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The probability of success in a binomial distribution is a key factor to understand when calculating probabilities in such scenarios. In the context of our exercise, "success" means a website visitor makes a purchase. The given probability of a customer making a purchase is stated to be 15%, or mathematically, \( p = 0.15 \). This probability is determined based on historical data or expected statistics. It helps us evaluate the likelihood of different numbers of "successes" (purchases) during multiple trials or "attempts." With each attempt (each hit on the website), there is a constant probability \( p \) of achieving a success, which is vital for the binomial probability calculations to follow. Understanding this probability aids us in predicting and analyzing the purchasing behavior of visitors and ultimately helps improve business strategies.

The expected number of purchases is an average estimate of how many successes, or purchases, we might expect in our set of trials. In a binomial distribution, the formula to find this expectation is \( E(X) = n \times p \). Here, \( n \) represents the total number of trials (in this case, site hits), which is 16; and \( p \) represents the probability of success on each trial, or 0.15. Substituting these values, we calculate:

• \( E(X) = 16 \times 0.15 = 2.4 \)

This means that, on average, we can expect 2.4 purchases from 16 website hits. Though you can't have a fraction of a purchase in reality, this number helps to understand the probability distribution and planning for resources and stock.

Calculating the probability of at least a certain number of successes is a common scenario. In our exercise, we seek the probability that four or more website hits result in a purchase. This is expressed as \( P(X \geq 4) \). Instead of computing this directly for all cases of \( k = 4 \) and greater, we use a complementary approach. First, compute \( P(X < 4) \)—the probability of fewer than 4 purchases—and then subtract this from 1, since probabilities sum to 1.

• \( P(X \geq 4) = 1 - P(X < 4) \)

Previously calculated, \( P(X < 4) \) is the sum of probabilities for 0, 1, 2, and 3 purchases. By performing these binomial calculations and subtracting from 1, we find \( P(X \geq 4) \approx 0.3064 \). This means there is about a 30.64% chance of getting four or more purchases from 16 hits.

The binomial probability formula is essential to find the likelihood of obtaining exactly \( k \) successes in \( n \) trials. The formula is:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]This formula includes:

• \( \binom{n}{k} \) which is the binomial coefficient, calculating the number of ways \( k \) successes can occur in \( n \) trials.
• \( p^k \) which accounts for the probability of getting exactly \( k \) successes.
• \( (1-p)^{n-k} \) which represents the probability of the remaining \( n-k \) trials resulting in failure.

Applying this to the problem, with \( n = 16 \) and \( p = 0.15 \) for 4 purchases, we plug in these values:

• \( P(X = 4) = \binom{16}{4} \times (0.15)^4 \times (0.85)^{12} \)

By calculating these components, we find the probability of exactly four purchases to be approximately 14.74%. This formula is a powerful tool in predicting outcomes based on known success probabilities for repeated trials.