Question

Godfrey plays a game in which he throws two fair six sided dice. If he rolls two sizes, he wins 20p, if he rolls one six, he wins 10p, otherwise he wins nothing. He pays 5p to enter.

Write out the probability distribution of X, the amount Godfrey gains in one turn.


(3 marks)

Answer 1

See explanantion

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

Answer 2

The Probabilty distribution for the amount Godfrey gains in one turn is then given as

X ||| P(X)

15p | 0.0278

5p | 0.278

-5p | 0.6942

Explanation:

If random variable X represents the amount Godfrey gains in one turn.

There are 3 different possible outcomes for X.

- Godfrey pays 5p to enter the game and gets two sixes and wins 20p.

Net gain = 15p

Probability of getting two sixes from two fair dice

= (number of outcomes with two sixes) ÷ (total number of outcomes)

number of outcomes with two sixes = 1

total number of possible outcomes = 36

Probability of getting two sides from two fair dice = (1/36) = 0.0278

- Godfrey pays 5p to enter the game and gets only one six and wins 10p.

Net gain = 5p

Probability of getting one six from either of two fair dice

= (number of outcomes with one six) ÷ (total number of outcomes)

number of outcomes with one six = 2 × n[(6,1), (6,2), (6,3), (6,4), (6,5)] = 2 × 5 = 10

total number of possible outcomes = 36

Probability of getting two sides from two fair dice = (10/36) = 0.278

- Godfrey pays 5p to enter the game and doesn't win anything

Net gain = -5p

Probability of not getting two sixes or one six.

= 1 - [(Probability of getting two sixes) + (Probability of getting one six on.wither dice)]

= 1 - 0.0278 - 0.278 = 0.6942

Probability of getting not getting two sixes or one six = 0.6942

The Probabilty distribution for the amount Godfrey gains in one turn is then given as

X ||| P(X)

15p | 0.0278

5p | 0.278

-5p | 0.6942

Hope this Helps!!!