2 A sample of two items is selected without replacement from a batch. Describe the (ordered) sample space for each of the following batches: a. The batch contains the items {a, b, c, d}. b. The batch contains the items {a, b, c, d, e, f , g}. c. The batch contains 4 defective items and 20 good items. d. The batch contains 1 defective item and 20 good items.

a.{a b, a c, a d, b c, b d, c d, b a, d c ,d b, c a, d a, c b}

b.{a b, a c, a d, a e ,a f, a g, b c, bd,be,bf,bg,cd,ce,cf,cg,fg,ba,ca,da,ea,fa,ga,cd,db,eb,fb,gb,dc,ec,fc,gc,fe,ge,gf,de,df,dg,ed,fd,gd,ef,eg}

c.{d d, dg, g d, g g}

d.{dg, g d, gg}

Step-by-step explanation:

We have to find the sample space

a. We are given items

{a, b, c, d}

Therefore, the sample space

{a b, a c, a d, b c, b d, c d, b a, d c ,d b, c a, d a, c b}

b. Items

Therefore, the sample space

{a b, a c, a d, a e ,a f, a g, b c, bd,be,bf,bg,cd,ce,cf,cg,fg,ba,ca,da,ea,fa,ga,cd,db,eb,fb,gb,dc,ec,fc,gc,fe,ge,gf,de,df,dg,ed,fd,gd,ef,eg}

c.

Defective items (d)=4

Good items(g)=20

Therefore, the sample space

{d d, dg, g d, g g}

d.

Defective items (d)=1

Good items(g)=20

Therefore, the sample space

{d g, g d, g g}

the sum of the lengths of any two sides of a triangle is greater than the length of the third side. if you take the three sides of a triangle and add them in pairs, the sum is greater than (not equal to) the third side. if that is not true, then it is not possible to construct a triangle with the given side lengths.

step-by-step explanation: