To find the odds of winning a lottery game where players choose 5 out of 40 numbers, we first need to calculate the total number of ways to choose 5 numbers from 40. This can be calculated using the combination formula:
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
Where:
- \( n \) is the total number of items (in this case, 40),
- \( k \) is the number of items to choose (in this case, 5).
So, we calculate \( C(40, 5) \):
\[ C(40, 5) = \frac{40!}{5!(40-5)!} = \frac{40!}{5! \cdot 35!} \] \[ = \frac{40 \times 39 \times 38 \times 37 \times 36}{5 \times 4 \times 3 \times 2 \times 1} = \frac{40 \times 39 \times 38 \times 37 \times 36}{120} \]
Calculating the numerator:
- \( 40 \times 39 = 1560 \)
- \( 1560 \times 38 = 59280 \)
- \( 59280 \times 37 = 2195760 \)
- \( 2195760 \times 36 = 78969600 \)
Now divide by \( 120 \):
\[ C(40, 5) = \frac{78969600}{120} = 657280 \]
So, there are 657280 different combinations of 5 numbers from 40.
Given that there are 3 winning tickets, the odds of winning would be calculated as follows:
Odds of winning = Number of ways to win / Total combinations
\[ \text{Odds of winning} = \frac{3}{657280} \]
To express this in the form of odds against winning, we take the opposite, which is:
\[ \text{Odds against winning} = \text{Total combinations} - \text{Winning combinations} : \text{Winning combinations} \] \[ = (657280 - 3) : 3 = 657277 : 3 \]
However, we want the standard format of expressing odds in the simplest format. But looking at the answer choices, let's calculate the overall odds correctly:
Winning odds = 1: (657280 / 3) = 1: 219093.33 (approximately).
As none of these match, we need to find the odds per winning ticket, given there are 3 total winning tickets:
1 winning ticket out of 657280 combinations:
Thus, the probability of winning with 3 tickets would be approximately expressed as:
1: 219093.33 dividing by total tickets from winning tickets divided by total combinations.
Given the answer choices, we align more closely to:
To summarize:
- You have 3 winning combinations out of 657280.
- 1 winner per 219093 situations.
This doesn't align directly to any of the presented answers, so we look again.
Given further info on format and understanding, if we calculate against:
Thus the nearest option by logical proportions go and we see indeed: Answer likely more proximate to C: 1:658,005 as the probable extraction odds-wise calculation fit for standards.
Given the complexity, I structured stepwise to see where exact odds are framed compared to choices from a winner ticket proportion across what most platforms would acknowledge within. If deeper computational detailing on combinations, please advise!
1 answer