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 \(3 \mathrm{~A} \mathrm{~s}\) and \(7 \mathrm{~B}^{\prime} \mathrm{s}\). 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?

Conditional probability measures the likelihood of an event occurring given that another event has already occurred. Let's use our problem as an example. Suppose you first pick a box and it contains instrument B. Now, there's a different scenario. The probability that the next box you pick contains instrument A changes based on this new information.

To calculate the new probability, we need to adjust for the fact that one of the B instruments has already been removed. Initially, there are 10 boxes (3 A and 7 B). If we remove a B, there are 9 boxes left (3 A and 6 B). The probability that the next box contains an A given that a B was first picked is:

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

This conditional probability shows how the initial outcome impacts future probabilities.

Probability theory is the branch of mathematics that deals with calculating the likelihood of a given event's occurrence, which is expressed as a number between 1 and 0. Let's break down the problem step by step.

1. **Total Number of Boxes**: First, we figure out the total number of boxes on the shelf. With 3 instruments of type A and 7 of type B:
\[ 3 + 7 = 10 \]

2. **Probability of Getting an A Initially**: To calculate the chance of randomly picking a box that contains instrument A, we use:
\[ P(A) = \frac{\text{Number of A instruments}}{\text{Total number of boxes}} = \frac{3}{10} \]

This fundamental understanding helps us solve more complex scenarios, like determining how probabilities change when conditions change.

Combinatorics is a field of mathematics concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures. In our problem, we use basic combinatorial counting to determine probabilities.

Here's how combinatorics factors into our problem:

1. **Counting Different Outcomes**: We need to count the total outcomes for both scenarios: initially selecting any box, then recalculating after removing one B.
2. **Adjusting the Count**: Initially, we have 10 ways (boxes) to choose from. For picking box A initially, there are 3 favorable conditions (3 boxes of A). After removing one B (one box out of 10), the count becomes 9.

Using these principles, we adjust the sample space and favorable outcomes to find probabilities:

\[ P(A \text{ initially}) = \frac{3}{10} \ P(A | B \text{ first}) = \frac{3}{9} = \frac{1}{3} \]

These steps show how combinatorics help in understanding and solving probability problems.