Question

Two coins are tossed and a (6-sided) die is rolled. Describe a sample space(probability space), together with the probability, on which such a situation can be modeled. Find the probability mass function of the random variable whose value is the sum of the number on the die and the total number of heads.

Answer 1

We can have the die outcomes of 1 -6 and the following for coins (HH, HT, TH, TT)

Coins	Die	Outcome (Sample Space)
HH	1	1HH
HT	1	1HT
TH	1	1TH
TT	1	1TT
HH	2	2HH
HT	2	2HT
TH	2	2TH
TT	2	2TT
HH	3	3HH
HT	3	3HT
TH	3	3TH
TT	3	3TT
HH	4	4HH
HT	4	4HT
TH	4	4TH
TT	4	4TT
HH	4	4HH
HT	4	4HT
TH	4	4TH
TT	4	4TT
HH	5	5HH
HT	5	5HT
TH	5	5TH
TT	5	5TT
HH	6	6HH
HT	6	6HT
TH	6	6TH
TT	6	6TT

There is total of 24 elements

If we let the events of getting heads, then we can have 0 ,1 or 2 heads

Having '0' heads has 6 possibilites = 6 / 24

= 0.25

TT1

TT2

TT3

TT4

TT5

TT6

Having 1 head has 12 possibilties = 12 / 24

= 0.5

We can easily get this by looking at no heads. First we look at the scenario of getting head on fisrt toss and tail on 2nd. This will a have 6 possibilites if we roll the die as well.

Then we look at the scenario of getting tail on fisrt toss and head on 2nd. This will a have 6 possibilites if we roll the die as well.

Getting 2 heads = 6 / 24

=0.25

Like no heads. Instead of TT we wil have HH

So the pmf is as follows

x (heads)	P(x)
0	0.25
1	0.5
2	0.25
Total	1

If we were to look at the sum of die we could have 6 possibilites 1 - 6 and for each sum we would have 4 out of 24 possibilites for the toss (HH, HT, TH, TT)

So each sum would have the possiblity of 4 / 24 = 1 / 6

x (sum on die)	P(x)
1	1/6
2	1/6
3	1/6
4	1/6
5	1/6
6	1/6
Total	1