UGC NET CSE | December 2014 | Part 2 | Question: 04

Question

A computer program selects an integer in the set $\left\{k:1\leq k \leq 10,00,000\right\}$ at random and prints out the result. This process is repeated 1 million times. What is the probability that the value k = 1 appears in the printout atleast once ? 

• A. $0.5$
• B. $0.704$
• C. $0.632121$
• D. $0.68$ 

Answer 1

Answer C

Probability of k=1 is 1/1000000= 10^-6

Probability of k!= 1 is 1- 10^-6 = 0.999

Probability that k=1 is never printed in all 10⁶ print outs = 0.999*0.999*.......0.9999 (10 ⁶ times)

= 0.999^{10^6}

Probability that 1 is printed at least once = 1- probability that 1 is never printed

= 1-0.999^{10^6}

=0.6321

Comments

as calci is not allowed in NET how u calculated 0.99910^6 

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No sir, i used calculator. I don't know any shortcut is there to solve without calculator.

Answer 2

Here, we have:

n=1000000, p=1/1000000

Then Mean:
m=np
m=1000000*(1/1000000)
m=1

As m=1, hence this question can be solved by Poisson Distribution method.

Therefore, required probability:
P=(e^-mm^r)r!
P=1-P(0)
P=1-(e^-11⁰)/0!
P=0.6321

Answer is C.

           

Comments

"As m=1, hence this question can be solved by Poisson Distributionmethod. "

 

Is this true for every question