Question

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

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

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: \[ 3 + 7 = 10 \]

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: \[ P(A) = \frac{3}{10} \]

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: \[ P(B) = \frac{7}{10} \]

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: \[ 10 - 1 = 9 \]

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: \[ P(A|B \, \text{removed}) = \frac{3}{9} = \frac{1}{3} \]

Key concepts

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.