "-1/4" mark.if the test is so designed that average score is '0',then p equals---
a)(100-N)/(3n+100)
b)(100+n)/(3n-100)
c)n/(100+n)
d)(n-100)/(n+100)
Should we assume here that with probability p, there are n correct answers and '(100-n)' INCORRECT answers (all attempted)??
Because then, the average score will be:
p( n*1 + (100-n)*(-0.25) ) + (1-p)( (100-n)*1 + n*0.25 ) = 0
In GMAT verbal section there are 41 questions. Each question has 5 options out of which only one is correct.
If someone clicks answers at random, what is the probability that he will get 20 out of the 41 questions correct ?
In GMAT verbal section there are 41 questions. Each question has 5 options out of which only one is correct.
If someone clicks answers at random, what is the probability that he will get 20 out of the 41 questions correct ?
Its been already solved by Ross (I mean Sagar ) here. Refer Post
Well Query on Query..
what do you mean by getting 20 correct out of 41..? exactly 20 or at least 20 (means 20 and above) as such teh question is quiet simple..
hi
i think the answer is 1/5*20/41=20/205=4/41
getting 20 out of 41 is 20/41
getting 1 right from 5 is 1/5
so the probability is 4/41
pls let me know whether it is right or not
I feel answer should be C(41,20) * (1/5)^20 * (4/5)^21............
I agree with that solution.
If we went ahead and modified that problem to " atleast 20 correct" what would be the solution?
To be, or not to be: that is the question: Whether 'tis nobler in the mind to suffer The slings and arrows of outrageous fortune, Or to take arms against a sea of troubles, And by opposing end them? To die: to sleep;
Help
) here. Refer Post
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