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Chapter 17: Problem 5

A box contains 8 balls: 5 are green and 3 are white, red and yellow.Three balls are chosen at random without replacement and the number of green balls Y is recorded. a) Explain why \(Y\) is not a binomial random variable. b) Explain why, when we repeat the experiment with replacement, then Y is a binomial. c) Give the values of \(n\) and \(p\) and display the probability distribution in tabular form. d) What is the probability that at most 2 green balls are drawn? e) What is the expected number of green balls drawn? f) What is the variance of the number of balls drawn? g) What is the probability that some green balls will be drawn?

Short Answer

Expert verified
a) Dependency due to no replacement; b) Independent with replacement; c) \(n = 3, p = \frac{5}{8}\); d) Use cumulative probability; e) \(1.875\); f) \(0.703\); g) \(1 - P(Y=0)\).

Step by step solution

01

Understanding Non-Binomial Conditions

When drawing without replacement, the probability of selecting a green ball changes after each draw. This dependency violates the condition that each trial should be independent, which is necessary for a binomial random variable.
02

Repeating Experiment with Replacement

With replacement, each draw is independent because the probability of selecting a green ball remains constant at each trial. This satisfies the conditions for a binomial distribution.
03

Determining 'n' and 'p' Values

In a repeated experiment with replacement, the total number of trials is 3 (since three balls are drawn, i.e., \(n = 3\)), and the probability of drawing a green ball remains \( p = \frac{5}{8} \) each time.
04

Binomial Probability Distribution Table

The probability distribution of Y can be computed using the binomial formula \( P(Y = k) = \binom{n}{k}p^k(1-p)^{n-k} \): \[\begin{array}{c|c}Y & P(Y) \\hline0 & \binom{3}{0}(\frac{5}{8})^0(\frac{3}{8})^3 \1 & \binom{3}{1}(\frac{5}{8})^1(\frac{3}{8})^2 \2 & \binom{3}{2}(\frac{5}{8})^2(\frac{3}{8})^1 \3 & \binom{3}{3}(\frac{5}{8})^3(\frac{3}{8})^0 \\end{array}\]
05

Calculating at Most 2 Green Balls

The probability for at most 2 green balls is given by \( P(Y \leq 2) = P(Y = 0) + P(Y = 1) + P(Y = 2) \). Calculate each probability using the binomial formula and sum them.
06

Calculating Expected Value

The expected number of green balls can be calculated using the formula \( E(Y) = np \). Substituting the values gives \( E(Y) = 3 \times \frac{5}{8} \).
07

Calculating Variance

The variance of the number of green balls is given by \( \sigma^2 = np(1-p) \). Substitute the values to find \( \sigma^2 = 3 \times \frac{5}{8} \times \frac{3}{8} \).
08

Probability of Drawing Some Green Balls

The probability of drawing some green balls (at least one) is \( 1 - P(Y = 0) \), where \( P(Y = 0) \) can be found using the binomial formula.

