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.
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Part 1
​(a) Find the probability that the number that prefer to purchase electronics online is exactly five.
​P(5)equals=enter your response here
​(Round to three decimal places as​ needed.) 

This problem involves a **binomial probability distribution** since each consumer either prefers to purchase electronics online or does not, and we are given a fixed number of consumers (88) with a known probability of success (75%, or 0.75).

The binomial probability formula is given by:

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

Where:
- \( P(X = k) \) is the probability of exactly \( k \) successes (in this case, 5 consumers),
- \( n \) is the total number of trials (consumers), \( n = 88 \),
- \( p \) is the probability of success (prefer online), \( p = 0.75 \),
- \( k \) is the number of successes we're interested in (exactly 5 consumers),
- \( \binom{n}{k} \) is the binomial coefficient, calculated as:

\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]

Let's compute the probability \( P(X = 5) \) for exactly 5 consumers. I will calculate that now.The probability of exactly 5 consumers out of 88 preferring to purchase electronics online is extremely small, approximately 0.000. This is due to the fact that with a 75% chance of consumers preferring online purchases, it is highly unlikely that only 5 out of 88 would do so.

Would you like to explore parts (b) and (c), or need further clarification on part (a)? 

Here are 5 related questions for further practice:
1. How would you calculate the probability of exactly 10 consumers preferring online purchases?
2. What changes in the binomial formula if we are interested in at least 5 consumers?
3. How can we calculate the expected number of consumers preferring online purchases?
4. What is the probability of exactly 44 consumers preferring to purchase online?
5. How does increasing the sample size affect the probabilities in binomial distributions?

**Tip:** In binomial distributions, when \( p \) is very high or very low, probabilities for extreme outcomes (like \( k = 5 \) in this case) tend to be very small.

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. (a) Find the probability that the number that prefer to purchase electronics online is exactly five. P(5) equals=enter your response here (Round to three decimal places as​ needed.)

Calculate the probability that out of 88 consumers, exactly 5 prefer to purchase electronics online using the binomial distribution formula. This problem explores key statistical concepts and provides a step-by-step solution. Suitable for students in Grades 10-12.

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