Acoin is tossed two times. the set of outcomes is
given as s = {hh, ht, th, tt}
let x represent the number of times heads
occurs.
complete the table.
a=
b =
c=
done

Step-by-step explanation:

Each probability can be written as y/4, since there are four different possible outcomes.  Also, it's worth noting that if you add the three  up they will equal 4/4 or 1.

Put super simply, of HH, HT, TH, TT how many have zero Hs, how many have 1 H and how many have 2 Hs?  These are the answers to a, b and x respectively.  

show us the graphs dude

step-by-step explanation:

the answer to the expression is 83