Answer to Question #160739 in Statistics and Probability for Soumyo

Question

i) In a hotel, an intercom system master station provides music to six rooms. The probability that any one room will be switched on and draw power at anytime is 0.4. When on, a room draws 0.5 W.

What is the probability that the master station amplifier is overloaded, if the amplifier is overloaded when more than 2W is demanded?

ii) If the amplifier is replaced by a 4W unit which is supposed to supply 12 rooms now, then what would be the probability of its overload?

iii) If instead two independent 2W units are used to supply 6 rooms each, will the probability of the overload be better?

Answer 1

Solution:

i) First of all, it does not matter how the rooms are connected to the amplifier (parallel or sequential connection), as the total power of the system is the sum of powers of each part of the system in any case:

$P_{total} = P_1+P_2+...+P_n$

That means that the overload happens when more than $2/0.5=4$ rooms are in usage.

$P(\xi \geq4) = P(\xi=4) + P(\xi = 5) + P(\xi=6)$

We will firstly find the probability that 6 rooms are turned on, there is only one such case:

$P(\xi=6) = 0.4^6 = 0.004096$

Then, we need to find the probability that 5 rooms are turned on. Remembering combinatorics, the number of such cases is:

$\begin{pmatrix} n \\ k \end{pmatrix} = \frac{n!}{k!(n-k)!}$

$\begin{pmatrix} 6 \\ 5 \end{pmatrix} = \frac{6!}{5!(6-5)!} = 6$

The probability of a single case is:

$0.4^5 = 0.01024$

And the probability is:

$P(\xi=5) = 6*0.01024 = 0.06144$

Finally, let us find the probability that 4 of the rooms are turned on:

$\begin{pmatrix} 6 \\ 4 \end{pmatrix} = \frac{6!}{4!(6-4)!} = \frac{6*5}{2} = 15$

$0.4^4=0.0256$

$P(\xi=4)=15*0.0256=0.384$

Finally, we get the answer:

$P(\xi\ge4)=0.004096+0.06144+0.384=0.449536$

ii) Using the same logic as in the start of i), we get that overload happens if the number of rooms that are turned on is greater or equal to $4/0.5=8$ rooms. But since we have only 6 rooms:

$P(\xi\ge8)=0$

iii) Since the power is added up despite the type of connection we can conclude that amplifiers are overloaded when the power is more than 2W. That means that the probability remains the same as in i)

Answer:

i) $0.449536$

ii)

iii) $0.449536$