Question

Let \(x\) be the number of successes observed in a sample of \(n=5\) items selected from \(N=10 .\) Suppose that, of the \(N=10\) items, 6 are considered "successes." a. Find the probability of observing no successes. b. Find the probability of observing at least two successes. c. Find the probability of observing exactly two successes

Short answer

Answer: The probabilities are as follows: - No successes: 0 - At least two successes: 246/252 - Exactly two successes: 60/252

Step-by-step solution

Set up the hypergeometric distribution formula for 0 successes

We want to find the probability of observing no successes (k=0) in a sample of n=5 items selected from N=10, where 6 items are considered successes (K=6). Set up the formula: \(P(X=0)=\frac{{6 \choose 0}{4 \choose 5}}{{10 \choose 5}}\)

Calculate the probability

Calculate the combinations and probabilities: \({6 \choose 0}=1\) \({4 \choose 5}=0\) \({10 \choose 5}=252\) \(P(X=0)=\frac{1 \cdot 0}{252}=0\) The probability of observing no successes is 0. #b. Find the probability of observing at least two successes.#

Set up the hypergeometric distribution formula for at least 2 successes

We want to find the probability of observing at least two successes, which means 2, 3, 4, or 5 successes. We will have to calculate the probability for each of these cases and then sum up the probabilities. \(P(X \geq 2)=P(X=2)+P(X=3)+P(X=4)+P(X=5)\)

Calculate the probabilities and sum them up

Calculate the probabilities for each case and sum them up: \(P(X=2)=\frac{{6 \choose 2}{4 \choose 3}}{{10 \choose 5}}=\frac{15 \cdot 4}{252}=\frac{60}{252}\) \(P(X=3)=\frac{{6 \choose 3}{4 \choose 2}}{{10 \choose 5}}=\frac{20 \cdot 6}{252}=\frac{120}{252}\) \(P(X=4)=\frac{{6 \choose 4}{4 \choose 1}}{{10 \choose 5}}=\frac{15 \cdot 4}{252}=\frac{60}{252}\) \(P(X=5)=\frac{{6 \choose 5}{4 \choose 0}}{{10 \choose 5}}=\frac{6 \cdot 1}{252}=\frac{6}{252}\) \(P(X \geq 2)=\frac{60}{252}+\frac{120}{252}+\frac{60}{252}+\frac{6}{252}=\frac{246}{252}\) The probability of observing at least two successes is \(\frac{246}{252}\). #c. Find the probability of observing exactly two successes#

Use the hypergeometric distribution formula for 2 successes

We want to find the probability of observing exactly two successes (k=2) in a sample of n=5 items selected from N=10, where 6 items are considered successes (K=6). We've already calculated this probability in the previous part: P(X=2)

Recall the probability

The probability of observing exactly two successes is \(P(X=2)=\frac{60}{252}\).

Key concepts

Probability of Success

The probability of success in the context of hypergeometric distribution is about figuring out the likelihood of achieving a certain number of successes in a sample taken from a larger population. Since hypergeometric distribution deals with scenarios of sampling without replacement, the probability of success differs from that in cases where events are independent, such as in the binomial distribution. This means each success or failure changes the odds for next draws due to the diminishing pool of available outcomes. To calculate such probabilities, we leverage combinatorial analysis, specifically combinations, to arrange the possible successful outcomes over the entire set. For example, if we want to calculate the probability of exactly two successes when drawing five items from ten, where six are successes, we analyze from various angles:

• How many ways we can choose two successful items from a total of six.
• How many ways we can choose the remaining items to form the total sample from the non-successful set.

This helps in understanding how to manipulate odds in finite populations when outcomes are mutually dependent due to the nature of the sampling.

Combinatorial Analysis

Combinatorial analysis is integral to solving hypergeometric distribution problems. It involves determining the number of different ways to select a group of items from a larger pool, which is crucial when dealing with scenarios of sampling without replacement. In these cases, we use the combination formula \({n \choose k}\) to calculate the ways to choose \(k\) successes from \(n\) available successes, as well as the failures. Combination formulas work because we are interested in grouping items, and the order does not matter.

• The formula \({n \choose k}\) = \(\frac{n!}{k!(n-k)!}\) where \'!\' represents factorial, meaning the product of all positive integers up to that number.
• For instance, if six items are considered successes out of ten, \({6 \choose 2}\) shows how to select two success items from these.

Applying this idea of combinations, we can precisely determine probabilities in a hypergeometric framework. Each calculation reflects how components of the selection process furnish the complete picture for scenario-based probabilities, thus explaining why probabilities are connected to specific outcomes and total sampling.

Sampling Without Replacement

Sampling without replacement is a key concept in understanding hypergeometric distribution because it affects the probability calculation by altering the sample space with each draw. Unlike sampling with replacement, where the odds remain constant because you 'put back' and redraw, sampling without replacement changes the composition of the pool you are sampling from after every draw. This concept significantly influences the process and results of finding probabilities since it involves selecting from a shrinking pool. When an item is chosen, it isn't replaced, leading to a dynamic probability calculation. For example, if choosing from a group of ten items, once an item is drawn, only nine are left for subsequent draws. This directly impacts the combinations that can be formed, sustaining its relevance in problems like calculating the chances of successes in non-replenished sample sets:

• As we progress through the sampling, both the number of potential successes and failures diminish.
• The likelihood of further successes generally decreases with every success removed unless the pool favors more successes.

Thus, understanding how this sampling affects individual odds is vital for accurately solving probability problems where items aren’t replaced after each draw.