Latest Tech Mahindra Aptitude Question SOLUTION: There are 600 pages. there is 1 error on average per page. what is the probability that there are n errors per page.

• Placement Papers
• /
• Tech Mahindra
• /
• Logical Reasoning
• /
• Mathematical Reasoning

Tech Mahindra Question Papers: Tech Mahindra Interview Experience(11) TECH MAHINDRA Company Profile

(#M40034090) TECH MAHINDRA Logical Reasoning question Keep an EYE

There are 600 pages. there is 1 error on average per page. what is the probability that there are n errors per page.

Asked In Tech Mahindra shamma mohammad (10 years ago)

Solved ravi jha Read Solution (6)

Is this Puzzle helpful?   (35)   (6) Submit Your Solution Mathematical Reasoning

Read Solution (6) :

ravi jha  (10 years ago)

According to Poisson distribution theorem,

Probability of n errors per page =1/(en!)
where e = 2.71828 and n! means n*(n-1)*(n-2) ... 1

So ,
For finding probabilty 0f 5 errors per page,
Answer: 1/(2.71828*5!)=1/(2.71828*120)

Is this solution Helpfull? Yes (20) | No (2)  

Vamsi Krishna  (10 years ago)

This can be solved using a formula based from the Poisson distribution:

p = [ (λ ^ k)(e ^ -λ) ] / k!

Here, k = n, λ = 1, so p = (1^n) (e^-1) / n!

So the answer is

1
---------
(e) (n!)

Is this solution Helpfull? Yes (10) | No (7)  

deepa gudivada  (10 years ago)

According to Poisson distribution theorem,

Probability of n errors per page =1/(en!)
where e = 2.71828 and n! means n*(n-1)*(n-2) ... 1

So ,
For finding probabilty 0f 5 errors per page,
Answer: 1/(2.71828*5!)=1/(2.71828*120)

Is this solution Helpfull? Yes (5) | No (2)  

MD   (10 years ago)

probability for the one error in page=1/600
prob. for the n errors in page=n/600.
so ans should be=n/600.

Is this solution Helpfull? Yes (1) | No (28)  

sharmi saha  (9 years ago)

it is 11/36

Is this solution Helpfull? Yes (1) | No (3)  

Davinder singh  (8 years ago)

You need higher math for this one, but I put it in because of how nifty the answer is. It turns out that the probability of n errors per page is 1/(en!) where e = 2.71828 and n! means n x (n-1) x (n-2) ... 1.

For example, the probability it zero errors per page is 1/e or 1/2.7, which is also the probabily of 1 error per page. The chance of 2 errors per page is half that and of 3 is 1(3x2) or 1/6 of the chance of 1. If you add up all of the possible probabilities of up to all 600 errors being on one page, the all add up to 1 (or very close to one. Technically, if you all them all the way to infinity, they add to one.)

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.