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Rhubarb Suppose you purchase a bundle of 10 bare-root rhubarb plants. The sales clerk tells you that on average you can expect 5\(\%\) of the plants to die before producing any rhubarb. Assume that the bundle is a random sample of plants. Let Y = the number of plants that die before producing any rhubarb. Use the binomial probability formula to find \(P(Y=1)\) . Interpret this result in context.

Short Answer

Expert verified
The probability that exactly one plant dies is approximately 0.3151.

Step by step solution

01

Understand the Binomial Setting

This exercise involves a random sample of rhubarb plants where each plant can either die or survive to produce rhubarb. We can model this as a binomial distribution: each trial (plant) has two possible outcomes (dies or survives). The total number of trials (plants) is 10, and the probability of success (a plant dying) is \(0.05\) because on average 5\(\%\) of the plants are expected to die.
02

Define the Variables and Parameters

Let \( n = 10 \) be the number of rhubarb plants, \( p = 0.05 \) be the probability of a plant dying before producing any rhubarb, and \( Y \) be the random variable representing the number of plants that die. Our goal is to find \( P(Y = 1) \), the probability that exactly one plant dies.
03

Use the Binomial Probability Formula

The binomial probability formula is given by: \[P(Y = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where \( \binom{n}{k} \) is the binomial coefficient, \( n \) is the number of trials, \( p \) is the probability of success, and \( k \) is the number of successful trials. For this problem, substitute \( n = 10 \), \( k = 1 \), and \( p = 0.05 \).
04

Calculate the Binomial Coefficient

Calculate the binomial coefficient \( \binom{10}{1} \):\[\binom{10}{1} = \frac{10!}{1!(10-1)!} = 10\]
05

Substitute into the Probability Formula

Substitute \( n = 10 \), \( k = 1 \), and \( p = 0.05 \) into the binomial probability formula:\[P(Y = 1) = \binom{10}{1} (0.05)^1 (0.95)^{10-1} = 10 \times 0.05 \times 0.95^9\]
06

Calculate the Probability

Calculate \( 0.95^9 \) and then find \( P(Y = 1) \):\[\ 0.95^9 \approx 0.6302 \]\[P(Y = 1) = 10 \times 0.05 \times 0.6302 \approx 0.3151\]
07

Interpret the Result

The probability that exactly one out of the 10 rhubarb plants will die before producing any rhubarb is approximately 0.3151. This means there is a 31.51\(\%\) chance that one plant in the bundle will die without producing rhubarb.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Variable
In statistics, a random variable is a variable that can take on different values based on the outcome of a random event. In our context, we are looking at the number of rhubarb plants that might die before producing any rhubarb, which is denoted by the random variable \( Y \).
This random variable can theoretically take on values from 0 to 10, as there are 10 plants in total. So if we describe \( Y \):
  • \( Y = 0 \): None of the plants die.
  • \( Y = 1 \): Exactly one plant dies.
  • \( Y = 10 \): All 10 plants die.
The randomness here arises because which plants, if any, actually die is not something we can predict with certainty, hence \( Y \) is random.
Binomial Distribution
A binomial distribution is a probability distribution that summarizes the likelihood of a value coming from a set of independent events, each with the same probability of occurrence. In this case, we have 10 independent trials (plants), each trial having two possible outcomes: dying or surviving.
The distribution is termed 'binomial' because it models scenarios that have precisely two outcomes per trial. Suppose you fix the number of success trials defined by a random variable, and you have a consistent probability of success for each trial. In that case, you are dealing with a binomial distribution—it fits this rhubarb example perfectly.
Mathematically, a binomial distribution can be represented using the parameters \( n \) (number of trials) and \( p \) (probability of success/the event occurring). Using these elements, we determine the probability of a particular number of successes (dead plants in this scenario) from these trials.
Probability of Success
The probability of success in a binomial distribution refers to the probability that a single trial results in a 'success'. This can be somewhat counterintuitive, as 'success' is merely the term used for the outcome we are focused on measuring. In our exercise, a plant dying is deemed a 'success' since that's the outcome of interest that we are examining probability calculations for.
For these rhubarb plants, the probability of a plant dying, or success, is 0.05, as 5\% of the plants on average might die. This single probability value \( p = 0.05 \) is essential for calculating probabilities regarding any number of plant deaths or survival. Given this percentage, it allows us to calculate other potential outcomes using the binomial probability formula.
Binomial Coefficient
The binomial coefficient is a critical component in calculating binomial probabilities. It is denoted as \( \binom{n}{k} \) in mathematical formulas, and it represents the number of ways to pick \( k \) successes from \( n \) trials.
The binomial coefficient is computed using factorials and can be written as:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Where \( n! \) ("n factorial") is the product of all positive integers up to \( n \). In the rhubarb plant exercise, the binomial coefficient \( \binom{10}{1} \) calculates to 10, representing the 10 different possibilities for one plant to die out of 10.
This component informs us how many sequences exist for achieving exactly \( k \) successes in \( n \) trials, important for understanding the full formula used for probability calculations in binomial distributions.

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Most popular questions from this chapter

A sample survey contacted an SRS of 663 registered voters in Oregon shortly after an clection and asked respondents whether they had voted. Voter records show that 56\(\%\) of registered voters had actually voted. We will see later that in repeated random samples of size 663 , the proportion in the sample who voted (call this proportion With vary according to the Normal distribution with mean \(\mu=0.56\) and standard deviation \(\sigma=0.019\) (a) If the respondents answer truthfully, what is \(\mathrm{P}(0.52 \leq V \leq 0.60)\) ? This is the probability that the sample proportion \(V\) estimates the population proportion 0.56 within \(\pm 0.04\) (b) In fact, 72\(\%\) of the respondents said they had voted \((V=0.72) .\) If respondents answer truthfully, what is \(P(V \geq 0.72) ?\) This probability is so small that it is good evidence that some people who did not vote claimed that they did vote.

Pair-a-dice Suppose you roll a pair of fair, six-sided dice. Let \(T=\) the sum of the spots showing on the up-faces. (a) Find the probability distribution of \(T\) (b) Make a histogram of the probability distribution. Describe what you see. (c) Find \(P(T \geq 5)\) and interpret the result.

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In which of the following situations would it be appropriate to use a Normal distribution to approximate probabilities for a binomial distribution with the given values of \(n\) and \(p ?\) (a) \(n=10, p=0.5\) (b) \(n=40, p=0.88\) (c) \(n=100, p=0.2\) (d) \(n=100, p=0.99\) (c) \(n=1000, p=0.003\)

Roulette Marti decides to keep placing a \(\$ 1\) bet on number 15 in consecutive spins of a roulette wheel until she wins. On any spin, there's a 1 -in- 38 chance that the ball will land in the 15 slot. (a) How many spins do you expect it to take until Marti wins? Justify your answer. (b) Would you be surprised if Marti won in 3 or fewer spins? Compute an appropriate probability to support your answer.

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