## Can you help? Probability of failure given hypotheticals

Hi, I've volunteered to try work something out, but my stats is 40 years out of date.
A company called "WMD" has fitted four wheels to 100,000 toy carts, but it seems like they had a pile of good old wheels, but also mixed that pile up with some new wheels with potential problems. I volunteered to work out the chance of a failure. Can you check my maths? Did I do it the right way?
Any help appreciated.

All numbers are hypothetical. What follows is what I have worked out, but I am not really that confident I have got the logic right.

Risk assessments are based on knowns, but all we have available at this time are unknowns. Every basis in this formula, which is loosely based on “Drakes Formula” is unknown. Numbers are purely hypothetical, with the aim of replacing hypothetical numbers with drawn observations and discoveries.

For clarity, old wheels are good wheels, and new wheels are flawed wheels.

For example, there is the suspicion that WMD have stocks of both new wheels and old wheels, and so that a calculation can be made, we here assume that they have 40% correct wheels, and 60% new wheels. Hence the chance of WMD fitting a new wheel is 60%. Lets pretend that there is unlimited stock, and we don't have to take into account depletion of, say, good wheels, when one is taken from the pile.

As information comes to light, we can replace hypothetical numbers with real numbers, just as “Drake’s Formula” has had hypothetical numbers replaced with real data recently, as indirect observation of astronomical objects has become possible.

Hypothetical numbers:

Total number of wheels available – infinite (ie we do not have to calculate from the actual number of good or bad wheels left in the bucket)

Risk of fitting incorrect wheel – 60%

Failure rate of new wheels in, say, a year’s use – 10%

Number of items produced - 100,000

Number of items distributed to people – 50%

Mathematical risk assessment based on hypothetical numbers:

When WMD fit a wheel, there is a 60% chance they fit an incorrect wheel.
With a failure rate of 10%, The risk of fitting a wheel that will fail is calculated according to the chances of a good wheel being fitted:

The risk of a piece of equipment failing, given there are 4 wheels per device, and any one failing is catastrophic (per year):

(1) Four correct wheels fitted:
(one possible configuration)
.6 x .6 x .6 x .6 = .1296 These items are safe. Nearly 13% of items are safe.

(2) Three incorrect wheels fitted:
(four possible configurations)
.4 x .6 x .6 x .6 = 0864 x4 = .3456 34.5% of sales will have one incorrect wheel. With a failure rate of 10%, 3.4% of items will fail.

(3) Two correct wheels fitted:
(six possible configurations)
.4 x .4 x.6 x .6 = .0576 x6 = .3456. 34.5 % of sales will have two correct wheels.
With a failure rate of 20% (two wheels to fail), 6.9% of items will fail.

(4) Three correct wheels fitted:
(four possible configurations)
.4 x .4 x .4 x .6 = .0384 x 4 = .1536 15.4 % of sales will have three incorrect wheels.
With a failure rate of 30% (three wheels to fail), 4.05% of sales will fail.

(5) Four incorrect wheels fitted:
(one possible configuration)
.6 x .6 x .6 x .6 = .1296. 13 % of sales will have four incorrect wheels.
With a failure rate of 10%, 1.3% of items will fail.

Chance of a sold item failing = (1) + (2) + (3) + (4) + (5)

For any item, 0 + 3.4 + 6.9 + 4.05 + 1.3

For all sales, 15.65 % of items will catastrophically fail in the first year.

Number of items in the field that will fail in the first year.

50,000 @ 15.65%

7,825 items will fail catastrophically in the first year of use.

If the remaining 50% in storage are distributed to patients

100,000 @ 15.65%

15,650 items will catastrophically fail in the first year of use.