An experiment consists of tossing a die and then flipping a coin once if the number on the die is even. If the number on the die is odd, the coin is flipped twice. Using the notation 4HH, for example, to denote the outcome that the die comes up 4 and then the coin comes up heads, and 3HTHT to denote the outcome that the die comes up 3 followed by a head and then a tail on the coin, construct a tree diagram to show the 18 elements of the sample space SS. For the sample space SS (a) list the elements corresponding to the event AA that
a number less than 3 occurs on the die;
(b) list the elements corresponding to the event BB that
two tails occur;
(c) list the elements corresponding to the event A′A
(d) list the elements corresponding to the event A′∩BA ∩B;
(e) list the elements corresponding to the event A∪BA∪B.

Answer and Step-by-step explanation:

Possible outcomes from dice are:   Even = 2,4,6 and odd = 1,3,5

For every even number coin is flipped once possible outcomes are = (H, T)

For every odd number coin is flipped twice possible outcomes are = (HH, HT, TH, TT)

Total outcomes from dice and coins = even = (2H, 2T, 4H, 4T, 6H, 6T)

Odd= [1(HH, HT, TH, TT) , 3(HH, HT, TH, TT), 5(HH, HT, TH, TT)]

Total sample space = 18

             (a) List the elements corresponding to the event AA that a number less than 3 occurs on the die;

        A = (1HH, 1HT, 1TH, 1TT, 2H, 2T)

(b) List the elements corresponding to the event BB that two tails occur;

         B= (1TT, 3TT, 5TT)

(c) List the elements  to the event A′A

          A’= ( 3HH, 3HT, 3TH, 3TT, 4H, 4T, 5HH, 5HT, 5TH, 5TT, 6H, 6T)

(d) List the elements to the event A′∩BA ∩B;

            A’∩B= (3TT, 5TT)

(e) List the elementsto the event A∪BA∪B.

                AUB=  (1HH, 1HT, 1TH, 1TT, 2H, 2T, 3TT, 5TT)

umumumumum

im not sure

c - they will have the same distribution

step-by-step explanation:

this is the only one that could make sense. : )