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Part 1
A state lottery game requires that you pick 6 different numbers from 1 to 43.
​(a) Assuming that order is​ unimportant, what is the probability of picking all 6 numbers correctly to win the big​ prize?
​(b) What is the probability of picking exactly 5 of the 6 numbers​ correctly?

​(Use scientific notation. Use the multiplication symbol in the math palette as needed. Round to three decimal places as​ needed.)




To solve this problem, we'll use combinatorics to calculate the probabilities.

### Part (a): Probability of Picking All 6 Numbers Correctly

1. **Total combinations of picking 6 numbers from 43:**
   The number of ways to choose 6 numbers from 43 (order does not matter) is given by the binomial coefficient:

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

   where \( n = 43 \) and \( k = 6 \).

   \[
   \binom{43}{6} = \frac{43!}{6!(43-6)!} = \frac{43!}{6! \cdot 37!}
   \]

   Calculating this:

   \[
   \binom{43}{6} = \frac{43 \times 42 \times 41 \times 40 \times 39 \times 38}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 609,645
   \]

2. **Probability of picking all 6 numbers correctly:**
   Since there is only 1 winning combination, the probability \( P \) is:

   \[
   P = \frac{1}{\text{Total combinations}} = \frac{1}{609,645} \approx 1.638 \times 10^{-6}
   \]

### Part (b): Probability of Picking Exactly 5 of the 6 Numbers Correctly

1. **Choosing 5 winning numbers:**
   You can choose 5 winning numbers from the 6 winning numbers:

   \[
   \binom{6}{5} = 6
   \]

2. **Choosing 1 losing number:**
   You need to choose 1 losing number from the remaining 37 numbers (since 43 total - 6 winning = 37 losing):

   \[
   \binom{37}{1} = 37
   \]

3. **Total ways to pick exactly 5 correct:**
   Therefore, the total number of ways to pick exactly 5 correct and 1 incorrect is:

   \[
   \text{Total ways} = \binom{6}{5} \cdot \binom{37}{1} = 6 \cdot 37 = 222
   \]

4. **Probability of picking exactly 5 correct:**
   The probability \( P \) is then:

   \[
   P = \frac{\text{Total ways to pick 5 correct}}{\text{Total combinations}} = \frac{222}{609,645} \approx 3.636 \times 10^{-4}
   \]

### Summary of Results

- **(a)** Probability of picking all 6 numbers correctly: \( \approx 1.638 \times 10^{-6} \)
- **(b)** Probability of picking exactly 5 numbers correctly: \( \approx 3.636 \times 10^{-4} \)

Would you like any further details or explanations on these calculations? Here are some related questions you might consider:

1. How do the probabilities change if more numbers are picked?
2. What is the expected number of winning tickets sold in a typical lottery draw?
3. How would the probabilities differ if the total range of numbers increased?
4. What strategies might increase the odds of winning a lottery?
5. How do combinations differ from permutations in probability calculations? 

**Tip:** Always verify your calculations step by step to catch any possible errors!

A state lottery game requires that you pick 6 different numbers from 1 to 43. (a) Assuming that order is unimportant, what is the probability of picking all 6 numbers correctly to win the big prize? (b) What is the probability of picking exactly 5 of the 6 numbers correctly?

Explore the probabilities of a state lottery game where you must choose 6 numbers from a range of 1 to 43. This guide covers the calculations for winning the big prize and picking exactly 5 correct numbers. Learn about combinatorics and the binomial coefficient with precise examples.

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