Question

A peony plant with red petals was crossed with a peony plant having streaky petals. The probability that an offspring from this cross has red flowers is .75. Let \(x\) be the number of plants with red petals resulting from 10 seeds from this cross that were collected and germinated. a. Does the random variable \(x\) have a binomial distribution? If not, why not? If so, what are the values of \(n\) and \(p\) ? b. Find \(P(x \geq 9)\). c. Find \(P(x \leq 1)\). d. Would it be unusual to observe one plant with red petals and the remaining nine plants with streaky petals? If these experimental results actually occurred, what conclusions could you draw?

Short answer

Answer: Yes, it is unusual to observe such results, considering that the probability of observing one red petal plant out of 10 seeds is approximately 0.0027, which is low. This could imply a need to re-evaluate certain assumptions or factors affecting the crossbreeding process, or it could mean that this specific case is an anomaly. Further experiments with a larger number of seeds would be helpful to draw more definitive conclusions.

Step-by-step solution

a. Identify the distribution of the random variable \(x\).

We are given probabilities of success i.e., the probability of a plant having red petals (0.75), and we have a fixed number of trials (10 seeds). The random variable \(x\) records the number of successful trials (plants with red petals) out of the 10 seeds. Thus, the random variable \(x\) follows a binomial distribution with parameters \(n\) and \(p\). Here, \(n\) represents the number of trials (seeds) and \(p\) represents the probability of success (plants having red petals). \(n = 10\) (number of seeds) \(p = 0.75\) (probability of plants having red petals)

b. Find \(P(x \geq 9)\).

To find the probability of \(x \geq 9\), we can calculate the sum of the probabilities \(P(x=9)\) and \(P(x=10)\). \(P(x \geq 9) = P(x=9) + P(x=10) = \binom{10}{9}(0.75)^9(1-0.75)^{10-9} + \binom{10}{10}(0.75)^{10}(1-0.75)^{10-10}\) \(P(x \geq 9) = \binom{10}{9}(0.75)^9(0.25)^1 + \binom{10}{10}(0.75)^{10}(0.25)^0\) Now, calculating the probabilities, we get \(P(x \geq 9) \approx 0.563 + 0.0563 = 0.619\)

c. Find \(P(x \leq 1)\).

To find the probability of \(x \leq 1\), we can calculate the sum of the probabilities \(P(x=0)\) and \(P(x=1)\). \(P(x \leq 1) = P(x=0) + P(x=1) = \binom{10}{0}(0.75)^0(1-0.75)^{10-0} + \binom{10}{1}(0.75)^1(1-0.75)^{10-1}\) \(P(x \leq 1) = \binom{10}{0}(0.75)^0(0.25)^{10} + \binom{10}{1}(0.75)^{1}(0.25)^{9}\) Now, calculating the probabilities, we get \(P(x \leq 1) \approx 0.0001 + 0.0027 = 0.0028\)

d. Analyze the unusualness of the observed result.

We are asked if observing one plant with red petals and the remaining nine with streaky petals is unusual. In other words, we want to know if \(P(x=1)\) is an unusual probability. We already calculated \(P(x=1) \approx 0.0027\), which is a quite low probability. This means that the given observed results are unusual. If these experimental results indeed occurred, it may imply that we should re-evaluate our assumptions, including the probability of a plant having red petals or some other factors affecting the crossbreeding process. It could also mean that this specific case is an anomaly, considering the calculated probability is low but not zero. Further experiments with a larger number of seeds would be helpful to draw more definitive conclusions.

Key concepts

Probability of Success

Understanding the probability of success is essential when utilizing the binomial distribution in probability exercises. In the context of our peony plants, the probability of success, denoted as p, is the chance that a single trial results in a desired outcome—in this case, a plant having red petals after a crossbreed.

For example, if we plant 10 seeds with the expectation that each one has a 75% chance, or a probability of 0.75, of resulting in red petals, this number becomes our value for p. Therefore, p is a crucial figure that influences the entire distribution; higher probabilities indicate a greater chance that more seeds will yield red petals.

It's crucial to note that within the framework of a binomial distribution, the probability of success needs to be constant across each trial and that each trial is independent, meaning the result of one seed sprouting does not affect the outcome of the others. These characteristics ensure that our random variable x, which counts the number of red-petaled plants out of 10, indeed follows a binomial distribution.

Random Variable

A random variable is a vital concept in probability theory and represents a quantity whose outcome is subject to chance. In our exercise with peony plants, the random variable x represents the number of plants with red petals that result from the germination of 10 seeds.

Random variables, such as x in this scenario, can take on a range of numerical values, each associated with a different probability. For x to qualify as following a binomial distribution, two criteria must be met: there must be a fixed number of independent trials, and each trial must have only two outcomes—success (red petals) or failure (streaky petals).

In the given exercise, because the trials (germinating seeds) are independent and the number of trials (10 seeds) is fixed, with each seed having the same probability of success (p = 0.75), x is indeed a binomial random variable.

Cumulative Probability

The term cumulative probability refers to the likelihood that a random variable will take a value less than or equal to a specified amount. Cumulative probabilities are particularly valuable when we wish to understand the ‘at most’ or ‘at least’ scenarios within our binomial distribution context.

Considering our peony plant problem, the cumulative probability can answer questions like, 'What is the probability that at most one of the seeds will sprout a red-petaled plant?' or 'What is the chance of getting at least nine red-petaled plants from ten seeds?' To calculate these, one adds up the individual probabilities for each relevant outcome.

For instance, to compute P(x ≤ 1), we add the probabilities that x will be 0 or 1. Conversely, to find P(x ≥ 9), we sum the probabilities of x equaling 9 and 10. These calculations reveal not just the probability of individual outcomes but also provide a bigger picture of the distribution, helping us understand the range and likelihood of possible results from our binomial experiment.