Probability question

Question

My name is AJEETH.By mistake, I interchange 2 letters in my name while filling a form. What is the probability that the name remains unchanged?

puneetj · Answer

Let's denote the 2 E's in AJEETH as E1 and E2 for better understanding
AJEETH can be arranged in 6!/2! ways = 360 ways

Name will remain unchanged only when its written in the following 2 ways:
1) AJE1E2TH
2)AJE2E1TH

So the probability = 2/360 = 1/180

OA pls??/

fluke · Answer

Question is asking:
What's the probability of choosing 2 E's in two picks.

\frac{C^2_2}{C^6_2}=\frac{1}{15}

puneetj · Answer

Is there any other way you can explain fluke, I still didn't understand

cvsmech · Answer

2 Es can be replaced with 2C2 which means theres only way to interchange their place in the word AJEETH.

2 letters can be chosen in as many as 6C2 ways.

Therefore probability = 1/(6C2) =1/15

fluke · Answer

The name will remain unchanged if he coincidentally interchanges the 2 E's present in AJEETH. Let's consider those E's as  E_1  and  E_2

So,
 AJE_1E_2TH

AJE_2E_1TH

The above two are same:
 AJEETH

Had the two interchanged letters been A and H, look what happens:
 AJEETH

HJEETA

Thus, we just need to find the probability of selecting 2 E's.

Probability=\frac{Desired \hspace{2} Outcomes}{Total \hspace{2} Outcomes}

Desired Outcome= Ways we can select 2 E's from 2 E's
 C^2_2=1

Total outcomes=Ways we can select any 2 letters from 6 letters(AJEETH)
 C^6_2=15

Thus,
 P=\frac{1}{15}

Another approach:
Pick 1 E(total E's=2) from 6 letters AND again pick E from the rest of the 5 letters.

P=\frac{2}{6}*\frac{1}{5}=\frac{1}{15}

Spidy001 · Answer

Here the desired outcome would be to change 2 letters and still name remains unchanged.

probability = desired outcomes/ total outcomes

i.e 2c2/6c2 = 1/15

puneetj · Answer

Question is asking:
What's the probability of choosing 2 E's in two picks.

\frac{C^2_2}{C^6_2}=\frac{1}{15}

Is there any other way you can explain fluke, I still didn't understand

The name will remain unchanged if he coincidentally interchanges the 2 E's present in AJEETH. Let's consider those E's as  E_1  and  E_2

So,
 AJE_1E_2TH

AJE_2E_1TH

The above two are same:
 AJEETH

Had the two interchanged letters been A and H, look what happens:
 AJEETH

HJEETA

Thus, we just need to find the probability of selecting 2 E's.

Probability=\frac{Desired \hspace{2} Outcomes}{Total \hspace{2} Outcomes}

Desired Outcome= Ways we can select 2 E's from 2 E's
 C^2_2=1

Total outcomes=Ways we can select any 2 letters from 6 letters(AJEETH)
 C^6_2=15

Thus,
 P=\frac{1}{15}

Another approach:
Pick 1 E(total E's=2) from 6 letters AND again pick E from the rest of the 5 letters.

P=\frac{2}{6}*\frac{1}{5}=\frac{1}{15}

Thanks Fluke....i need to brush up probability

pkmme · Answer

The answer is 1/15.

Probability question

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