A system is comprised of 3 components, each of which is either "working", "iﬀy", or "failed". The system will work if any two components are working or if none of the components are failed. Let W be the event that the system is working. Let A be the event that at least one component is iﬀy. a). Deﬁne carefully a sample space, S, for this experiment. How many elements does it have? b). In terms of your sample space, give the events W, A, and W ∩ A.fa

I already know that there are 27 elements, but don't understand how did they do part b) for events W?

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MrRoyal MrRoyal · Answer 1 · 2020-02-14T15:12:31+01:00

Answer:

a. 27 elements

b. W= 11 elements

A = 19 elements

W n A = 19 elements

Step-by-step explanation:

(a).

Let W= Working

Let I = Iffy

Let F = Failed

Let SS = { WWW, WWI, WWF, WIW, WII, WIF, WFW, WFI, WFF, IWW, IWI, IWF, ..., FFF}

This represents the total samples of elements that has a 3--lettered spelling of words with letters W,I and F in no particular order.

Each of the elements has 3 * 3 total elements

For the 3 alphabets, there are 3 * (3 * 3) elements

Thus, there are |SS| = 3³ = 27 elements of SS

(Eaach of 3 positions in the spelling can be any of 3 letters).

(b).

The event W is the set W = { WWW, WWI, WIW, IWW, WWF, WFW, FWW, IIW, IWI, WII,III }.

Thus, W has |W| = 11 elements.

The event A is the set A = { III, IIW, IWI, WII, IIF, IFI, FII, IFF, FIF, FFI, IWW, WIW, WWI, IFW, FIW, FWI, IWF, WIF, WFI }.

This, A has |A| = 19.

Since there are 2³ elements having only letters W and F

2³ = 8

Thus, there are 27 −8 = 19 elements of Sample Space that have at least one “I”