In the above scenario, when there were 53 numbers to choose from, and players pick 6 numbers, the total number of combinations was 22,957,480.  The calculations for this are shown step by step in the above image using the Combination formula. Use the information above to answer the following questions: 1. What are the odds of winning in this scenario with one winning ticket? 2. What would the odds of winning be if there are 54 numbers instead of 53 with one winning ticket?3. Suppose the Lottery Commission offers a Pick 5 game, where the players choose 5 out of 40 numbers.  What would the odds of winning be if there are three winning tickets in this drawing?

1. What are the odds of winning in this scenario with one winning ticket?

To find the odds of winning, we need to calculate the probability of picking the correct 6 numbers out of 53. The total number of combinations is given as 22,957,480. The probability of winning with one ticket is the reciprocal of the total number of combinations.

$\text{Probability of winning} = \frac{1}{22,957,480}$

The odds of winning are typically expressed as the ratio of the number of successful outcomes to the number of unsuccessful outcomes. Therefore, the odds of winning are:

$\text{Odds of winning} = \frac{1}{22,957,479}$

2. What would the odds of winning be if there are 54 numbers instead of 53 with one winning ticket?

First, we need to calculate the total number of combinations when there are 54 numbers to choose from and players pick 6 numbers. We use the combination formula:

$\binom{54}{6} = \frac{54!}{6!(54-6)!}$

Calculating this:

$\binom{54}{6} = \frac{54 \times 53 \times 52 \times 51 \times 50 \times 49}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 25,827,165$

The probability of winning with one ticket is the reciprocal of the total number of combinations:

$\text{Probability of winning} = \frac{1}{25,827,165}$

The odds of winning are:

$\text{Odds of winning} = \frac{1}{25,827,164}$

3. Suppose the Lottery Commission offers a Pick 5 game, where the players choose 5 out of 40 numbers. What would the odds of winning be if there are three winning tickets in this drawing?

First, we need to calculate the total number of combinations when there are 40 numbers to choose from and players pick 5 numbers. We use the combination formula:

$\binom{40}{5} = \frac{40!}{5!(40-5)!}$

Calculating this:

$\binom{40}{5} = \frac{40 \times 39 \times 38 \times 37 \times 36}{5 \times 4 \times 3 \times 2 \times 1} = 658,008$

If there are three winning tickets, the probability of winning with one ticket is the number of winning tickets divided by the total number of combinations:

$\text{Probability of winning} = \frac{3}{658,008}$

The odds of winning are:

$\text{Odds of winning} = \frac{3}{658,005}$