Suppose you have 3 nickels and 4 dimes in your right pocket and 2 nickels and a quarter in your left pocket. You pick a pocket at random and from it select a coin at random. If it is a. nickel, what is the probability that it came from your right pocket?

Short Answer

Expert verified
\(P(R|N) = \frac{9}{23} \approx 0.39\)

Step by step solution

01

- Define the Events

Let R be the event of choosing the right pocket and L be the event of choosing the left pocket. Also, let N be the event of picking a nickel.
02

- Calculate Pocket Probabilities

Since a pocket is chosen at random, the probability of choosing the right pocket (R) or the left pocket (L) is \(P(R) = \frac{1}{2}\) and \(P(L) = \frac{1}{2}\).
03

- Determine the Probability of Picking a Nickel from Each Pocket

Calculate the probability of picking a nickel from the right pocket:\[P(N | R) = \frac{3 \text{ nickels}}{3 \text{ nickels} + 4 \text{ dimes}} = \frac{3}{7}\]and from the left pocket:\[P(N | L) = \frac{2 \text{ nickels}}{2 \text{ nickels} + 1 \text{ quarter}} = \frac{2}{3}\]
04

- Apply Bayes' Theorem

Bayes' Theorem states:\[P(R | N) = \frac{P(N | R) P(R)}{P(N)}\]We need to determine \(P(N)\), the total probability of picking a nickel:
05

- Calculate the Total Probability of Picking a Nickel

\(P(N)\) is calculated using the Law of Total Probability:\[P(N) = P(N | R) P(R) + P(N | L) P(L)\]Substitute the values from previous steps:\[P(N) = \left( \frac{3}{7} \right) \left( \frac{1}{2} \right) + \left( \frac{2}{3} \right) \left( \frac{1}{2} \right) = \frac{3}{14} + \frac{1}{3} = \frac{3}{14} + \frac{14}{42} = \frac{9}{42} + \frac{14}{42} = \frac{23}{42}\]
06

- Compute the Final Probability with Bayes' Theorem

Now, substitute \(P(N)\) back into Bayes' Theorem:\[P(R | N) = \frac{P(N | R) P(R)}{P(N)} = \frac{\left( \frac{3}{7} \right) \left( \frac{1}{2} \right)}{\frac{23}{42}} = \frac{\frac{3}{14}}{\frac{23}{42}} = \frac{3}{14} \times \frac{42}{23} = \frac{3 \times 42}{14 \times 23} = \frac{126}{322} = \frac{9}{23} \approx 0.39\]

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Key Concepts

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

Conditional Probability
Conditional probability helps us find the likelihood of an event occurring, given that another event has already taken place. In this problem, we're interested in the probability that a coin came from the right pocket given that we've picked a nickel. To find this, we first identify the probability of picking a nickel (event N) from each pocket. We then update this information based on additional data using Bayes' Theorem. This way, conditional probability enables us to revise our probabilities as new information comes in. In simpler terms, it tells us “how likely something is to happen, given that something else has happened.”
Law of Total Probability
The Law of Total Probability is crucial when we need to calculate the overall probability of an event based on different scenarios. Here, we use the law to determine the total probability of picking a nickel from either pocket. The formula is: \[ P(N) = P(N|R)P(R) + P(N|L)P(L) \] This means we add up the probabilities of picking a nickel from the right pocket and the left pocket, each multiplied by the probability of choosing the respective pocket. This law allows us to consider all probable paths that lead to the event we care about (picking a nickel). It’s like thinking of all possible ways something could happen and combining those probabilities.
Nickels and Dimes Problem
Let's dive into the details of the nickels and dimes problem. You have three nickels and four dimes in your right pocket and two nickels and a quarter in your left pocket. You randomly choose a pocket and select a coin. If you pick a nickel, what’s the chance it’s from the right pocket? By setting up the problem, we find the probabilities of picking a nickel from each pocket first. Then, we use Bayes' Theorem to adjust these probabilities based on our condition that the picked coin is a nickel. Here's the key takeaway: even when different outcomes could lead to the same event (like picking a nickel from either pocket), understanding and carefully applying conditional probability and the law of total probability helps us get accurate results.

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Most popular questions from this chapter

Suppose a coin is tossed three times. Let \(x\) be a random variable whose value is 1 if the number of heads is divisible by 3 , and 0 otherwise. Set up the sample space for \(x\) and the associated probabilities. Find \(\hat{x}\) and \(\sigma\).

Two dice are thrown. Use the sample space (2.4) to answer the following questions. (a) What is the probability of being able to form a two-digit number greater than 33 with the two numbers on the dice? (Note that the sample point 1, 4 yields the two-digit number 41 which is greater than 33, etc.) (b) Repeat part (a) for the probability of being able to form a two-digit number greater than or equal to 42 (c) Can you find a two-digit number (or numbers) such that the probability of being able to form a larger number is the same as the probability of being able to form a smaller number? [See note, part (a).]

Recall that two events \(A\) and \(B\) are called independent if \(p(A B)=p(A) \cdot p(B) .\) Similarly, two random variables \(x\) and \(y\) with probability functions \(f(x)\) and \(g(y)\) are called independent if the probability of \(x=x_{i}\) and \(y=y_{j}\) is \(f\left(x_{i}\right) \cdot g\left(y_{j}\right)\) for every pair of values of \(x\) and \(y\), that is, if the joint probability function for \(x, y\) is \(f(x) g(y)\). Show that if \(x\) anct \(y\) are independent, then the expectation or average of \(x y\) is \(E(x y)=E(x) \cdot E(y)=\mu_{x} \mu_{y}\).

Define the sample variance by \(s^{2}=(1 / n) \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2} .\) Show that the expected value of \(s^{2}\) is \([(n-1) / n] \sigma^{2} .\) Hints: Write $$ \begin{aligned} \left(x_{i}-\bar{x}\right)^{2} &=\left[\left(x_{i}-\mu\right)-(\bar{x}-\mu)\right]^{2} \\ &=\left(x_{i}-\mu\right)^{2}-2\left(x_{i}-\mu\right)(\bar{x}-\mu)+(\bar{x}-\mu)^{2} \end{aligned} $$ Find the average value of the first term from the definition of \(\sigma^{2}\) and the average value of the third term from Problem 2, To find the average value of the middle term write $$ (\bar{x}-\mu)=\left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}-\mu\right)=\frac{1}{n}\left[\left(x_{1}-\mu\right)+\left(x_{2}-\mu\right)+\cdots+\left(x_{n}-\mu\right)\right] $$ show by Problem \(7.12\) that $$ E\left[\left(x_{i}-\mu\right)\left(x_{j}-\mu\right)\right]=E\left(x_{i}-\mu\right) E\left(x_{j}-\mu\right)=0 \quad \text { for } \quad t \neq j $$ and evaluate \(E\left[\left(x_{i}-\mu\right)^{2}\right]\) (same as the first term). Collect terms to find $$ E\left(s^{2}\right)=\frac{n-1}{n} \sigma^{2} $$

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