Answer:
See explanantion
Step-by-step explanation:
Solution:-
- Let the random variable X be defined as the amount Godfrey gains in one turn.
- He wins by getting ( 6 , 6 ) = 20 p
- He wins by getting either ( 6 & any number) = 10 p
- Otherwise he wins nothing..
- Entry fee = 5p
- The entire sample space (S) is defined as:
1 2 3 4 5 6
1 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
3 . . . . . .
4 . . . . . .
5 . . . . . .
6 . . . . . .
- The total number of outcome in sample space = 36
Godfrey wins, 20p, X = (20-5) p:
Wins this amount when he gets double sixes i.e ( 6 , 6 ). There is only one outcome out of 36 outcomes in the sample space given above.
So the probability of Godfrey gaining X = 15 p would be:
p ( X = 15p) = ( 6 , 6 ) / S
= 1 / 36
Godfrey wins, 10p, X = (10-5) p:
Wins this amount when he gets following number:
( 1 , 6 ) , ( 2 , 6 ) , ( 3, 6) , ( 4 ,6 ) , ( 5 , 6 )
( 6 , 1 ) , ( 6 , 2 ) , ( 6 , 3 ) , ( 6 , 4 ) , ( 6 , 5 ) = 10 outcomes
There are ten outcome out of 36 outcomes in the sample space given above.
So the probability of Godfrey gaining X = 5 p would be:
p ( X = 5p) = ( 10 / 36 )
Godfrey wins nothing, X = ( 0 - 5 )p:
All the other possibilities = 15 from the sample space which exclude both a single "6" and double " 6 " gives him a loss of entry fee = 5.
So the probability of Godfrey loosing X = -5 p would be:
p ( X = -5 ) = 15 / 36
- The probability distribution for random variable "X" is:
X : -5 5 15
P(X): 15/36 10/36 1/36
