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!