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Difficult Math Problem #115 - Probability

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by 800guy » Fri Apr 06, 2007 8:51 am
My name is AJEET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged?

a) 5%
b) 10%
c) 20%
d) 25%
e) 30%

oa coming after some people respond w/explanations. from diff math doc
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Re: Difficult Math Problem #115 - Probability

by gabriel » Fri Apr 06, 2007 9:55 am
800guy wrote:My name is AJEET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged?

a) 5%
b) 10%
c) 20%
d) 25%
e) 30%

oa coming after some people respond w/explanations. from diff math doc

my answer is b 10% ... there are 5 letters so a pair can be formed in 5c2 ways .. that is 10 ways ... but in only one of these cases that is when E & E is chosen does the name remain unchanged ... so prob = 1/10 = 10% ..
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by Cybermusings » Sat Apr 07, 2007 12:12 am
My name is AJEET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged?

a) 5%
b) 10%
c) 20%
d) 25%
e) 30%

The only way that the letters of the name can be interchanged without changing the name is 1; i.e. when the position of the two
E's
is exhanged.

Now 5 letters can be arranged in 5P2 ways = 20 ways

Thus the probability that the name remains unchanged = 1/20 * 100 = 5%

Hence my answer is A
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by gabriel » Sat Apr 07, 2007 9:46 am
Cybermusings wrote:My name is AJEET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged?

a) 5%
b) 10%
c) 20%
d) 25%
e) 30%

The only way that the letters of the name can be interchanged without changing the name is 1; i.e. when the position of the two
E's
is exhanged.

Now 5 letters can be arranged in 5P2 ways = 20 ways

Thus the probability that the name remains unchanged = 1/20 * 100 = 5%

Hence my answer is A

ok cyber .. look at the darkened part .. that i think is the key .. suppose u choose A , J ... u can interchange their positon in only one way ( i.e. J ,A )... but by taking the permutation u also count cases in which the position is not interchanged .. so according to me it shuld be 5c2 and not 5p2 .. but lets wait for the OA ..
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by jayhawk2001 » Sat Apr 07, 2007 10:52 am
gabriel wrote:
ok cyber .. look at the darkened part .. that i think is the key .. suppose u choose A , J ... u can interchange their positon in only one way ( i.e. J ,A )... but by taking the permutation u also count cases in which the position is not interchanged .. so according to me it shuld be 5c2 and not 5p2 .. but lets wait for the OA ..
I agree with Gabriel. If you take permutation into account i.e. 5P2,
you also have to use A.J.E1.E2.T and A.J.E2.E1.T towards the numerator

So, you'll get 2 / 20 which is 10%

OA will ofcourse confirm :-)
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oa

by 800guy » Mon Apr 09, 2007 7:41 am
oa:

there are actually 20 ways to interchange the letters, namely, the first letter could be one of 5, and the other letter could be one of 4 left. So total pairs by product rule = 20.

Now, there are two cases when it wouldn’t change the name. First, keeping them all the same. Second, interchanging the two EEs together. Thus 2 options would leave the name intact.

Prob = 2/20 = 0.1, or 10%.
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