In this problem, you are placing a $1 straight bet on a lottery and need to find two probabilities:

1. $P(A)$: the probability of winning the lottery (event $A$).
2. $P(\bar{A})$: the probability of not winning the lottery (the complement of $A$).

Finding $P(A)$ (Probability of winning):

• There are 5,040 different possible ways to select the four digits (with repetition allowed).
• Only one of these four-digit numbers will win.

Therefore, the probability of winning $P(A)$ is:

$P(A) = \frac{1}{5040}$

Finding $P(\bar{A})$ (Probability of not winning):

The complement of event $A$ is not winning the lottery. The probability of not winning is the opposite of winning:

$P(\bar{A}) = 1 - P(A) = 1 - \frac{1}{5040}$

$P(\bar{A}) = \frac{5039}{5040}$

Final Results:

• $P(A) = \frac{1}{5040}$
• $P(\bar{A}) = \frac{5039}{5040}$

Would you like further details or have any questions?

Here are 5 related questions:

1. How would the probability change if there were 10,000 different combinations?
2. What is the probability of winning twice in a row in the same lottery?
3. What is the expected value of a single $1 bet in this lottery?
4. How do probabilities change if repetition is not allowed in choosing digits?
5. What is the probability of winning at least once if you buy 10 tickets?

Tip: Complementary events are a useful tool when calculating probabilities indirectly!