Assume that your are randomly picking  students to survey by phone. Let X = the number of students who answer their phone. Use some form of technology to find the probability distribution of X given a probability  of success on a single trial. 
(Report answers accurate to 4 decimal places.)

k	P(X = k)
0	
1	
2	
3	
4	
5	


To find the probability distribution of \( X \), where \( X \) is the number of students who answer their phone out of 5 randomly picked students, we can model this situation using a **binomial distribution**. The binomial distribution is appropriate because:

1. There are a fixed number of trials (5 students).
2. Each trial has two outcomes: "success" (student answers) or "failure" (student doesn't answer).
3. The probability of success on a single trial is constant.

The probability mass function (PMF) of a binomial distribution is given by:

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

where:
- \( n \) is the total number of trials (5 in this case),
- \( k \) is the number of successes (students who answer),
- \( p \) is the probability of success on each trial (probability that a student answers their phone),
- \( \binom{n}{k} \) is the binomial coefficient, also written as \( \frac{n!}{k!(n-k)!} \).

To complete the probability distribution, we need a value for \( p \), the probability of success on a single trial. Let's assume \( p = 0.7 \) (for example), which means each student has a 70% chance of answering the phone.

Now, let’s calculate the probability distribution \( P(X = k) \) for \( k = 0, 1, 2, 3, 4, 5 \) using this formula.

### Calculations for \( p = 0.7 \):
- For \( X = 0 \):
  \[
  P(X = 0) = \binom{5}{0} \cdot (0.7)^0 \cdot (0.3)^5 = 1 \cdot 1 \cdot 0.00243 = 0.0024
  \]
- For \( X = 1 \):
  \[
  P(X = 1) = \binom{5}{1} \cdot (0.7)^1 \cdot (0.3)^4 = 5 \cdot 0.7 \cdot 0.0081 = 0.0284
  \]
- For \( X = 2 \):
  \[
  P(X = 2) = \binom{5}{2} \cdot (0.7)^2 \cdot (0.3)^3 = 10 \cdot 0.49 \cdot 0.027 = 0.1323
  \]
- For \( X = 3 \):
  \[
  P(X = 3) = \binom{5}{3} \cdot (0.7)^3 \cdot (0.3)^2 = 10 \cdot 0.343 \cdot 0.09 = 0.3087
  \]
- For \( X = 4 \):
  \[
  P(X = 4) = \binom{5}{4} \cdot (0.7)^4 \cdot (0.3)^1 = 5 \cdot 0.2401 \cdot 0.3 = 0.3602
  \]
- For \( X = 5 \):
  \[
  P(X = 5) = \binom{5}{5} \cdot (0.7)^5 \cdot (0.3)^0 = 1 \cdot 0.16807 \cdot 1 = 0.1681
  \]

### Final Probability Distribution (rounded to 4 decimal places):

\[
\begin{array}{c|c}
k & P(X = k) \\
\hline
0 & 0.0024 \\
1 & 0.0284 \\
2 & 0.1323 \\
3 & 0.3087 \\
4 & 0.3602 \\
5 & 0.1681 \\
\end{array}
\]

Let me know if you'd like more details or have any questions!

Here are 5 related questions you might find interesting:
1. How does the probability distribution change if \( p \) is 0.5?
2. What is the expected number of students to answer their phone?
3. How do we calculate the variance of \( X \)?
4. What would happen to the distribution if we increased the number of students surveyed to 10?
5. Can you simulate the binomial distribution with a different \( p \)?

**Tip:** In binomial problems, the mean is \( \mu = np \) and the variance is \( \sigma^2 = np(1-p) \).

Assume that you are randomly picking students to survey by phone. Let X = the number of students who answer their phone. Use some form of technology to find the probability distribution of X given a probability of success on a single trial. (Report answers accurate to 4 decimal places.)

Find the probability distribution for the number of students answering their phones in a survey of 5 students, with a success probability of 0.7. Detailed calculations are provided using the binomial distribution formula, making it ideal for high school and early college students.

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