Question

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

The probability that the nickel came from the right pocket is \( \frac{21}{46} \).

Step-by-step solution

Define the Events

Define the events involved in the problem. Let:- Event R: Coin is picked from the right pocket.- Event L: Coin is picked from the left pocket.- Event N: Coin picked is a nickel.

Determine Total Probabilities

Calculate the total probability of picking from each pocket since you pick the pocket at random. Thus, \[ P(R) = P(L) = \frac{1}{2} \]

Calculate Conditional Probabilities

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

Use Total Probability Theorem

Calculate the total probability of picking a nickel: \[ P(N) = P(N|R) P(R) + P(N|L) P(L) = \frac{3}{7} \cdot \frac{1}{2} + \frac{2}{3} \cdot \frac{1}{2} = \frac{3}{14} + \frac{1}{3} \]Find a common denominator and sum up the fractions: \[ P(N) = \frac{3}{14} + \frac{14}{42} = \frac{9}{42} + \frac{14}{42} = \frac{23}{42} \]

Apply Bayes' Theorem

Calculate the probability that the nickel came from the right pocket using Bayes' Theorem: \[ P(R|N) = \frac{P(N|R)P(R)}{P(N)} = \frac{\frac{3}{7} \cdot \frac{1}{2}}{\frac{23}{42}} = \frac{3}{7} \cdot \frac{1}{2} \cdot \frac{42}{23} = \frac{21}{46} \]So, \[ P(R|N) = \frac{21}{46} \]

Key concepts

conditional probability

Conditional probability is a fundamental concept in probability theory. It helps us find the probability of an event occurring, given that another event has already occurred. This relationship is crucial when events are interconnected or influence each other.
In this exercise, we are asked to find the probability that the nickel came from the right pocket, given that we have already picked a nickel. This means that knowing the event of picking a nickel changes the probability of the pocket it came from.
By definition, conditional probability is expressed as:
\( P(A|B) = \frac{P(A \text{ and } B)}{P(B)} \)
Here, \( P(A|B) \) represents the probability of event A occurring given that B has occurred.
This formula lets us re-evaluate probabilities based on new information, which is what Bayes' Theorem fundamentally provides.

probability theory

Probability theory deals with the likelihood of different outcomes. It is the mathematical framework that we use for quantifying uncertainty.
In this problem, various aspects of probability theory are used:

• Calculating the likelihood of picking either pocket \(P(R) = P(L) = \frac{1}{2}\)
• Finding the chances of picking a nickel from each pocket using conditional probabilities \(P(N|R) = \frac{3}{7} \) and \(P(N|L) = \frac{2}{3}\)
• Using these probabilities to find the total probability of picking a nickel
  \( P(N) = P(N|R)P(R) + P(N|L)P(L) \)

Each aspect helps in breaking down and recombining individual probabilities to find the answer to our specific question.

random variables

Random variables are a key concept in probability theory. A random variable is a variable that can take on different values, each with a certain probability. They can be discrete or continuous.
In this exercise, we are dealing with discrete random variables:

• The pocket chosen can be seen as a random variable with two outcomes: right or left.
• The coin picked is another random variable with possible outcomes based on the types of coins present.

By using these random variables, we model and calculate the probabilities of different events occurring.
The fact that the coin chosen is a nickel, makes the event conditional based on the initial probabilities of picking from each pocket and the different types of coins in them. This illustrates how random variables help in the precise modeling of real-world scenarios in probability theory.