You are trying to find instrument \(A\) in a laboratory. Unfortunately, someone has put both instruments \(A\) and another kind (which we shall call \(B\) ) away in identical unmarked boxes mixed at random on a shelf. You know that the laboratory has \(3 A\) 's and \(7 B\) 's. If you take down one box, what is the probability that you get an \(A ?\) If it is a \(B\) and you put it on the table and take down another box, what is the probability that you get an \(A\) this time?

Short Answer

Expert verified
The probability of getting an A initially is \( \frac{3}{10} \). If the first pick is a B, the probability of getting an A next is \( \frac{1}{3} \).

Step by step solution

01

Identify the total number of instruments

The total number of instruments is the sum of type A and type B instruments. There are 3 A's and 7 B's. Therefore, the total number of instruments is: \[ 3 + 7 = 10 \]
02

Determine the initial probability of picking an A

The probability of picking an A on the first try is the number of A's divided by the total number of instruments. Thus, the probability is: \[ P(A) = \frac{3}{10} \]
03

Determine the probability of picking a B

The probability of picking a B on the first try is the number of B's divided by the total number of instruments. Thus, the probability is: \[ P(B) = \frac{7}{10} \]
04

Calculate the updated total instruments after removing a B

If a B is picked first and placed on the table, the total number of instruments on the shelf decreases by 1, leaving: \[ 10 - 1 = 9 \]
05

Determine the new probability of picking an A

After removing one B, there are still 3 A's but only 9 instruments in total. Thus, the new probability of picking an A is: \[ P(A|B \, \text{removed}) = \frac{3}{9} = \frac{1}{3} \]

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Key Concepts

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

Combinatorics
Combinatorics is a fundamental concept in probability theory dealing with counting methods to calculate probabilities. In our exercise, we use basic combinatorics to find total instruments and probabilities. We handle the set of instruments as a collection of objects where we are interested in particular arrangements and selections.

Key steps in this exercise that involve combinatorics are:
  • Identifying the total number of instruments by summing different types.
  • Calculating probabilities by considering the number of favorable outcomes over the total number of outcomes.
For example, the total number of instruments on the shelf is calculated by combining the number of type A (3) and type B (7):
\[ 3 + 7 = 10 \]
This simple use of combinatorics helps us to proceed with calculating the probability of different events.
Conditional Probability
Conditional probability is the probability of an event occurring given that another event has already occurred. In this exercise, after picking and removing a B, we need to adjust our calculations since the initial conditions have changed.

We start with an initial probability, calculated as:
\[ P(A) = \frac{3}{10} \]
If a B is picked and removed, the total number of instruments changes from 10 to 9. Therefore, the new probability of picking an A changes because we now condition on having removed one B:
\[ P(A|B \ \text{removed}) = \frac{3}{9} = \frac{1}{3} \]
Here, we've conditioned the probability of selecting an A on the event that a B has already been removed.
Random Variables
In probability theory, a random variable represents a possible outcome. In this scenario, the random variable can be defined as the type of instrument picked from the shelf.

Consider the random variable X which can take values 'A' or 'B'. Initially, the probability distribution can be described as:
\[ P(X = A) = \frac{3}{10} \]
\[ P(X = B) = \frac{7}{10} \]
After removing a B, the random variable distribution changes, as we now have 9 items, and the probability distribution must be updated. This provides a practical illustration of how random variables can change based on given conditions or previous outcomes.

Understanding random variables is crucial as they lay the foundation for more complex probability and statistical analyses.

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Most popular questions from this chapter

Two people are taking turns tossing a pair of coins; the first person to toss two alike wins. What are the probabilities of winning for the first player and for the second player? Hint: Although there are an infinite number of possibilities here (win on first turn, second turn, third turn, etc.), the sum of the probabilities is a geometric series which can be summed; see Chapter 1 if necessary.

Let \(m_{1}, m_{2}, \cdots, m_{n}\) be a set of measurements, and define the values of \(x_{i}\) by \(x_{1}=\) \(m_{1}-a, x_{2}=m_{2}-a, \cdots, x_{n}=m_{n}-a,\) where \(a\) is some number (as yet unspecified, but the same for all \(x_{i}\) ). Show that in order to minimize \(\sum_{i=1}^{n} x_{i}^{2},\) we should choose \(a=(1 / n) \sum_{i=1}^{n} m_{i} .\) Hint: Differentiate \(\sum_{i=1}^{n} x_{i}^{2}\) with respect to \(a .\) You have shown that the arithmetic mean is the "best" average in the least squares sense, that is, that if the sum of the squares of the deviations of the measurements from their "average" is a minimum, the "average" is the arithmetic mean (rather than, say, the median or mode).

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 \(\bar{x}\) and \(\sigma\).

Using both the binomial distribution and the normal approximation. A die is thrown 720 times. (a) Find the probability that 3 comes up exactly 125 times. (b) Find the probability that 3 comes up between 115 and 130 times.

It is shown in the kinetic theory of gases that the probability for the distance a molecule travels between collisions to be between \(x\) and \(x+d x\), is proportional to \(e^{-x / \lambda} d x,\) where \(\lambda\) is a constant. Show that the average distance between collisions (called the "mean free path") is \(\lambda\). Find the probability of a free path of length \(\geq 2 \lambda\).

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