Seventy-five percent of consumers prefer to purchase electronics online. You randomly select 88 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.


This problem involves a **binomial distribution**, which is characterized by two outcomes: success (a consumer prefers to purchase electronics online) and failure (a consumer does not prefer to purchase online).

### Given:
- The probability of success \( p = 0.75 \) (since 75% prefer online).
- The number of trials (consumers) \( n = 88 \).
- The probability of failure \( q = 1 - p = 0.25 \).

Let \( X \) be the random variable representing the number of consumers who prefer to purchase electronics online, so \( X \) follows a binomial distribution:
\[
X \sim \text{Binomial}(n = 88, p = 0.75)
\]

We need to compute the following probabilities:
- **(a) Exactly 5 consumers prefer to purchase online**: \( P(X = 5) \).
- **(b) More than 5 consumers prefer to purchase online**: \( P(X > 5) \).
- **(c) At most 5 consumers prefer to purchase online**: \( P(X \leq 5) \).

We'll use the binomial probability formula:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
where \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) successes out of \( n \) trials.

Let's calculate these probabilities.Here are the probabilities based on the binomial distribution:

- **(a) The probability that exactly 5 consumers prefer to purchase electronics online** is approximately \( 9.94 \times 10^{-44} \), which is extremely close to zero.
  
- **(b) The probability that more than 5 consumers prefer to purchase electronics online** is approximately \( 1 \) (essentially a certainty).
  
- **(c) The probability that at most 5 consumers prefer to purchase electronics online** is approximately \( 1.01 \times 10^{-43} \), again extremely close to zero.

The results show that it's highly unlikely for only 5 or fewer consumers to prefer purchasing electronics online, given the high preference rate (75%) and large sample size (88).

Do you want further clarification or details?

Here are 5 related questions:
1. How is the binomial distribution related to the normal distribution for large \( n \)?
2. How would the probabilities change if the sample size were larger?
3. How does changing the probability \( p \) affect the binomial distribution?
4. What is the expected number of consumers who prefer to purchase online?
5. How do you compute the standard deviation for a binomial distribution?

**Tip:** The binomial distribution approaches a normal distribution as the number of trials \( n \) increases, which can simplify calculations for large \( n \).

Seventy-five percent of consumers prefer to purchase electronics online. You randomly select 88 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.

Calculate the probability that 5 or fewer consumers out of 88 prefer to purchase electronics online, using binomial distribution. Learn to apply the binomial probability formula for exact values, greater than, and at most scenarios in this detailed example. Suitable for advanced high school and college students.

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