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Elemantary Probability

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Elemantary Probability

by Aman verma » Sat Feb 06, 2010 2:34 am
Q: If two consecutive letters are selected at random English Alphabet , the probability that they are both consonants is :

a) 2/5

b) 9/25

c) 16/25

d) 3/5

e) none of these

Now the answer given to this problem is c) 16/25 .However, no explanation is provided to this problem. I tried my best, but was unable to come up with 16/25 . My solution was 21/26 X 20/25 which is no way near to 16/25. Can anybody help me with this problem. Can anybody provide me the explanation to the given answer 16/25.
Last edited by Aman verma on Sat Feb 06, 2010 2:53 am, edited 1 time in total.
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by harsh.champ » Sat Feb 06, 2010 2:49 am
Aman verma wrote:Q: If two consecutive letters are selected at random English Alphabet , the probability that they are both consonants is :

a) 2/5

b) 9/25

c) 16/25

d) 3/5

e) none of these

Now the answer given to this problem is [spoiler]c) 16/25[/spoiler] .However, no explanation is provided to this problem. I tried my best, but was unable to come up with [spoiler]16/25 [/spoiler]. My solution was [spoiler]20/26 X 19/25 [/spoiler]which is no way near to [spoiler]16/25[/spoiler]. Can anybody help me with this problem. Can anybody provide me the explanation to the given answer [spoiler]16/25[/spoiler].
First of all,aman plz post the answer in the spoiler mode.

As for solving the question,
5 vowels are there - a,e,i,o,u.
21 consonants are there.

so,no first of the consecutive letters should be d,h,n,t,. [These are the consonants adjoining to the 5 vowels(front and back)]
Removing these 4 consonant choices.
so, nothing from (a,e,i,o,u,d,h,n,t)[9 in total],hence total left no is 26-9=17

The ans should be [spoiler]17/26[/spoiler] which would be E.

Hence, we are left
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by Aman verma » Sat Feb 06, 2010 2:59 am
As I am a beginer I don't know what a spoiler is or how it works or what are the advantages of a spoiler. So please tell me the benefits and uses of a spoiler. As regards to the answer it is specifically given as 16/25 but no explanation provided . My answer is (16.15)/25 which cannot be correct.
Last edited by Aman verma on Sat Feb 06, 2010 11:42 pm, edited 1 time in total.
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by ajith » Sat Feb 06, 2010 3:27 am
Aman verma wrote:Q: If two consecutive letters are selected at random English Alphabet , the probability that they are both consonants is :

a) 2/5

b) 9/25

c) 16/25

d) 3/5

e) none of these

Now the answer given to this problem is c) 16/25 .However, no explanation is provided to this problem. I tried my best, but was unable to come up with 16/25 . My solution was 21/26 X 20/25 which is no way near to 16/25. Can anybody help me with this problem. Can anybody provide me the explanation to the given answer 16/25.
There are 25 ways to select the first letter, since the second letter is the next one in alphabetical order we do not have much choice there.

Now there are 9 ways in which these contain one vowel (AB, DE, EF.....) [alternatively (1 +4*2)]

So the probability that both are consonants is 1-9/25 =16/25
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by ajith » Sat Feb 06, 2010 3:29 am
harsh.champ wrote: The ans should be [spoiler]17/26[/spoiler] which would be E.

Hence, we are left
There are only 25 choices for the two consecutive alphabets (first letter ranging from A to X)
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by harsh.champ » Sat Feb 06, 2010 4:08 am
Aman verma wrote:As I am a beginer I didn;t knew what a spoiler is or how it works or what are the advantages of a spoiler. So please tell me the benefits and uses of a spoiler. As regards to the answer it is specifically given as 16/25 but no explanation provided . My answer is (16.15)/25 which cannot be correct.
____________________________
Well, spoilers can be found on the "edit post" window where you can select the text(suppose ,the answer) and then click on the spoiler button.
After that , when you submit your post that text is marked by black.Only if you move the mouse over that text,will you see the answer.

As for why they are used,suppose I arrive at your thread.If the answer itself is written alongside the question ,the natural-thinking process of mind is obstructed .
If you use the spoiler,only after I solve the question,I will look at the OA and then compare if my answer is right and whether the correct problem approach is implemented.
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by harsh.champ » Sat Feb 06, 2010 4:09 am
I made a mistake in my upper post
As for solving the question,
5 vowels are there - a,e,i,o,u.
21 consonants are there.

so,no first of the consecutive letters should be d,h,n,t,. [These are the consonants adjoining to the 5 vowels(front and back)]
Removing these 4 consonant choices.
so, nothing from (a,e,i,o,u,d,h,n,t)[9 in total],hence total left no is 26-9=17

The ans should be 17/26 which would be E.
I also took the case when the consecutive letter will start from z,but only 5 cases,last case being yz.
Hence,subtracting 1 both from numerator and denominator,
the ans. is [spoiler]16/25[/spoiler].
Last edited by harsh.champ on Sat Feb 06, 2010 4:20 am, edited 1 time in total.
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by shashank.ism » Sat Feb 06, 2010 4:16 am
Q: If two consecutive letters are selected at random English Alphabet , the probability that they are both consonants is :

a) 2/5

b) 9/25

c) 16/25

d) 3/5

e) none of these

Now the answer given to this problem is c) 16/25 .However, no explanation is provided to this problem. I tried my best, but was unable to come up with 16/25 . My solution was 21/26 X 20/25 which is no way near to 16/25. Can anybody help me with this problem. Can anybody provide me the explanation to the given answer 16/25.
For this type of problem we consider consecutive letter as on entity. thus total no. of ways of selecting consecutive no.
25C1 = 25..
now no. of consecutive letters in which there is a vowel = 9 (1 for A as it is in beginning & 2 each for E,I,O,U)
so no. of consecutive letters with both consonant = 25-9 =16.

so probability = 16/25. Ans is C.
Last edited by shashank.ism on Sat Feb 06, 2010 11:55 am, edited 1 time in total.
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by Aman verma » Sat Feb 06, 2010 10:22 am
My deepest gratitude to Ajith and Shashank.ism as this problem seemed to me insurmountable. I look forward to the continued support of you people as my mathematical skills are pretty weak ; haven't done any math after my 10th standard. I am not a student of science or mathematics.
Last edited by Aman verma on Sat Feb 06, 2010 11:43 pm, edited 1 time in total.
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by shashank.ism » Sat Feb 06, 2010 11:53 am
Aman verma
My deepest gratitude to Ajith and Shashank.ism as this problem seemed to me unsurmountable. I look forward to the continued support of you people as my mathematical skills are pretty weak ; haven't done any math after my 10th standard. I am not a student of science or mathematics.
Dear Aman, Thanks for these words. You can ask me any problem whenever required. I qualified IIT-JEE in 2005 and hopefully quite strong in mathematics. So can help you anytime with all the technicalities required. Feel free to post in forum or you can message me personally. If you are serious about GMAT thing, just solve problems by taking help and soon you will like to solve these problems. Its always enjoying and encouraging to solve these mathematical problems as you don't have a single method but a lots of method for a single problem. Never forget "Practice makes a man perfect" & " this saying is applicable in every sphere of life".
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