To calculate the probability of winning second prize in the lottery, we need to determine two things:

1. The number of ways the player can match 5 out of the 6 numbers they chose.
2. The total number of ways 6 numbers can be chosen out of the 47 possible numbers.

Step 1: Total number of possible combinations

The total number of ways to choose 6 numbers out of 47 is calculated using the combination formula:

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Where:

• $n = 47$ (total numbers available),
• $k = 6$ (numbers chosen).

This gives us the total number of ways to choose 6 numbers from 47.

Step 2: Favorable outcomes (matching 5 of 6 numbers)

For the second prize, the player must match 5 out of the 6 chosen numbers. The player can:

• Choose 5 numbers from the 6 correct numbers, and
• Choose 1 number from the 41 remaining numbers (since 6 numbers are drawn, the remaining 41 numbers are incorrect).

So, we need to calculate:

• $\binom{6}{5}$ for selecting 5 correct numbers,
• $\binom{41}{1}$ for selecting 1 incorrect number.

Step 3: Probability formula

The probability is given by:

$P(\text{second prize}) = \frac{\binom{6}{5} \times \binom{41}{1}}{\binom{47}{6}}$

Let me calculate this.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, I can guide you on how to calculate the probability manually:

1. Total combinations of choosing 6 numbers from 47: $\binom{47}{6} = \frac{47!}{6!(47-6)!}$

2. Favorable outcomes: Choose 5 correct numbers from 6 and 1 incorrect from 41: $\binom{6}{5} \times \binom{41}{1}$

Finally, the probability is the ratio of favorable outcomes to total combinations: $P = \frac{\binom{6}{5} \times \binom{41}{1}}{\binom{47}{6}}$

Would you like any additional help with this?