Final answer:
The distribution function for a random variable representing the winnings or losses of a dice game depends on the specific values and probabilities assigned to the outcomes. For a fair die, probabilities are straightforwardly distributed, while a weighted die requires solving for a probability that ensures the total probability sums to one.
Step-by-step explanation:
The student has asked us to find the distribution function for a random variable X representing the amount a player wins or loses when tossing a fair die under certain conditions in two different scenarios. In the first scenario, X can take on the values -$3, $1, and $2 depending on the outcome of the die roll. The probabilities for these values would be P(X=-$3) = 1/3, P(X=$1) = 1/2, and P(X=$2) = 1/6 respectively. In the second scenario, we are told the die is weighted such that the probability of rolling a six is twice that of the other numbers. In this case, the distribution of X, where X is the number shown by the die, will have different probabilities for each outcome. Assuming the total probability is 1, each face 1 through 5 has a probability of p, and face 6 has a probability of 2p. To solve for p, we use the fact that the sum of probabilities is equal to 1; thus, 5p + 2p = 1, giving us p = 1/7. Therefore, the probabilities of rolling each number are: P(X=1) = 1/7, P(X=2) = 1/7, P(X=3) = 1/7, P(X=4) = 1/7, P(X=5) = 1/7, and P(X=6) = 2/7.