This is a binomial probability problem. In this case, the probability of success (a consumer preferring to purchase electronics online) is $p = 0.40$, and the number of trials (consumers selected) is $n = 8$. We want to find the probability of exactly 5 successes, i.e., $P(X = 5)$, where $X$ is the binomial random variable representing the number of consumers who prefer to purchase electronics online.

The binomial probability formula is:

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

Where:

• $n = 8$ (the number of consumers),
• $k = 5$ (the number of successes we are looking for),
• $p = 0.40$ (the probability of success),
• $\binom{n}{k}$ is the binomial coefficient.

First, calculate $\binom{8}{5}$, which is the number of ways to choose 5 successes from 8 trials:

$\binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!} = \frac{40320}{120 \cdot 6} = 56$

Now, apply the formula:

$P(X = 5) = 56 \times (0.40)^5 \times (0.60)^3$

Let’s calculate each part:

$(0.40)^5 = 0.01024, \quad (0.60)^3 = 0.216$

Now, multiply all the parts together:

$P(X = 5) = 56 \times 0.01024 \times 0.216 = 0.124$

So, the probability that exactly 5 consumers prefer to purchase electronics online is approximately:

$P(5) \approx 0.124$

Do you want details on parts (b) and (c)? Feel free to ask! Also, here are five related questions:

1. How is the binomial coefficient $\binom{n}{k}$ calculated in general?
2. What is the formula for the expected value in a binomial distribution?
3. How would the probabilities change if $p = 0.60$ instead of 0.40?
4. What if the number of trials were increased to 10? How would the probability change?
5. How is the binomial distribution related to the normal distribution?

Tip: The binomial distribution can be approximated by the normal distribution when $n$ is large and $p$ is not too close to 0 or 1.