Combo or Perm.?: A network engineer is examining a set of 10 wireless access points.

[seiliez]

A network engineer is examining a set of 10 wireless access points. At any given time, an access pointcan be in one of the following three states: (on), stand-by (sb) or error (err). Assume that all states areequally probable and independent of each other.

a) How many possible state patterns are available for these 10 access points at any given time (e.g. 1stpoint on, 2nd point err, 3rd point on, 4th point sb, …10th point on is one possible state pattern)?

b) What is the probability that the engineer will find the 10 access points in this configuration: any 3access points in the ‘on’ state, any 5 in the ‘sb’ state and any 2 in the ‘err’ state?

a - 3^10 = 59049

b - on on on sb sb sb sb sb err err = 1 / 59049

Is this correct? I'm not sure.

 

[stapel]

Question: A network engineer is examining a set of 10 wireless access points. At any given time, an access pointcan be in one of the following three states: (on), stand-by (sb) or error (err). Assume that all states are equally probable and independent of each other.

a) How many possible state patterns are available for these 10 access points at any given time (e.g. 1stpoint on, 2nd point err, 3rd point on, 4th point sb, …10th point on is one possible state pattern)?

b) What is the probability that the engineer will find the 10 access points in this configuration: any 3access points in the ‘on’ state, any 5 in the ‘sb’ state and any 2 in the ‘err’ state?

My answers:
a) 3^10 = 59049

b) on on on sb sb sb sb sb err err = 1 / 59049

Is this correct? I'm not sure.

Part (a): You have ten slots (similar to when you're flipping ten different coins):

. . . . .__..__..__..__..__..__..__..__..__..__

Each of these slots can take on any of three different values (so this is more like a three-sided fair coin, if such a thing existed). Each of these values is equally likely, and no one slot has any effect on any other slot. As such, the computations are:

. . . . . 3 .. 3 .. 3 .. 3 .. 3 .. 3 .. 3 .. 3 .. 3 .. 3

Multiplying, you get 3¹⁰. I agree with your answer.

Part (b): Since each outcome is equally likely, the probability of any one particular result is of the form "(one particular result) / (total number of possible results)", or 1/3¹⁰. I agree with your answer.

 

[pka]

A network engineer is examining a set of 10 wireless access points. At any given time, an access pointcan be in one of the following three states: (on), stand-by (sb) or error (err). Assume that all states areequally probable and independent of each other.
a) How many possible state patterns are available for these 10 access points at any given time (e.g. 1stpoint on, 2nd point err, 3rd point on, 4th point sb, …10th point on is one possible state pattern)?

b) What is the probability that the engineer will find the 10 access points in this configuration: any 3access points in the ‘on’ state, any 5 in the ‘sb’ state and any 2 in the ‘err’ state?

a - 3^10 = 59049 CORRECT

b - on on on sb sb sb sb sb err err = 1 / 59049 INCORRECT

The number of ways to arrange the string \(\displaystyle \mathbf{OOOSSSSSEE}\) is \(\displaystyle \dfrac{10!}{3!\cdot 5!\cdot 2!}\)

 

[stapel]

Part (b): Since each outcome is equally likely, the probability of any one particular result is of the form "(one particular result) / (total number of possible results)", or 1/3¹⁰.

Clarification: The answer given above to part (b) assumes that the access points (like the positions of letters and numbers on a license plate, thinking back to when you did simpler exercises of that type) are "distinguishable", there being a "first" access point, a "second", etc. If the access points are indistinguishable, then I'm not sure that the answer to part (a) is correct, since "OOOSSSEEEE" would be the same as (that is, indistinguishable from) "EEEESSSOOO", and one would need to compute the number of ways each indistinguishable outcome could occur.

This would be similar to asking the number of outcomes of flipping four "fair" two-sided (that is, normal) coins. If the coins are numbered (by, say, marking the numbers "1" through "10" on the coins), then the outcome "HHTT" would be distinguishable from the outcome "TTHH". Assuming distinguishable coins, we would then have 2*2*2*2 = 16 possible outcomes:

HHHH
HHHT
HHTH
HTHH
THHH
HHTT
HTHT
THHT
HTTH
THTH
TTHH
HTTT
THTT
TTHT
TTTH
TTTT

Any given outcome would have a probability of 1/16.

However, if we assume that the coins are not distinguishable, then we have to account for the numbers of manners for each outcome:

zero tails: one way
one tail: four ways
two tails: six ways
three tails: four ways
four tails: one way

Then the total number of non-distinguishable outcomes is five. The probability of obtaining any particular outcome will vary with the particular outcome chosen, as illustrated in the formula contained in the second reply, as explained here.