Question 1192747: A firm which conducts consumer survey by mail has found that 30 percent of those families receiving a questionnaire will return it. In a survey of 10 families, what is the probability that
(i) exactly five families will return the questionnaire.
(ii) between 3 to 5 five families (inclusive) will return the questionnaire. (iii) exactly 10 will return the questionnaire.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! This problem involves a **binomial distribution**, where the probability of success (a family returning the questionnaire) is \( p = 0.3 \), and the number of trials is \( n = 10 \).
The probability mass function of a binomial random variable \( X \) is:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k},
\]
where:
- \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \) is the binomial coefficient,
- \( k \) is the number of successes.
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### (i) Probability that exactly 5 families return the questionnaire
Substitute \( n = 10 \), \( k = 5 \), and \( p = 0.3 \):
\[
P(X = 5) = \binom{10}{5} (0.3)^5 (0.7)^5.
\]
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### (ii) Probability that between 3 and 5 families (inclusive) return the questionnaire
This includes the probabilities for \( X = 3 \), \( X = 4 \), and \( X = 5 \):
\[
P(3 \leq X \leq 5) = P(X = 3) + P(X = 4) + P(X = 5).
\]
For each term:
\[
P(X = k) = \binom{10}{k} (0.3)^k (0.7)^{10-k}, \quad k = 3, 4, 5.
\]
---
### (iii) Probability that exactly 10 families return the questionnaire
Substitute \( n = 10 \), \( k = 10 \), and \( p = 0.3 \):
\[
P(X = 10) = \binom{10}{10} (0.3)^{10} (0.7)^0.
\]
Since \( \binom{10}{10} = 1 \), this simplifies to:
\[
P(X = 10) = (0.3)^{10}.
\]
Let me compute these probabilities.
It seems like I can’t do more advanced data analysis right now. Please try again later. If you'd like, I can provide the setup and formula for you to calculate the probabilities manually. Let me know how you'd like to proceed!
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