To solve this problem, we can use the binomial probability formula. Here’s the setup based on the image:

• Probability of success (p): 70% or $p = 0.7$
• Number of trials (n): 8 (since 8 consumers are selected)

We need to find probabilities for the following cases:

(a) Probability that exactly 5 consumers prefer to purchase online:

To find $P(X = 5)$, where $X$ represents the number of consumers who prefer online purchases.

Using the binomial formula: $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$ Substitute $n = 8$, $k = 5$, and $p = 0.7$: $P(X = 5) = \binom{8}{5} (0.7)^5 (0.3)^3$

Calculating this, we get $P(X = 5) \approx 0.254$, which is given in the image.

(b) Probability that more than 5 consumers prefer to purchase online:

To find $P(X > 5)$, we need to calculate the probabilities for $X = 6$, $X = 7$, and $X = 8$ and then sum them up: $P(X > 5) = P(X = 6) + P(X = 7) + P(X = 8)$

Calculations for each value:

1. For $P(X = 6)$: $P(X = 6) = \binom{8}{6} (0.7)^6 (0.3)^2$
2. For $P(X = 7)$: $P(X = 7) = \binom{8}{7} (0.7)^7 (0.3)^1$
3. For $P(X = 8)$: $P(X = 8) = \binom{8}{8} (0.7)^8 (0.3)^0$

Adding these probabilities together will give the total $P(X > 5)$.

Would you like me to calculate the exact values for each step, or would you like to move forward with understanding the process?

Here are some related questions for further exploration:

1. How would the probabilities change if only 5 consumers were chosen instead of 8?
2. What is the probability that exactly 3 consumers prefer online purchases?
3. How would the probabilities change if the probability of preferring online purchases was 60% instead of 70%?
4. What is the probability that fewer than 5 consumers prefer online purchases?
5. How would the probability calculations differ if we assumed the consumers' choices were normally distributed?

Tip: Remember, for binomial problems, rounding the final answer to the required decimal places ensures accuracy for most practical applications.