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In a group of 12 persons, 3 are left-handed. Suppose that 2 persons are randomly selected from this group. Let \(x\) denote the number of left-handed persons in this sample. Write the probability distribution of \(x\). You may draw a tree diagram and use it to write the probability distribution. (Hint: Note that the selections are made without replacement from a small population. Hence, the probabilities of outcomes do not remain constant for each selection.)

Short Answer

Expert verified
Hence, the probability distribution of \(x\) (number of left-handed persons in a random sample of two individuals) is as follows: \(P(x=0)\) = 0.545, \(P(x=1)\) = 0.409, and \(P(x=2)\) = 0.045.

Step by step solution

01

Determining the total possible outcomes

Among the 12 persons, if two are selected at random, the total possible outcomes or combinations would be \({{12}\choose{2}} = 66\). This uses the combination formula 'n choose r', which is denoted as \(nCr = n! / [(n−r)! r!]\), where \(n\) is the total number of items, and \(r\) is the number of items to choose.
02

Calculating probability when \(x=0\)

This represents the case where none of the selected individuals are left-handed. The number of ways this could happen would be to choose 2 right-handed people out of 9 (total - left handed = 12 - 3 = 9). So, \({{9}\choose{2}} = 36\). The probability would be that number divided by the total possible outcomes. So \(\frac{36}{66} = \frac{0.545}{1}\).
03

Calculating probability when \(x=1\)

This signifies that only one of the selected persons is left-handed. We could choose 1 left-handed and 1 right-handed individual. It would be \({{3}\choose{1}} * {{9}\choose{1}} = 3*9 = 27\). The probability becomes \(\frac{27}{66} = \frac{0.409}{1}\).
04

Calculating probability when \(x=2\)

For this scenario, both the chosen individuals would be left-handed. The number of ways would be \({{3}\choose{2}} = 3\). The probability would then be \(\frac{3}{66} = \frac{0.045}{1}\).

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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 a fundamental part of mathematics that deals with the likelihood of different outcomes. It provides tools to analyze random events and quantify the uncertainty. In the context of our example, we are dealing with selecting individuals from a group without replacement, meaning once an individual is selected, they are not put back into the pool.

When calculating probabilities, one needs to determine the total number of possible outcomes and then identify the number of favorable outcomes that meet the criteria of interest. This is essential for understanding the probability distribution, which is a list of all possible outcomes and their associated probabilities. In this instance, probabilities are calculated for scenarios where 0, 1, or 2 left-handed individuals are selected from the group.
Combinatorics
Combinatorics is a branch of mathematics that studies counting, arrangement, and combination. It underpins the solution to our original exercise, where we calculated the total possible outcomes. Here, we used the combination formula, often noted as 'n choose r.'

This formula, denoted by \(nCr = \frac{n!}{(n - r)!r!}\), helps us determine how many ways we can choose \(r\) items from \(n\) total items, where order doesn't matter. In the original exercise, this was used to compute the number of ways to select 2 people from a pool of 12, and to separate different numbers of left-handed individuals from right-handed ones.
  • Finding total possible combinations is computed by \(^{12}C_{2} = 66\).
  • When choosing 2 right-handed people out of 9, the calculation is \(^{9}C_{2} = 36\).
  • For selecting one from each group, left and right-handed, it's \(^{3}C_{1} \times ^{9}C_{1} = 27\).
  • Finally, for both left-handed, use \(^{3}C_{2} = 3\).
Random Sampling
Random sampling refers to the process of selecting individuals from a population in such a way that each individual has an equal chance of being chosen. However, the original problem involves sampling without replacement, meaning probability changes with each selection.

In simple terms, once a person is selected, they're not placed back into the pool, altering the probabilities of subsequent selections. This means each choice affects the pool's composition, and the probabilities for combinations have to account for these changes. It's important because it represents a real-world approach where selections often affect the pool, unlike with replacement where probabilities remain constant.
Discrete Random Variable
A discrete random variable is an important concept when dealing with probability distributions. It is a variable that can take on a countable number of distinct values, as part of an experiment or random process.

In our case, the random variable \(x\) represents the number of left-handed individuals chosen from the group. It's discrete because it can only take on specific values: 0, 1, or 2. Understanding this helps in constructing the probability distribution, which describes how probabilities are spread across possible values of the random variable.
  • \(x = 0\): No left-handers selected, probability calculated as \(\frac{36}{66} = 0.545\).
  • \(x = 1\): One left-hander selected, probability is \(\frac{27}{66} = 0.409\).
  • \(x = 2\): Both selected are left-handed, probability is \(\frac{3}{66} = 0.045\).
These probabilities sum up to 1, confirming that they represent a full probability distribution for the variable \(x\).

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