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Let a bowl contain 10 chips of the same size and shape. One and only one of these chips is red. Continue to draw chips from the bowl, one at a time and at random and without replacement, until the red chip is drawn. (a) Find the pmf of \(X\), the number of trials needed to draw the red chip. (b) Compute \(P(X \leq 4)\).

Short Answer

Expert verified
\[P(X \leq 4) = 0.4\] and the pmf \(P(X=i) = \frac{1}{10}\) for \(i = 1,2,...,10.\)

Step by step solution

01

Determine the Probability Mass Function

To calculate the pmf, consider that the red chip can be drawn in any of the 10 trials. For each trial \(i\), the probability is determined by the number of ways to pick the red chip on the \(i\)th draw and the total number of ways to draw the chips. Remember, some chips are chosen before the red chip (therefore not including the red chip), then the red chip is chosen. This results in a geometric probability distribution.The pmf is given by:\[P(X=i) = \frac {1}{10}, \quad i = 1,2,...,10\]
02

Compute \(P(X \leq 4)\)

We can calculate the cumulative probability for \(X \leq 4\) by summing the probabilities of individual events for \(X = 1, X = 2, X = 3, and X =4\). \[P(X\leq 4) = P(X=1) + P(X=2) + P(X=3) + P(X=4) \]Substituting the values of \(P(X=i)\) from the pmf,\[ P(X \leq 4) = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10}\]

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Key Concepts

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

Geometric Probability Distribution
Understanding the geometric probability distribution is crucial when dealing with the likelihood of a specific event occurring for the first time. In the exercise provided, the event in question is drawing the red chip from a bowl containing 10 uniformly sized and shaped chips.

Geometric probability distribution is a model applicable for scenarios with a sequence of trials, where each trial has only two possible outcomes, often termed as 'success' and 'failure'. In this context, success is drawing the red chip, and failure is drawing any other chip. The trials are independent; the outcome of one trial does not influence another. Moreover, the probability of success remains constant with each trial.

For example, the probability mass function (pmf) for our exercise's geometric distribution is expressed as:\[\begin{equation} P(X=i) = \frac{1}{10}, \ foreachdrawdenoted\ by \ i,whereeach i representsthetrialnumberfrom 1to10.\end{equation}\]The pmf tells us that no matter which draw we are considering, the chance of drawing the red chip on that particular draw remains at 10%. This description of the pmf aligns with the characteristics of a geometric distribution.
Cumulative Probability
Cumulative probability is a measure that captures the probability of a discrete random variable taking on a value less than or equal to a specific value. It's the sum of probabilities of the outcomes up to that value. It helps in understanding the likelihood of an event happening by a particular point in a process.

In the provided solution, to compute the cumulative probability of drawing the red chip in the first four trials, you sum the individual probabilities for each of the trials from one through four, as each trial is an independent event. This process is shown by:\[\begin{equation} P(X \leq 4) = P(X=1) + P(X=2) + P(X=3) + P(X=4), \end{equation}\]and hence it simplifies to:\[\begin{equation} P(X \leq 4) = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{4}{10}. \end{equation}\]This result indicates that there is a 40% chance of drawing the red chip within the first four trials.
Discrete Random Variable
A discrete random variable is a type of variable that takes on a countable number of distinct outcomes. It is an essential concept in the realm of probability and statistics. These variables can often be enumerated and typically arise from scenarios such as flips of a coin, rolls of a dice, or, as in our exercise, drawing chips from a bowl.

In our case, the discrete random variable is 'X', which represents the number of trials needed to draw the red chip. Since there are ten chips and a single red chip is drawn without replacement, 'X' can have the values 1, 2, ..., up to 10—each representing one of the ten possible scenarios in which the red chip could be drawn. Defining 'X' this way allows for the precise calculation of probabilities for different scenarios related to drawing the red chip, which is fundamental to solving problems that revolve around discrete random variables.

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Most popular questions from this chapter

Let \(p_{X}(x)=x / 15, x=1,2,3,4,5\), zero elsewhere, be the pmf of \(X\). Find \(P(X=1\) or 2\(), P\left(\frac{1}{2}

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Each bag in a large box contains 25 tulip bulbs. It is known that \(60 \%\) of the bags contain bulbs for 5 red and 20 yellow tulips, while the remaining \(40 \%\) of the bags contain bulbs for 15 red and 10 yellow tulips. A bag is selected at random and a bulb taken at random from this bag is planted. (a) What is the probability that it will be a yellow tulip? (b) Given that it is yellow, what is the conditional probability it comes from a bag that contained 5 red and 20 yellow bulbs?

Generalize Exercise \(1.2 .5\) to obtain $$ \left(C_{1} \cup C_{2} \cup \cdots \cup C_{k}\right)^{c}=C_{1}^{c} \cap C_{2}^{c} \cap \cdots \cap C_{k}^{c} $$ Say that \(C_{1}, C_{2}, \ldots, C_{k}\) are independent events that have respective probabilities \(p_{1}, p_{2}, \ldots, p_{k} .\) Argue that the probability of at least one of \(C_{1}, C_{2}, \ldots, C_{k}\) is equal to $$ 1-\left(1-p_{1}\right)\left(1-p_{2}\right) \cdots\left(1-p_{k}\right) $$

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