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Part 1
43​% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is ​ (a) exactly​ five, (b) at least​ six, and​ (c) less than four

This problem can be modeled using the **binomial probability distribution** since we have a fixed number of trials (selecting 10 adults), two possible outcomes for each trial (having little confidence in newspapers or not), and a constant probability of success (43%).

Let:
- \( n = 10 \) (the number of trials or selected adults),
- \( p = 0.43 \) (the probability of success, i.e., an adult having very little confidence),
- \( q = 1 - p = 0.57 \) (the probability of failure, i.e., an adult not having very little confidence),
- \( X \) = number of adults with little confidence in newspapers, which follows a binomial distribution \( X \sim \text{Binomial}(n=10, p=0.43) \).

The probability mass function for a binomial distribution is given by:

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

where:
- \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \) is the binomial coefficient, and
- \( k \) is the number of successes (adults having little confidence).

### (a) Probability that exactly five adults have very little confidence in newspapers:
We need to find \( P(X = 5) \).

\[
P(X = 5) = \binom{10}{5} (0.43)^5 (0.57)^{5}
\]

First, compute the binomial coefficient \( \binom{10}{5} = \frac{10!}{5!(10-5)!} = 252 \).

Now compute the probability:

\[
P(X = 5) = 252 \cdot (0.43)^5 \cdot (0.57)^5 \approx 252 \cdot 0.005184 \cdot 0.037865 \approx 0.0494
\]

So, \( P(X = 5) \approx 0.0494 \).

### (b) Probability that at least six adults have very little confidence in newspapers:
We need to find \( P(X \geq 6) \).

This is the sum of probabilities from \( X = 6 \) to \( X = 10 \):

\[
P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)
\]

Each of these terms can be calculated using the binomial probability formula as in part (a). Alternatively, this can be found using a cumulative binomial probability table or software (e.g., a binomial calculator).

Using a calculator:

\[
P(X \geq 6) \approx 0.2324
\]

### (c) Probability that fewer than four adults have very little confidence in newspapers:
We need to find \( P(X < 4) \).

This is the sum of probabilities from \( X = 0 \) to \( X = 3 \):

\[
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]

Using a calculator:

\[
P(X < 4) \approx 0.3686
\]

### Summary of answers:
(a) \( P(X = 5) \approx 0.0494 \)

(b) \( P(X \geq 6) \approx 0.2324 \)

(c) \( P(X < 4) \approx 0.3686 \)

Would you like further explanation on any part of this? 

Here are 5 related questions to expand your understanding:

1. How would the probabilities change if the sample size increased to 15?
2. What is the expected number of U.S. adults who have very little confidence in newspapers?
3. How do you calculate the variance of a binomial distribution?
4. What would the probability be if exactly two adults had very little confidence in newspapers?
5. How would the results differ if the confidence level in newspapers was only 30%?

**Tip:** Always remember to check if your binomial problem satisfies the conditions: fixed number of trials, two possible outcomes, and constant probability per trial.

43% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four.

Solve a probability problem using the binomial distribution where 43% of U.S. adults have very little confidence in newspapers. Calculate the probability for exactly five, at least six, and fewer than four adults out of 10. Suitable for Grades 10-12.

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