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Chapter 5: Problem 20

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
The probability distribution of 'X' is as follows: \(P(X=0)\) = \(\frac{{9 \choose 2}}{{12 \choose 2}} \); \(P(X=1)\) = \(\frac{{3 \choose 1} \cdot {9 \choose 1}}{{12 \choose 2}}\); \(P(X=2)\) = \(\frac{{3 \choose 2}}{{12 \choose 2}}\).

Step by step solution

01

Identify the Problem

Here, we are interested in the situation where two individuals are selected without replacement from a group of 12 persons, including 3 left-handed. We are to find the probability distribution of left-handed persons in these two selections.
02

Steps to Calculate Probability

We calculate probabilities for possible occurrences of left-handed persons which can be either 0, 1, or 2 among the two selected individuals. The probability that none of the two selected persons is left-handed: \(P(X=0)\) is calculated by the ratio of the number of ways 2 persons can be selected from 9 right-handed persons to the total number of ways 2 persons can be selected from a group of 12. Similarly, we calculate \(P(X=1)\) and \(P(X=2)\), denoting the probabilities of selecting one left-handed and two left-handed individuals respectively.
03

Construct the Probability Distribution

The probability distribution of 'X' (the number of left-handed persons in the sample) is then constructed with the possible number of left-handed persons which are 0, 1, or 2, and their corresponding probabilities calculated in the previous step.

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