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Chapter 4: Problem 8

A sample of 4 items is drawn at a random without replacement from a lot of 10 items. Containing 3 defective. If \(X\) denotes the number of defective items in the sample then \(P(0

Short Answer

Expert verified
The answer is (b) \(\frac{4}{5}\).

Step by step solution

01

Understand the Problem

We have 10 items, including 3 defective ones, and we are drawing a sample of 4 items without replacement. We need to find the probability that there are more than 0 but fewer than 3 defective items in the sample.
02

Define Random Variable and Event

Define the random variable \(X\) as the number of defective items in the sample of 4. We need to calculate \(P(0<X<3)\), which includes \(P(X=1)\) and \(P(X=2)\).
03

Calculate Probability for \(X=1\)

The probability \(P(X=1)\) is the number of ways to choose 1 defective item from 3 and 3 non-defective items from 7. Using the formula, \(P(X=1) = \frac{{^3C_1 \times ^7C_3}}{{^{10}C_4}}\).
04

Calculate Combinations and Probability (\(X=1\))

Compute \(^3C_1\) as 3, \(^7C_3\) as 35, and \(^{10}C_4\) as 210. Then, \(P(X=1) = \frac{3 \times 35}{210} = \frac{105}{210}\).
05

Calculate Probability for \(X=2\)

The probability \(P(X=2)\) is the number of ways to choose 2 defective items from 3 and 2 non-defective items from 7. Use the formula: \(P(X=2) = \frac{{^3C_2 \times ^7C_2}}{{^{10}C_4}}\).
06

Calculate Combinations and Probability (\(X=2\))

Compute \(^3C_2\) as 3, \(^7C_2\) as 21, and \(^{10}C_4\) as 210. Then, \(P(X=2) = \frac{3 \times 21}{210} = \frac{63}{210}\).
07

Sum Probabilities for \(0

Add the probabilities \(P(X=1)\) and \(P(X=2)\): \(P(0<X<3) = \frac{105}{210} + \frac{63}{210} = \frac{168}{210}\).
08

Simplify the Probability

Simplify \(\frac{168}{210}\) to get \(\frac{4}{5}\) by dividing both the numerator and the denominator by their greatest common divisor, which is 42.

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics that involves counting, arranging, and combining items. In problems like ours, where we draw a sample of items from a larger group, combinatorics provides the tools we need to calculate possible outcomes. When we talk about combinations, we are specifically interested in figuring out how many ways we can choose a certain number of items from a larger set, irrespective of the order in which we choose them. For instance, the combination symbol \(^nC_k\) denotes the number of ways to choose \(k\) items from a total of \(n\) items.
  • \(^nC_k = \frac{n!}{k!(n-k)!}\), where \(!\) means factorial, the product of all positive integers up to that number.
  • For our exercise, we used combinations to figure out the number of ways to pick defective and non-defective items from our given sample size.
Understanding these basics of combinatorics allows us to solve probability problems by accurately counting all the possible events that can occur.
Random Variables
A random variable is a concept to understand and model random events in probability. It assigns numbers to outcomes of a random experiment to analyze its behavior mathematically.In our problem, we use a random variable \(X\) to represent the number of defective items in our sample of 4. Since we can have 0, 1, 2, or 3 defective items, \(X\) can take these integer values. With this, we calculate the likelihood of different outcomes within our given conditions.
  • Important aspect: The function \(P(X=x)\) represents the probability \(X\) equals a particular value \(x\).
  • For probabilities, we calculate considering each feasible outcome: \(P(X=1)\), \(P(X=2)\), etc.
This approach allows us to quantify uncertain scenarios and answer questions like "What is the probability of having exactly 1 or 2 defective items in the sample?"
Defective Items
Understanding 'defective items' is essential in calculating how they affect the probability of different outcomes in a set.In this exercise, "defective items" refer to the items that are not functioning as intended. We have 3 defective items in our total of 10. Items being defective influences our random drawing possible outcomes substantially. When drawing items:
  • If we draw a defective item, it's taken out of the pool, leaving fewer defective options for subsequent draws. This highlights why the sampling method 'without replacement' can significantly impact the outcome probabilities.
  • In calculating \(P(0
By understanding which items are defective, we target specific scenarios and probabilities, such as drawing exactly 1 or 2 defective items, to solve real-world inspection and quality control problems efficiently.

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Most popular questions from this chapter

A sample of 4 items is drawn at a random without replacement from a lot of 10 items. Containing 3 defective. If \(X\) denotes the number of defective items in the sample then \(P(0

In a test an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is \(1 / 3\) and the probability that he copies the answer is \(1 / 6\). The probabilitythat his answer is correct given that he copied it, is \(1 / 8\). The probability that he knew the answer to the question given that he correctly answered it is [IIT-1991] (a) \(24 / 29\) (b) \(25 / 24\) (c) \(29 / 24\) (d) \(24 / 25\) Solution (a) Let \(A_{1}\) be the event that the examinee guesses the answer; \(A_{2}\) the event that he copies the answer and \(A_{3}\) the event that he knows the answer. Also let \(A\) be the event that he answers correctly. Then as given, we have $$ P\left(A_{1}\right)=\frac{1}{3}, P\left(A_{2}\right)=\frac{1}{6}, P\left(A_{3}\right)=1-\frac{1}{3}-\frac{1}{6}=\frac{1}{2} $$ (We have assumed here that the events \(A_{1}, A_{2}\) and \(A_{3}\) are mutually exclusive and totally exhaustive.) Now \(P\left(A / A_{1}\right)=\frac{1}{4}, P\left(A / A_{2}\right)=\frac{1}{8}\) (as given) Again it is reasonable to take the probability of answering correctly given that he knows the answer as 1 , that is, \(P\left(A / A_{3}\right)=1\) We have to find \(P\left(A_{3} / \mathrm{A}\right)\) By Bayes' theorem, we have $$ \begin{aligned} &P\left(A_{3} / A\right)=\frac{P\left(A_{3}\right) P\left(A / A_{3}\right)}{P\left(A_{1}\right) P\left(A / A_{1}\right)+P\left(A_{2}\right) P\left(A / A_{2}\right)+} \\ &\quad P\left(A_{3}\right) P\left(A / A_{3}\right) \\ &=\frac{(1 / 2) \times 1}{(1 / 3)(1 / 4)+(1 / 6)(1 / 8)+(1 / 2) \times 1}=\frac{24}{29} \end{aligned} $$

The first twelve letters of the alphabet are written at random. Find the probability that there are exactly four letters between \(A\) and \(B\). (a) \(7 / 66\) (b) \(8 / 66\) (c) \(7 / 56\) (d) None of these

For a binomial variate \(X\) if \(n=5\) and \(P(X=1)=\) \(8 P(\mathrm{X}=3)\), then \(P\) is (a) \(4 / 5\) (b) \(1 / 5\) (c) \(1 / 3\) (d) \(2 / 3\) tion $$ \begin{aligned} &\text { (b) }{ }^{5} C_{1} q^{4} p^{1}=8 \times{ }^{5} C_{3} q^{2} p^{3} \\ &\Rightarrow q=4 P \\ &\Rightarrow 1-p=4 P \\ &\Rightarrow P=1 / 5 \end{aligned} $$

The probability distribution of a random variable \(X\) is given below: \(\begin{array}{llll}X=x_{i} & 2 & 3 & 4 \\ P:\left(X=x_{i}\right) & 1 / 4 & 1 / 8 & 5 / 8\end{array}\) Then its mean is (a) \(27 / 8\) (b) \(5 / 4\) (c) 1 (d) \(4 / 5\)
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