[catasp2005]

Hi,
Can anyone please solve ths question for me?
In an examination, there are 20 questions with 4 alternatives. There is only 1 correct
alternative. One gets 4 marks for every correct answer & 1 negative mark for every wrong answer.50% are the passing marks. If a student attempts the questions on guesswork, find the probability of his passing.
Avi.

[TheAkshaT]

Originally Posted by catasp2005 @ Thu Dec 02, 2004 11:11 pm

Hi,
Can anyone please solve ths question for me?
In an examination, there are 20 questions with 4 alternatives. There is only 1 correct
alternative. One gets 4 marks for every correct answer & 1 negative mark for every wrong answer.50% are the passing marks. If a student attempts the questions on guesswork, find the probability of his passing.
Avi.

To pass he should have a total of 40 marks, that means anything less than 40 is fail :P

to get more than 40 he has to get atleast 12 correct....and at max 8 wrong
if 8 are wrong then 20C8 ways...
if 7 are wrong 20C7 ways...
if 6 are wrong 20C6....
and so on...

sum them up and get the ans



-Akshat

[ramnik]

u should specify pass marks

[GauravShah]

Originally Posted by TheAkshaT @ Thu Dec 02, 2004 11:37 pm

To pass he should have a total of 40 marks, that means anything less than 40 is fail :P

to get more than 40 he has to get atleast 12 correct....and at max 8 wrong
if 8 are wrong then 20C8 ways...
if 7 are wrong 20C7 ways...
if 6 are wrong 20C6....
and so on...

sum them up and get the ans



-Akshat

Hey Akshat,

I guess u missed out to consider that there are 4 alternatives to each Q...

So now P(Pass) = P(1w) + P(2w) + .... + P(8w)

Where
P(1w) = (20C1) * [(1/4)^19] * [(3/4)^1]
till
P(8w) = (20C * [(1/4)^12] * [(3/4)^8]

I feel there wud be a shorter method.....

Gaurav.

PG Rocks -->

[TheAkshaT]

Originally Posted by GauravShah @ Fri Dec 03, 2004 9:27 am

Hey Akshat,

I guess u missed out to consider that there are 4 alternatives to each Q...

So now P(Pass) = P(1w) + P(2w) + .... + P(8w)

Where
P(1w) = (20C1) * [(1/4)^19] * [(3/4)^1]
till
P(8w) = (20C * [(1/4)^12] * [(3/4)^8]

I feel there wud be a shorter method.....

Ooops!...

Thanks for correcting

-Akshat

[catasp2005]

Originally Posted by GauravShah

Hey Akshat,

I guess u missed out to consider that there are 4 alternatives to each Q...

So now P(Pass) = P(1w) + P(2w) + .... + P(8w)

Where
P(1w) = (20C1) * [(1/4)^19] * [(3/4)^1]
till
P(8w) = (20C * [(1/4)^12] * [(3/4)^8]

I feel there wud be a shorter method.....

Gaurav.

PG Rocks --> *

Hey Gaurav,
i got this answer, bt its quite a long method........
Do u knw of any short-cut???????
avi.

[supervish]

Originally Posted by catasp2005

Hi,
Can anyone please solve ths question for me?
In an examination, there are 20 questions with 4 alternatives. There is only 1 correct
alternative. One gets 4 marks for every correct answer & 1 negative mark for every wrong answer.50% are the passing marks. If a student attempts the questions on guesswork, find the probability of his passing.
Avi.

The way I look at it, if the student is plain guessing, we can assume, he's attempting all the 20 questions.

Let C be the number of correct answers. He needs to score more than 50% ( of maximum marks I assume ).

so, score >= 40

4C - 1*(20-C) >=40

4C - 20 + C >= 40

5C >= 60

C >= 12

Which can happen in 9 ways ( 12 correct, 13 correct , ..... all 20 correct )

So probablity of passing = 9 / 21



21 because, we have to take the zero correct eventuality.

Vish

Disclaimer :I came up with this "logic" on the spur of the moment. It doesnt seem right to me as I've ignored the 4 correct alternatives condition. I'm leaving it here just for ppl to munch some brains over it. Or dissect it.

Shortcuts may arise if you can provide the answer options. I cant think of any shortcut to the solution. Might think about it if I have nothing ele to do.

Vish

[ritika_j]

In how many ways can the letter of the word "daughter" be arranged such that no vowels are together?

answer this question with explanation

[GauravShah]

Originally Posted by ritika_j

In how many ways can the letter of the word "daughter" be arranged such that no vowels are together?

answer this question with explanation

Edit : Please ignore.... solution incorrect

Not sure if my approach is correct still here is my solution.....

Lets place this first V-C-V-C-V...here three vowels can be arrange in 3! = 6 ways and consonants in 5P2 = 5.4 = 20 ways...

hence total = 6*20 = 120 ways...

Now consider the spaces between these that is...

_ - V - _ - C - _ - V - _ - C - _ - V - _

In this stepp we fill these spces will the remaining consonants i.e. 3

This can be done in 6C3 * 3! ways = 20*6 = 120 ways

Hence total no. of ways = 120*120 = 14400 ways

If I missed something please reply...

Also i don't think i was able to convey the solution properly so if anyone understood this and can simplify please do

Cheers

Gaurav.

[QuintEssence]

Originally Posted by GauravShah

Not sure if my approach is correct still here is my solution.....

Lets place this first V-C-V-C-V...here three vowels can be arrange in 3! = 6 ways and consonants in 5P2 = 5.4 = 20 ways...

hence total = 6*20 = 120 ways...

Now consider the spaces between these that is...

_ - V - _ - C - _ - V - _ - C - _ - V - _

In this stepp we fill these spces will the remaining consonants i.e. 3

This can be done in 6C3 * 3! ways = 20*6 = 120 ways

Hence total no. of ways = 120*120 = 14400 ways

If I missed something please reply...

Gaurav.

well champ, u did miss something..u see "daughter" has 3 vowels nd 5 consonants..so its not necessary to have a vc-vc or -v-c sort of repetitive relatonship..actually the solution is much simpler

step1: calculate all possible permutations of "daughter"= 8!

step2: calculate all cases in which three vowels are always together=3!x6!

step3: calculate cases of two vowels being together..this is slightly tricky, so a diagram follows

_ _ _ _ _ _ _ _
1 2 3 4 5 6 7 8 lets consider two vowels at 1-2 position..3C2=3 arrangements
=> rest arrangements can be made in 6! ways

out of these 6! arrangements 5! arrangements will have a vowel at place 3

so we have 3x(6!-5!) ways

now this 1-2 group can be taken in 7 ways

so arrangements for two vowels together are= 7x3x(6!-5!)

hence, we have total arrangements= 8!- [3!x6! + 7x3x(6!-5!)]