80% of all families own a television set. If 10 families are interviewed at random, then the probability that at most three families own a television set is

Options

A. `(8401)(0.8)^10`
B. `(8401)(0.2)^10`
C. `(8441)(0.8)^10`
D. `(8441)(0.2)^10`

Answer

To solve the problem, we need to find the probability that at most 3 out of 10 families own a television set, given that 80% of families own a TV. This is a binomial probability problem.

### Step-by-Step Solution:

1. **Identify the parameters of the binomial distribution**:
- Let \( n = 10 \) (the number of families interviewed).
- Let \( p = 0.8 \) (the probability that a family owns a television).
- Therefore, the probability that a family does not own a television is \( q = 1 - p = 0.2 \).

2. **Define the event**:
- We want to find the probability that at most 3 families own a television set. This can be expressed mathematically as:
\[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \]
where \( X \) is the random variable representing the number of families that own a television.

3. **Use the binomial probability formula**:
The binomial probability formula is given by:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
where \( \binom{n}{k} \) is the binomial coefficient.

4. **Calculate each probability**:
- For \( k = 0 \):
\[ P(X = 0) = \binom{10}{0} (0.8)^0 (0.2)^{10} = 1 \cdot 1 \cdot (0.2)^{10} = (0.2)^{10} \]

- For \( k = 1 \):
\[ P(X = 1) = \binom{10}{1} (0.8)^1 (0.2)^9 = 10 \cdot 0.8 \cdot (0.2)^9 = 10 \cdot 0.8 \cdot (0.000000512) = 0.000004096 \]

- For \( k = 2 \):
\[ P(X = 2) = \binom{10}{2} (0.8)^2 (0.2)^8 = 45 \cdot (0.64) \cdot (0.000000256) = 0.000007680 \]

- For \( k = 3 \):
\[ P(X = 3) = \binom{10}{3} (0.8)^3 (0.2)^7 = 120 \cdot (0.512) \cdot (0.0000128) = 0.000786432 \]

5. **Sum the probabilities**:
Now we add the probabilities for \( k = 0, 1, 2, 3 \):
\[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \]
\[ P(X \leq 3) = (0.2)^{10} + 0.000004096 + 0.000007680 + 0.000786432 \]
\[ = 0.0000001024 + 0.000004096 + 0.000007680 + 0.000786432 \]
\[ = 0.0007983104 \]

6. **Final Probability**:
Thus, the final probability that at most 3 families own a television set is approximately:
\[ P(X \leq 3) \approx 0.000798 \]