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Chapter 2: Problem 20

A television store owner figures that 50 percent of the customers entering his store will purchase an ordinary television set, 20 percent will purchase a color television set, and 30 percent will just be browsing. If five customers enter his store on a certain day, what is the probability that two customers purchase color sets, one customer purchases an ordinary set, and two customers purchase nothing?

Short Answer

Expert verified
The probability that two customers purchase color sets, one customer purchases an ordinary set, and two customers purchase nothing is approximately 0.768 or 76.8%.

Step by step solution

01

Find the binomial coefficient

Since there are 5 customers, and we need to find out how many ways we can have 2 buying color TV sets, 1 buying an ordinary TV set, and 2 not buying anything, we will need to use the binomial formula which is: \( C(n, k) = \frac{n!}{k!(n-k)!} \), where \(C(n,k)\) is the number of combinations of n things taken k at a time, n is the total number of customers, and k is the number of customers buying color or ordinary TV sets. In our case, we will need to find: 1. The number of ways 2 customers can buy color TVs out of 5: C(5,2) 2. The number of ways 1 customer can buy an ordinary TV out of the remaining 3 customers: C(3,1)
02

Calculate the binomial coefficients

Now we can calculate the binomial coefficients mentioned in step 1 by using the formula given: 1. C(5,2) = \(\frac{5!}{2!(5-2)!} = \frac{120}{(2)(6)} = 10\) 2. C(3,1) = \(\frac{3!}{1!(3-1)!} = \frac{6}{(1)(2)} = 3\)
03

Calculate the probability

Now that we have the number of ways 2 customers can buy color TVs and 1 customer can buy an ordinary TV, we can calculate the probability. First, we will calculate the probabilities of each type of customer buying a TV set. The probability that 2 of the 5 customers buy color sets is \(10 * 0.2^2 * 0.8^3 = 2.048\). The probability that 1 of the remaining 3 customers buys an ordinary set is \(3 * 0.5^1 * 0.5^2 = 0.375\). To get the probability that 2 customers buy color TVs, 1 customer buys an ordinary TV, and the remaining 2 customers don't buy anything (browsing), we multiply the two probabilities together: \(2.048 * 0.375 = 0.768\). So, the probability that 2 customers purchase color sets, 1 customer purchases an ordinary set, and 2 customers purchase nothing is approximately 0.768 or 76.8%.

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Key Concepts

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

Combination Formula
When dealing with problems involving selections or arrangements, the combination formula is a valuable tool. Unlike permutations which consider order,combinations only care about the selection. The formula used is:
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
where:
  • \( n \) represents the total number of items,
  • \( k \) represents the number of items being chosen at a time.
This formula calculates how many ways you can select \( k \) items from a group of \( n \) without regard to the order of selection. Utilizing the factorial operation,which multiplies all positive integers up to a specified number, you can solve these types of problems efficiently.
For instance, if you have 5 customers and you want to see how many ways you can choose 2 of them to buy a color TV, you'd compute it as:\[C(5, 2) = \frac{5!}{2!(5-2)!} = 10\]which gives you 10 different combinations.
Probability Calculations
Probability calculations involve determining the likelihood of different outcomes occurring. This requires understanding the fundamental concept of probability,which is the ratio of the favorable outcomes to the total possible outcomes.
In the context of binomial probability, you multiply the probability of one specific outcome by the number of ways that outcome can occur.Let's break it down further.
If we have scenario probabilities like 0.2 (buying a color TV), 0.5 (buying an ordinary TV), and 0.3 (browsing or buying nothing),we perform probability calculations by raising each probability to the power of the number of customers doing these actions, and then multiply by the combination calculated before.
For example:
  • For 2 customers buying color TVs: \[ 10 \times (0.2)^2 \times (0.8)^3 = 2.048 \]
  • For 1 customer buying an ordinary TV: \[ 3 \times (0.5)^1 \times (0.5)^2 = 0.375 \]
Combining both probabilities together to represent our desired outcomes:
\[ 2.048 \times 0.375 = 0.768 \]
Binomial Coefficient
The binomial coefficient, also known as \( C(n, k) \), is a coefficient appearing in the binomial theorem or binomial expansions.It tells us the number of ways to choose \( k \) successes (or failures) out of \( n \) trials and is integral to solving binomial probability problems.
This coefficient is computed using the same combination formula mentioned earlier, highlighting its role in both permutations and probability calculations.
In a real-world context, understanding binomial coefficients can help predict the likelihood of a combination of events鈥攍ike how many customers will purchase or just browse in a store.For example, determining the number of ways to get 2 purchasing color TVs and 1 purchasing an ordinary TV from 5 customers involves these coefficients.
Recognizing the importance of binomial coefficients in probability and statistics is crucial for career fields such as economics, healthcare, and anyindustry that analyzes consumer behavior or predicts trends.

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Most popular questions from this chapter

Suppose that we want to generate a random variable \(X\) that is equally likely to be either 0 or 1 , and that all we have at our disposal is a biased coin that, when flipped, lands on heads with some (unknown) probability \(p .\) Consider the following procedure: 1\. Flip the coin, and let \(0_{1}\), either heads or tails, be the result. 2\. Flip the coin again, and let \(0_{2}\) be the result. 3\. If \(0_{1}\) and \(0_{2}^{-}\) are the same, return to step 1 . 4\. If \(0_{2}\) is heads, set \(X=0\), otherwise set \(X=1\). (a) Show that the random variable \(X\) generated by this procedure is equally likely to be either 0 or 1 . (b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different, and then sets \(X=0\) if the final flip is a head, and sets \(X=1\) if it is a tail?

Let \(c\) be a constant. Show that (i) \(\operatorname{Var}(c X)=c^{2} \operatorname{Var}(X)\). (ii) \(\operatorname{Var}(c+X)=\operatorname{Var}(X)\) \(\because\)

An individual claims to have extrasensory perception (ESP). As a test, a fair coin is flipped ten times, and he is asked to predict in advance the outcome. Our individual gets seven out of ten correct. What is the probability he would have done at least this well if he had no ESP? (Explain why the relevant probability is \(P\\{X \geqslant 7\\}\) and not \(P\\{X=7\\} .)\)

For the multinomial distribution (Exercise 17), let \(N_{i}\) denote the number of times outcome \(i\) occurs. Find (i) \(E\left[N_{i}\right]\) (ii) \(\operatorname{Var}\left(N_{i}\right)\) (iii) \(\operatorname{Cov}\left(N_{i}, N_{j}\right)\) (iv) Compute the expected number of outcomes that do not occur.

Suppose that \(X\) and \(Y\) are independent binomial random variables with parameters \((n, p)\) and \((m, p)\). Argue probabilistically (no computations necessary) that \(X+Y\) is binomial with parameters \((n+m, p)\).
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