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An employee of the records office at a university currently has ten forms on his desk awaiting processing. Six of these are withdrawal petitions and the other four are course substitution requests. a. If he randomly selects six of these forms to give to a subordinate, what is the probability that only one of the two types of forms remains on his desk? b. Suppose he has time to process only four of these forms before leaving for the day. If these four are randomly selected one by one, what is the probability that each succeeding form is of a different type from its predecessor?

Short Answer

Expert verified
(a) \( \frac{1}{105} \), (b) \( \frac{6}{7} \)

Step by step solution

01

Understanding the Problem (Part a)

We need to determine the probability that six randomly selected forms include only one type of form remaining on the desk. We can approach this by finding cases where all the forms given to the subordinate are either withdrawal petitions or course substitution requests.
02

Combinatorial Selections (Part a)

There are two scenarios to consider: 1) All six forms are withdrawal petitions, or 2) All six forms are course substitution requests. We calculate the number of ways to select all forms for these cases.
03

Calculate Combinations (Part a)

The number of ways to select 6 withdrawal petitions is \( \binom{6}{6} = 1 \). The number of ways to select 6 course substitutions is \( \binom{4}{4} = 1 \). Thus, there are only 2 favorable outcomes.
04

Total Selection Ways (Part a)

The total number of ways to select any 6 forms from 10 is \( \binom{10}{6} = 210 \).
05

Compute Probability (Part a)

The probability that only one form type remains is given by the ratio of favorable outcomes to total outcomes: \( \frac{2}{210} = \frac{1}{105} \).
06

Understanding the Problem (Part b)

We now need the probability of selecting 4 forms such that each is of a different type compared to the preceding one. Forms must alternate between the two types starting from any type.
07

Scenario Enumeration (Part b)

Identify possible alternating sequences. The sequences starting from either a withdrawal (W) or substitution (S) can take the form W-S-W-S or S-W-S-W.
08

Compute Probability for a Sequence (Part b)

Consider the sequence W-S-W-S. The number of ways to choose 2 Ws and 2 Ss can be calculated and repeated for the opposite sequence.
09

Calculate Favorable Sequences (Part b)

The number of ways to choose 2 withdrawal courses from 6 forms is \( \binom{6}{2} = 15 \), and 2 substitutions from 4 forms is \( \binom{4}{2} = 6 \). So, for sequence W-S-W-S, there are \( 15 \times 6 = 90 \) ways, and similar for S-W-S-W.
10

Compute Total Outcomes (Part b)

The total number of ways to select any 4 forms from 10 is \( \binom{10}{4} = 210 \).
11

Calculate Total Probability (Part b)

The sequences W-S-W-S and S-W-S-W together contribute \( 2 \times 90 = 180 \) favorable outcomes. Thus, the probability is \( \frac{180}{210} = \frac{6}{7} \).

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Key Concepts

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

Probability Theory
Probability theory is all about predicting how likely events are to occur. It's like guessing, but with math! In our exercise, we're trying to figure out the likelihood of different form scenarios on the employee's desk.
For part (a), the problem asks us to calculate the probability that the forms left on the desk are all of the same type after some are given away. The theory shows us that probability is just a fraction, where we put the number of desired outcomes on top and the total possible outcomes on the bottom.
So, if there are only 2 ways for one form type to be left on the desk out of 210 possible ways of choosing forms, the probability is the fraction \( \frac{2}{210} \), which simplifies to \( \frac{1}{105} \). This means there's a pretty slim chance for this to happen. Probability theory helps us understand how likely random events, like the selected forms, occur.
Combinatorial Analysis
Combinatorial analysis gives us tools to count how many ways things can be arranged or selected. This is super useful when we want to evaluate different scenarios, like in our exercise.
We use combinations when the order doesn't matter, like choosing forms regardless of how they're lined up on a table. For example, in part (a), we considered two main combinations: all withdrawal petitions and all course substitutions.
A combination is counted using the formula \( \binom{n}{k} \), which is read as 'n choose k'. It calculates the number of ways to select k items from n total items without considering order. In our problem, to choose 6 forms from 10, we use \( \binom{10}{6} \). This is because arrange doesn't matter. Combinatorics is the key to solving many probability problems.
Mathematical Statistics
Mathematical statistics involves interpreting and analyzing data to help us make decisions. It's like being a detective to find patterns and insights. In the exercise, we're looking at probabilities (calculated through various statistical methods) to make predictions.
For example, in part (b), determining the probability of randomly selecting forms that alternate in type involves creating and evaluating specific patterns or sequences of data.
Statistics help us quantify uncertainty and variance. By considering all potential ways forms can be selected and processed, we can derive probabilities like \( \frac{6}{7} \), indicating how likely such an alternating sequence is. This approach translates messy, unknown real-world situations into clear insights and decisions through structured data analysis.

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