Some transistors of two different kinds (call them \(N\) and \(P)\) are stored in two boxes. You know that there are \(6 \mathrm{~N}^{\text {'s }}\) in one box and that \(2 \mathrm{~N}^{\prime} \mathrm{s}\) and \(3 \mathrm{P}^{\prime} \mathrm{s}\) 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 \(N\) ?

Understanding conditional probabilities is essential for solving problems involving dependencies between events. A conditional probability, denoted as \(P(A|B)\), is the probability of event \(A\) occurring given that event \(B\) has already occurred. It's calculated by dividing the probability of both events happening together by the probability of the given event. In our transistor problem, we need to find the probability of picking a specific type of transistor from a specific box, given certain initial conditions. Here's a breakdown:

• \(P(N|A)\): Probability of picking an N transistor from Box A. Here, Box A only contains N transistors, so \(P(N|A) = 1\).


• \(P(N|B)\): Probability of picking an N transistor from Box B. Box B contains 2 N transistors out of a total of 5 transistors (2 N and 3 P), so \(P(N|B) = 2/5 = 0.4\).

Understanding these conditions helps us calculate further probabilities using Bayes' Theorem.

Bayes' Theorem is a fundamental theorem in probability, which allows us to update our beliefs based on new evidence. It's used for calculating the probability of an event based on prior knowledge of conditions related to the event. The theorem states:
\ P(A|B) = \frac{P(B|A) * P(A)}{P(B)} \
In our transistor example, we use Bayes' Theorem to find the probability that a picked N transistor comes from Box A or Box B:

• \(P(A|N)\): Probability that the selected transistor came from Box A given that it is an N transistor.


• \(P(B|N)\): Probability that the selected transistor came from Box B given that it is an N transistor.

These are calculated using the given probabilities:
\ P(A|N) = \frac{P(N|A) * P(A)}{P(N|A) * P(A) + P(N|B) * P(B)} = \frac{1 * 0.5}{(1 * 0.5) + (2/5 * 0.5)} = \frac{0.5}{0.7} = \frac{5}{7} \
Similarly, \(P(B|N) = \frac{2}{7}\). Bayes' Theorem is crucial for making these inferences.

The transistor selection problem involves understanding how to use probability to make decisions based on given data. In the problem, you are trying to determine the likelihood of which box an N transistor came from and predicting the type of subsequent transistors.

• Define the sample space: There are two boxes, Box A and Box B, with different compositions of transistors.
• Assign initial probabilities: We start assuming equal likelihood of selecting either box, so \(P(A) = 0.5\) and \(P(B) = 0.5\).
• Calculate conditional probabilities: These help provide information about the events within each box (e.g., \(P(N|A) = 1\), \(P(N|B) = 2/5\)).
• Utilize Bayes' Theorem: Update these probabilities with new information, i.e., finding an N transistor, to get \(P(A|N) = 5/7\) and \(P(B|N) = 2/7\).

Next, we need to determine the probability of drawing another N transistor from the same box. This depends on which box the first transistor came from:
\ P(\text{Second N}|\text{First N}) = P(A|N) * 1 + P(B|N) * \frac{1}{2} \
Substituting our values, we get:
\ P(\text{Second N}|\text{First N}) = \frac{5}{7} * 1 + \frac{2}{7} * \frac{1}{2} = \frac{5}{7} + \frac{1}{7} = \frac{6}{7} \
The overall probability is \(11/14\), indicating a higher likelihood of getting another N transistor if the first one was an N. This transistor selection problem exemplifies the practical application of conditional probabilities and Bayesian probability.