Difficult Math Problem #115 - Probability

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% ..

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

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 ..

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

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%.