Questions: Thirty percent of consumers prefer to purchase electronics online. You randomly select 8 consumers. Find the probability that the number of consumers who prefer to purchase electronics online is (a) exactly five, (b) more than five, and (c) at most five. (a) Find the probability that the number that prefer to purchase electronics online is exactly five. P(5)= (Round to three decimal places as needed.)

Solution Steps

We are given that \( p = 0.3 \) (the probability that a consumer prefers to purchase electronics online) and \( n = 8 \) (the number of consumers selected). We need to find the probability that exactly \( x = 5 \) consumers prefer to purchase electronics online.

The probability of exactly \( x \) successes in \( n \) trials for a binomial distribution is given by the formula:

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

where \( q = 1 - p = 0.7 \).

Substituting the values into the formula:

\[ P(5) = \binom{8}{5} \cdot (0.3)^5 \cdot (0.7)^{8-5} \]

Calculating \( \binom{8}{5} \):

\[ \binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]

Now substituting back into the probability formula:

\[ P(5) = 56 \cdot (0.3)^5 \cdot (0.7)^3 \]

Calculating \( (0.3)^5 \) and \( (0.7)^3 \):

\[ (0.3)^5 = 0.00243, \quad (0.7)^3 = 0.343 \]

Thus,

\[ P(5) = 56 \cdot 0.00243 \cdot 0.343 \approx 0.047 \]

Final Answer