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Key Concepts

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

Binomial Distribution
A binomial distribution is a specific type of probability distribution characterized by a sequence of independent and identically distributed trials. Each trial can result in one of two outcomes: success or failure.
For a distribution to be binomial, each trial must be independent, and the probability of a success must remain constant with each trial. This leads to the binomial distribution requirement: fixed number of trials, where each trial is independent with the same probability of success.
This becomes apparent in the exercise when drawing balls with replacement, as each draw is independent, and the probability of drawing a green ball is always \( \frac{5}{8} \). Without replacement, each draw affects the probability of subsequent draws, making it non-binomial.
Expected Value
The expected value, often referred to as the mean, is a measure of the central tendency of a probability distribution. It provides a weighted average of all possible values a random variable can take, where each value is weighted by its probability of occurrence.
In the context of a binomial distribution, the expected value is calculated by multiplying the number of trials \(n\) by the probability of success \(p\), expressed as \(E(Y) = np\).
In the exercise, since there are 3 trials and the probability of drawing a green ball is \(\frac{5}{8}\), the expected number of green balls is \(3 \times \frac{5}{8}\), highlighting the average outcome of the experiment.
Variance
Variance is a statistical measure of the dispersion or spread of a probability distribution. It describes how much the values of the random variable are expected to deviate from the expected value.
For a binomial distribution, variance is calculated using the formula \(\sigma^2 = np(1-p)\), where \(n\) is the number of trials and \(p\) is the probability of success.
In the exercise context, variance would tell us how much the number of green balls drawn deviates from the average number. Substituting the known values, \(n = 3\) and \(p = \frac{5}{8}\), the variance is \(3 \times \frac{5}{8} \times \frac{3}{8}\), providing insight into the variability of outcomes.
Probability Distribution
A probability distribution lists all possible outcomes of a random variable and their corresponding probabilities. It provides a complete description of the random variable鈥檚 possible behavior.
In a binomial distribution context, each outcome is associated with a probability that can be calculated using the formula \(P(Y = k) = \binom{n}{k}p^k(1-p)^{n-k}\).
For this exercise, a probability distribution table was computed using this formula for drawing green balls when \(n = 3\) and \(p = \frac{5}{8}\). Each value of \(Y\), the number of green balls drawn, has an associated probability, offering a detailed picture of all potential experimental outcomes.

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Most popular questions from this chapter

Medical research has shown that a certain type of chemotherapy is successful \(70 \%\) of the time when used to treat skin cancer. In a study to check the validity of such a claim, researchers chose different treatment centres and chose five of their patients at random. Here is the probability distribution of the number of successful treatments for groups of five: $$\begin{array}{|l|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline P(x) & 0.002 & 0.029 & 0.132 & 0.309 & 0.360 & 0.168 \\ \hline \end{array}$$ a) Find the probability that at least two patients would benefit from the treatment. b) Find the probability that the majority of the group does not benefit from the treatment. c) Find \(E(X)\) and interpret the result. d) Show that \(\sigma(x)=1.02\) e) Graph \(P(x)\). Locate \(\mu, \mu \pm \sigma\) and \(\mu \pm 2 \sigma\) on the graph. Use the empirical rule to approximate the probability that \(x\) falls in this interval. Compare this with the actual probability.

Consider the following binomial distribution: $$P(x)=\left(\begin{array}{l}5 \\\x\end{array}\right)(0.6)^{x}(0.4)^{5-x}, x=0,1, \ldots, 5$$ a) Make a table for this distribution. b) Graph this distribution. c) Find the mean and standard deviation in two ways: (i) by formula (ii) by using the table of values you created in part a). d) Locate the mean \(\mu\) and the two intervals \(\mu \pm \sigma\) and \(\mu \pm 2 \sigma\) on the graph. e) Find the actual probabilities for \(x\) to lie within each of the intervals \(\mu \pm \sigma\) and \(\mu \pm 2 \sigma\) and compare them to the empirical rule.

Consider the binomial random variable with \(n=6\) and \(p=0.3\) a) Fill in the probabilities below. $$\begin{array}{|l|l|l|l|l|l|l|l|}\hline k & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline P(x \leqslant k) & & & & & & & \\\\\hline\end{array}$$ b) Fill in the table below. Some cells have been filled for you to guide you. (TABLE CAN'T COPY)

In a large school, heights of students who are 13 years old are normally distributed with a mean of \(151 \mathrm{cm}\) and a standard deviation of \(8 \mathrm{cm} .\) Find the probability that a randomly chosen child is a) shorter than \(166 \mathrm{cm}\) b) within \(6 \mathrm{cm}\) of the average.

\(X \sim N\left(\mu, \sigma^{2}\right), P(x>162)=0.122\) and \(P(x<56)=0.0276 .\) Find \(\mu\) and \(\sigma\)
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