Question

Some transistors of two different kinds (call them \(N\) and \(P\) ) are stored in two boxes. You know that there are \(6 N\) 's in one box and that \(2 N\) 's and \(3 P\) 's got mixed in the other box, but you don't know which box is which. You select a box and a transistor from it at random and find that it is an \(N ;\) what is the probability that it came from the box with the \(6 \mathrm{N}\) 's? From the other box? If another transistor is picked from the same box as the first, what is the probability that it is also an \(N ?\)

Short answer

The probability it came from Box 1 is \(\frac{5}{7}\) and from Box 2 is \(\frac{2}{7}\). If another is picked from the same box, the probability it's an N is 1 for Box 1 and \(\frac{2}{5}\) for Box 2.

Step-by-step solution

- Define the Problem

Let Box 1 contain 6N and Box 2 contain 2N and 3P. We need to find the probability that a selected N transistor came from Box 1 and then find the probability that it came from Box 2.

- Define Hypotheses and Probabilities

Define hypotheses: - Box 1: Contains 6N - Box 2: Contains 2N and 3P.The prior probability of choosing either box is \( P(\text{Box 1}) = \frac{1}{2} \) and \( P(\text{Box 2}) = \frac{1}{2} \)

- Calculate Likelihood of Selecting an N

Calculate the likelihood of selecting an N transistor from each box: For Box 1: \( P(N | \text{Box 1}) = 1 For Box 2:\) \( P(N | \text{Box 2}) = \frac{2}{5} \)

- Apply Bayes’ Theorem

Use Bayes' Theorem to find the probability that the transistor came from Box 1 given that an N transistor was selected. \[ P(\text{Box 1} | N) = \frac{P(N | \text{Box 1}) \times P(\text{Box 1})}{P(N)} \]First, calculate the total probability of selecting an N, \[ P(N) = P(N | \text{Box 1}) \times P(\text{Box 1}) + P(N | \text{Box 2}) \times P(\text{Box 2}) \]\[ P(N) = 1 \times \frac{1}{2} + \frac{2}{5} \times \frac{1}{2} \]\[ P(N) = \frac{1}{2} + \frac{1}{5} \]\[ P(N) = \frac{5}{10} + \frac{2}{10} = \frac{7}{10} \]

- Final Probability for Box 1

Calculate the probability that it came from Box 1: \[ P(\text{Box 1} | N) = \frac{1 \times \frac{1}{2}}{\frac{7}{10}} = \frac{\frac{1}{2}}{\frac{7}{10}} = \frac{5}{7} \]

– Final Probability for Box 2

Calculate the probability that it came from Box 2: \[ P(\text{Box 2} | N) = \frac{\frac{2}{5} \times \frac{1}{2}}{\frac{7}{10}} = \frac{\frac{1}{5}}{\frac{7}{10}} = \frac{2}{7} \]

- Conditional Probability for Second Transistor

If a second transistor is picked from the same box as the first: If it's Box 1: \[ P(\text{Second N | Box 1}) = 1 \]If it's Box 2: \[ P(\text{Second N | Box 2}) = \frac{2}{5} \]

Key concepts

Bayes' Theorem

Bayes' Theorem is a powerful tool in probability theory. It helps us update our beliefs about the likelihood of an event given new evidence. In the context of our problem, we want to know the probability that a transistor came from a specific box after we've found it is of type N. This involves updating our initial hypothesis about which box it came from based on the evidence that the transistor is N.

Bayes' Theorem Formula:

The theorem states: \[ P(A|B) = \frac{P(B|A) \times P(A)}{P(B)} \]
Where:

• \( P(A|B) \) is the probability of event A occurring given that B has occurred.
• \( P(B|A) \) is the probability of event B occurring given that A has occurred.
• \( P(A) \) and \( P(B) \) are the independent probabilities of events A and B occurring.

Applying this to our problem, we calculate the probability of selecting a transistor from Box 1 or Box 2 given that it is N.

Conditional Probability

Conditional probability is at the heart of Bayes' Theorem. It's the probability of one event happening given that another event has already occurred. In our scenario, the event of interest is picking an N transistor, and the condition is choosing the box.

Understanding Conditional Probability:
Often written as \( P(A|B) \), it tells us how likely A is when we know B has happened. To find \( P(\text{Box 1} | N) \), we needed to consider \( P(N | \text{Box 1}) \) and \( P(N | \text{Box 2}) \). These are the probabilities of picking an N transistor from Box 1 or Box 2 respectively.

This calculation involves understanding that we are not just focused on each box independently but considering how the selection of N impacts our belief of which box was chosen.

Probability Theory

Probability theory provides the foundation for working with uncertainties and random events. It encompasses a variety of concepts, including events, outcomes, and likely occurrences.

In our example, we are interested in the overall probability of an event (selecting an N transistor) and how it changes based on different conditions (which box we pick from). This section covers some important points:

• Events: Different possible outcomes, like picking a N or P transistor.
• Probability: The chance of an event happening, like the probability of selecting Box 1 or Box 2.
• Combination: Bringing together these probabilities to update our beliefs using tools like Bayes' Theorem.

All this helps us move from a straightforward understanding of choosing transistors to a more nuanced view of updating probabilities based on new information.

Hypotheses

In the context of Bayesian probability, hypotheses are the different scenarios or possibilities that we want to evaluate. When we apply Bayes' Theorem, we start with prior probabilities (our initial beliefs about these hypotheses) and update them based on observed data.

Hypotheses in Our Problem:
We have two hypotheses to evaluate:

• Hypothesis 1 (Box 1): The transistor came from Box 1, which contains 6 N transistors.
• Hypothesis 2 (Box 2): The transistor came from Box 2, which contains 2 N transistors and 3 P transistors.

Prior Probabilities:

Before picking a transistor, we assume there's an equal chance of picking either box because we don't have other information (\( P(\text{Box 1}) = 0.5 \) and \( P(\text{Box 2}) = 0.5 \)).
These hypotheses and priors lay the groundwork for applying Bayes' Theorem and updating our beliefs based on the evidence we encounter.