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We can solve this problem using the binomial probability formula since it involves multiple independent trials (whether or not a specific person is selected) with two outcomes (selected or not selected).

The probability of being selected for jury duty is \( p = 0.01 \), or 1%. The probability of not being selected is \( 1 - p = 0.99 \). There are 3 adults in the Anderson family, and we are asked to calculate the probabilities for different cases regarding how many of them are selected for jury duty.

The binomial probability formula is:

\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]

Where:
- \( P(X = k) \) is the probability that exactly \( k \) people are selected.
- \( n \) is the total number of trials (here, 3 adults).
- \( k \) is the number of people selected.
- \( p \) is the probability of being selected.
- \( \binom{n}{k} \) is the binomial coefficient, which counts the number of ways to choose \( k \) successes from \( n \) trials.

### a. Probability that none of the Andersons will be chosen

This corresponds to \( k = 0 \):

\[
P(X = 0) = \binom{3}{0} (0.01)^0 (0.99)^3 = 1 \times 1 \times (0.99)^3 = (0.99)^3
\]

Calculating \( (0.99)^3 \):

\[
P(X = 0) = 0.970299
\]

So, the probability that none of the Andersons will be chosen is approximately 0.9703 or 97.03%.

### b. Probability that exactly one of the Andersons will be chosen

This corresponds to \( k = 1 \):

\[
P(X = 1) = \binom{3}{1} (0.01)^1 (0.99)^2 = 3 \times 0.01 \times (0.99)^2
\]

Calculating \( (0.99)^2 \) and the whole expression:

\[
P(X = 1) = 3 \times 0.01 \times 0.9801 = 0.029403
\]

So, the probability that exactly one of the Andersons will be chosen is approximately 0.0294 or 2.94%.

### c. Probability that exactly two of the Andersons will be chosen

This corresponds to \( k = 2 \):

\[
P(X = 2) = \binom{3}{2} (0.01)^2 (0.99)^1 = 3 \times 0.01^2 \times 0.99
\]

Calculating the values:

\[
P(X = 2) = 3 \times 0.0001 \times 0.99 = 0.000297
\]

So, the probability that exactly two of the Andersons will be chosen is approximately 0.000297 or 0.0297%.

### d. Probability that all three Andersons will be chosen

This corresponds to \( k = 3 \):

\[
P(X = 3) = \binom{3}{3} (0.01)^3 (0.99)^0 = 1 \times 0.01^3 \times 1 = 0.000001
\]

So, the probability that all three Andersons will be chosen is 0.000001 or 0.0001%.

### Summary of probabilities:
- a. None of the Andersons will be chosen: 97.03%
- b. Exactly one of the Andersons will be chosen: 2.94%
- c. Exactly two of the Andersons will be chosen: 0.0297%
- d. All three Andersons will be chosen: 0.0001%

Would you like more details or have any further questions? 

Here are 5 related questions to expand your understanding:
1. How would the probabilities change if the selection probability increased to 5%?
2. What is the probability that at least one Anderson will be chosen?
3. How does the binomial distribution apply to real-world events like jury selection?
4. What is the expected number of Andersons chosen for jury duty?
5. How would you calculate the probability for 4 adults in the Anderson family?

**Tip:** To find the probability of "at least one" event happening, subtract the probability of the complement (in this case, none of the Andersons being chosen) from 1.

In the town of Hickoryville, an adult citizen has a 1% chance of being selected for jury duty next year. If jury members are selected at random, and the Anderson family includes three adults, determine the probability that: 
a. none of the Andersons will be chosen. 
b. exactly one of the Andersons will be chosen. 
c. exactly two of the Andersons will be chosen. 
d. all three Andersons will be chosen.

Explore the probability of the Anderson family members being selected for jury duty in Hickoryville, using the binomial probability formula. Determine the likelihood of none, one, two, or all three family members being chosen, given a 1% chance per adult.

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