Suppose that there are 15 antennas in a store, of which 3 are defective. Assume all the defectives and all the functional antennas are indistinguishable. If we lay all the antennas down in a row, how many linear orderings are there in which no two defectives are consecutive?

Step-by-step explanation:

Total number of antenna is 15

Defective antenna is 3

The functional antenna is 15-3=12.

Now, if no two defectives are to be consecutive, then the spaces between the functional antennas must each contain at most one defective antenna.

So,

We line up the 13 good ones, and see where the bad one will fits in

__G __ G __ G __ G __ G __G __ G __ G __ G __ G __ G __ G __G __

Each of the places where there's a line is an available spot for one (and no more than one!) bad antenna.

Then,

There are 14 spot available for the defective and there are 3 defective, so the arrange will be combinational arrangement

ⁿCr= n!/(n-r)!r!

The number of arrangement is

14C3=14!/(14-3)!3!

14C3=14×13×12×11!/11!×3×2

14C3=14×13×12/6

14C3=364ways

P(x<10) =0

Step-by-step explanation:

M =1/μ

M= 1/15 =0.066667

X- Exp (0.066667)

Where: X is the random variable for number of days

P (X<x) = 1 - e^-mx

P(X - x) = 1 - e^-0.06666 x10

P(x<10) = 0