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

Expert verified
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

01

- 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.
02

- 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 \( P(\text{Box 2}) = \frac{1}{2} \)
03

- Calculate Likelihood of Selecting an N

Calculate the likelihood of selecting an N transistor from each box: For Box 1: \( P(N | \text{Box 1}) = 1 For Box 2:\) \( P(N | \text{Box 2}) = \frac{2}{5} \)
04

- 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, \[ P(N) = P(N | \text{Box 1}) \times P(\text{Box 1}) + P(N | \text{Box 2}) \times P(\text{Box 2}) \]\[ P(N) = 1 \times \frac{1}{2} + \frac{2}{5} \times \frac{1}{2} \]\[ P(N) = \frac{1}{2} + \frac{1}{5} \]\[ P(N) = \frac{5}{10} + \frac{2}{10} = \frac{7}{10} \]
05

- Final Probability for Box 1

Calculate the probability that it came from Box 1: \[ P(\text{Box 1} | N) = \frac{1 \times \frac{1}{2}}{\frac{7}{10}} = \frac{\frac{1}{2}}{\frac{7}{10}} = \frac{5}{7} \]
06

– Final Probability for Box 2

Calculate the probability that it came from Box 2: \[ P(\text{Box 2} | N) = \frac{\frac{2}{5} \times \frac{1}{2}}{\frac{7}{10}} = \frac{\frac{1}{5}}{\frac{7}{10}} = \frac{2}{7} \]
07

- 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: \[ P(\text{Second N | Box 2}) = \frac{2}{5} \]

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Key Concepts

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

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.

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Most popular questions from this chapter

Set up sample spaces for Problems 1 to 7 and list next to each sample point the value of the indicated random variable \(x,\) and the probability associated with the sample point. Make a table of the different values \(x_{i}\) of \(x\) and the corresponding probabilities \(p_{i}=f\left(x_{i}\right)\) Compute the mean, the variance, and the standard deviation for \(x\). Find and plot the cumulative distribution function \(F(x)\). Suppose that Martian dice are 4-sided (tetrahedra) with points labeled 1 to 4. When a pair of these dice is tossed, let \(x\) be the product of the two numbers at the tops of the dice if the product is odd; otherwise \(x=0\).

Recall that two events \(A\) and \(B\) are called independent if \(p(A B)=p(A) p(B) .\) Similarly two random variables \(x\) and \(y\) are called independent if the joint probability function \(f(x, y)=g(x) h(y) .\) Show that if \(x\) and \(y\) are independent, then the expectation or average of \(x y\) is \(E(x y)=E(x) E(y)=\mu_{x} \mu_{y}\).

Show that the expected number of heads in a single toss of a coin is \(\frac{1}{2}\). Show in two ways that the expected number of heads in two tosses of a coin is 1: (a) Let \(x=\) number of heads in two tosses and find \(\bar{x}\). (b) Let \(x=\) number of heads in toss 1 and \(y=\) number of heads in toss 2 ; find the average of \(x+y\) by Problem \(9 .\) Use this method to show that the expected number of heads in \(n\) tosses of a coin is \(\frac{1}{2} n\).

Two cards are drawn at random from a shuffled deck and laid aside without being examined. Then a third card is drawn. Show that the probability that the third card is a spade is \(\frac{1}{4}\) just as it was for the first card. Hint: Consider all the (mutually exclusive) possibilities (two discarded cards spades, third card spade or not spade, etc.).

Find the binomial probability for the given problem, and then compare the normal and the Poisson approximations. Out of 1095 people, what is the probability that exactly 2 were born on Jan. \(1 ?\) Assume 365 days in a year.

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