Maria is playing a game of chance at the Hibiscus festival, costing $1 for each game. In the game two fair dice are rolled and the sum of the numbers that turn up is found. If the sum is seven, then Maria wins $5. Otherwise Maria loses her money.

a) Please can you Construct Maria’s probability distribution for gaining.
Thank you for your help.

To construct Maria's probability distribution for gaining, we need to calculate the probability of each possible outcome and the corresponding gain or loss.

Outcome 1: The sum of the numbers rolled is 7.
There are 6 possible ways to get a sum of 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). So, the probability of getting a sum of 7 is 6/36 = 1/6. Since Maria wins $5 in this case, the gain is $5.

Outcome 2: The sum of the numbers rolled is not 7.
There are 36 possible outcomes when rolling two fair dice. We have already accounted for the 6 outcomes with a sum of 7, so there are 36 - 6 = 30 outcomes with different sums. The probability for each of these outcomes is 1/36. In all these cases, Maria loses $1.

Therefore, the probability distribution for gaining is as follows:
- Gain of $5 with a probability of 1/6 (sum of 7)
- Gain of -$1 with a probability of 30/36 (all other sums)

Remember that the probabilities in a probability distribution must add up to 1. Let's verify this:
Probability of gaining $5: 1/6
Probability of losing $1: 30/36 (which simplifies to 5/6)

1/6 + 5/6 = 6/6 = 1

So, the probability distribution is valid.

There are 36 possible outcomes which can be grouped into 11 events E represented by the totals of 2 to 12, each with probability P(E).

For example construct the probability distribution table:
E P(E)
2 1/36 (only one way to make a sum of 2)
3 2/36 (two ways to make a sum of 3)
4 3/36 (3 ways to make 4: 1+3,2+2,3+1)
...
12 1/36

The gain for each event is G(E)= -1 except for E=7 where the gain is G(7)=5.

The expected gain is ∑G(E)P(E) for all E.