Unacademy AIMT - 1

Question

In a shared Ethernet link, each user is active only 10% of the time. Suppose switching is used for 10 users and each user is equally likely to transmit at any point of time.
The probability P that at any given time exactly 8 users are transmitting simultaneously.
What is the value of $10^{10} * P$?

Solve for both:
1)Packet switching.
2)Circuit switching.

Answer 1

You can use Binomial Theorem to solve this:

P(each user being active) = $\frac{1}{10}$

P(each user not being active) = $\frac{9}{10}$

P(exactly $8$ users are active/ are transmitting at the same time):

1. Out of $10$ users we can select any $8$ users = $_{}^{10}\textrm{}C_{8}$

2. $8$ users transmitting successfully/ are active = $(\frac{1}{10})^8$

3. $2$ users not being active/ not transmitting = $(\frac{9}{10})^2$

P =  $_{}^{10}\textrm{}C_{8}$ x  $(\frac{1}{10})^8$ x $(\frac{9}{10})^2$

   = $45$ x $10^{-10}$ x 81

   = $3645$ x $10^{-10}$

So we need to find out $10^{10}$ x P = $3645$.

Both Packet and Circuit Switching will give the same answer.

Comments

@Pranavpurkar Probability in those cases should be 1 as only a particular station will transmit

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@Sunnidhya Roy

Yes i agree , But  in this question, Probability for circuit switching will be 0? as  packets cannot be sent simultaneously?

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@Pranavpurkar I am not sure about that.