To determine the probability of winning the jackpot in this lottery, we need to calculate two probabilities:

1. The probability of correctly matching all five distinct numbers drawn from the white balls (1 through 43).
2. The probability of correctly matching the number on the gold ball (1 through 34).

Step 1: Calculate the number of possible combinations of white balls.

• The number of ways to choose 5 distinct numbers from 43 white balls is given by the combination formula:

$\binom{43}{5} = \frac{43!}{5!(43-5)!} = \frac{43!}{5! \times 38!}$

Calculating this:

$\binom{43}{5} = \frac{43 \times 42 \times 41 \times 40 \times 39}{5 \times 4 \times 3 \times 2 \times 1} = 962598$

So, there are 962,598 possible combinations of the white balls.

Step 2: Calculate the probability of matching the gold ball.

• There are 34 possible numbers for the gold ball, so the probability of picking the correct one is:

$\frac{1}{34}$

Step 3: Calculate the overall probability.

• The probability of winning the jackpot is the product of the probability of matching all five white balls and the probability of matching the gold ball:

$\text{Probability of winning} = \frac{1}{\binom{43}{5}} \times \frac{1}{34} = \frac{1}{962598} \times \frac{1}{34}$

Simplifying the product:

$\text{Probability of winning} = \frac{1}{962598 \times 34} = \frac{1}{32728332}$

Final Answer:

The probability of winning the jackpot with one ticket is:

$\frac{1}{32,728,332}$

Would you like more details or have any questions?

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Related Questions:

1. How do you calculate the probability of winning with only the white balls matched?
2. What if the lottery had 6 white balls instead of 5? How would that change the probability?
3. How does the probability change if the number of white balls is reduced to 40?
4. What are combinations, and how are they different from permutations?
5. Can you provide an example of a similar probability problem involving dice?

Tip: Always remember that when calculating combinations, the order of selection does not matter, which simplifies many probability problems!