91Ó°ÊÓ

In the realm of binomial probability, the combinatorial factor is a crucial concept. It refers to the number of distinct ways in which you can arrange successes in a series of trials. This is mathematically represented by the binomial coefficient \( \binom{n}{k} \).
Understanding how to compute the combinatorial factor is vital. In simple terms, it involves calculating the possible ways to achieve exactly \( k \) successes (hits, in the context of our exercise) in \( n \) trials. The formula used is \( \frac{n!}{k!(n-k)!} \), where "!" denotes a factorial, meaning the product of all positive integers up to that number.
For the exercise involving Lauren Wible, where she tries to get 4 hits in 4 at-bats, the combinatorial factor is \( \binom{4}{4} \). This simplifies to \( \frac{4!}{4!(0)!} = 1 \). Thus, there is only one way to achieve 4 hits in 4 tries, indicating certainty in the arrangement when every trial results in success.

The probability of success is a fundamental piece of any probability problem, especially in binomial distributions.
It defines the likelihood of achieving a successful outcome in a single trial. Notably, this probability must remain constant across each independent trial within a series.
In our context, the term "success" refers to a hit at-bat. Based on the information given, the probability of Lauren Wible making a hit in one attempt is \( p = 0.524 \). This figure was derived from her remarkable batting average, translating to a 52.4% chance of a hit each time she steps to the plate.
In a binomial context, it’s crucial to also consider the probability of failure, which complements the probability of success. Calculated as \( 1-p \), this represents the likelihood that any given trial results in a non-hit. In Lauren's case, the probability of not getting a hit is \( 1 - 0.524 = 0.476 \). Understanding both probabilities is essential in determining the overall probability of an event using binomial distributions.

The binomial probability formula is a powerful tool that allows us to calculate the chances of a certain number of successes within a set number of trials. This formula is given as \[P(k; n, p) = \binom{n}{k} \, p^k \, (1-p)^{n-k},\]where:

• \( P(k; n, p) \): The probability of getting exactly \( k \) successes in \( n \) trials.


• \( \binom{n}{k} \): The combinatorial factor, or number of ways to choose \( k \) successes out of \( n \) trials.


• \( p^k \): The probability of success raised to the power of the number of successful trials.


• \((1-p)^{n-k}\): The probability of failure in the remaining trials.

Applying this to our scenario with Lauren Wible, where \( n \) is 4, \( k \) is also 4, and \( p \) is 0.524, we substitute these values into our formula. First, calculate the combinatorial factor as \( 1 \). Then, compute \( p^k = (0.524)^4 \approx 0.0759 \). Since Lauren’s desired number of hits equals her number of attempts, each failure term is raised to the zero power, providing a value of one. Therefore, the resulting probability is \( 0.0759 \), signifying the likelihood that she gets a hit each of the four times at bat.