Question

A player throws an ordinary dice with faces numbered from 1 to 6, if he throws 1 he has second throw. Let n denote his total score find p.m.f of random variables N and find E(n)

Answer 1

Solution:

The person gets to throw the dice once which is numbered from 1 -6. The score on first throw will be

S = { 2, 3, 4, 5 ,6 }

Each will have a probability = 1/ 6.

We do not include 1 because it will be added in the second throw

If the person rolls a 1 then he /she gets another throw Then the total score will be 1+ above all the listed scores

S = {2,3,4,5,6,7}

Here the probability of getting any score will be again 1/6.

But together getting 1 and another = 1/6 * 1/ 6 = 1 / 36 ............since each throw is independent.

Therefore score from 2 -6 can happens twice with difference possiblities and 7 only once.

P(total 2) = P(getting 2 on 1st )+ P(getting 1 on both rolls)

=1 / 6 + 1/6 *1/6

= 0.1944

P(total 7) = P(getting 1 on 1st and 6 on 2nd)

= 1 / 36

	Scores	No. of times sco	P(first time	P(second time)	Total P
2   & 1,1	2	2	0.167	0.028	0.1944
3   & 1,2	3	2	0.167	0.028	0.1944
4   & 1,3	4	2	0.167	0.028	0.1944
5   & 1,4	5	2	0.167	0.028	0.1944
6   & 1,5	6	2	0.167	0.028	0.1944
- & 1,6	7	1	-	0.028	0.0278
					1

N	P(N)	N *P(N)
2	0.1944	0.3889
3	0.1944	0.5833
4	0.1944	0.7778
5	0.1944	0.9722
6	0.1944	1.1667
7	0.0278	0.1944
Total	1	4

E(X) =

E(X) = 4

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