IIT JEE , AIEEE Forum - Mathematics , Algebra

sharmadon

please let me know the method of solving this question

in an examination in an objective test there are two sections containing 10 questions each.In section 1,each objective question has 5 choices and only one choice is correct.Section 2 has 4 choices for each question and more than one choice may be correct.Marks for section 2 is awarded only if a student gives all choices correct.There is no negative marking for any section.For each question in section1 and section 2, 1 and 3 marks are will be awarded respectively.

1. if a student attempts only two questions,one from section 1 and the other from section 2, then what is the probability that he is awarded marks in both questions?

2.what is the probability of getting marks less than 40 for a student ?

3. if a student attempts 4 questions, 2 from each section,then what is the probability that he gets 4 marks ?

4.if a student attempts all the questions then what is the probability that he gets 5 marks in section 1 and 15 in section 2?

answers:

1)1/5 * 1/15

2) 1-{(1/5)^10}*{(1/15)^10}

3)(1/5)*(1/15)*(14/15)*(4/5)*4

4)10C5*(1/5)^5*(4/5)^5*10C5*(1/15)^5*(14/15)^5

sboosy

computer001

pl ref link i posted the ans there

http://www.goiit.com/posts/list/algebra-probability-47181.htm#234450

sboosy

There are 10 questions in section A .and each question carries 1 mark ...thus max marks is 10
There are 10 questions in section B and each question carries 3 mark ...thus max marks is 30
Thus total max marks obtainable is 40 ...and there is only way to do so ..that is get all questions right ...P of which is (1/5)^10 * (1/15)^10 ...Thus P of getting mark less than 40 is
1-[(1/5)^10 * (1/15)^10]

sharmadon

hey u computer001...there is a big difference bw posting the answers and giving the solution...okk...after getting irritated from u i have asked the question again here..

computer001

i think ritin values in terms of pdts like a*b*c etc is enuf reasoning..in tht problem..so i said refer..thts it buddy

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