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Solve each problem.The probability that a male will be color-blind is 0.042 . Find the probabilities that in a group of 53 men, the following are true. (a) Exactly 5 are color-blind. (b) No more than 5 are color-blind. (c) None are color-blind. (d) At least 1 is color-blind.

Short Answer

Expert verified
(a) 0.161. (b) 0.973. (c) 0.103. (d) 0.897.

Step by step solution

01

Define the Probability and Parameters

The probability of a male being color-blind is given as 0.042. Let this be represented as p = 0.042. The group size is 53 men, so n = 53. This can be solved using the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \], where \(X\) is the number of color-blind men, \(k\) is the specific number of successes, and \(n\) and \(p\) are as defined.
02

Compute Binomial Coefficient

For part (a), exactly 5 are color-blind. Compute the binomial coefficient: \[ \binom{53}{5} = \frac{53!}{5!(53-5)!} \]
03

Apply Probability Formula for Part (a)

Substitute the values into the binomial probability formula: \( P(X = 5) = \binom{53}{5} (0.042)^5 (1-0.042)^{53-5} \). Calculate the value.
04

Calculate Cumulative Probability for Part (b)

For part (b), no more than 5 are color-blind. Calculate \( P(X \leq 5) \) which is the sum of the probabilities from 0 to 5 color-blind men: \[ P(X \leq 5) = \sum_{k=0}^{5} \binom{53}{k} (0.042)^k (1-0.042)^{53-k} \]
05

Compute Probability for Part (c)

For part (c), none are color-blind. Use the binomial formula where k = 0: \[ P(X = 0) = \binom{53}{0} (0.042)^0 (1-0.042)^{53} \].
06

Calculate Cumulative Probability for Part (d)

For part (d), at least 1 is color-blind. This is the complement of none being color-blind: \[ P(X \geq 1) = 1 - P(X = 0) \].

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Key Concepts

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

Understanding Color Blindness Probability
Color blindness is a condition where a person is unable to distinguish between certain colors. The probability of a male being color-blind is 0.042, or 4.2%. This means, if we randomly select a male, there is a 4.2% chance that he will be color-blind.
This probability can be used to predict outcomes in larger groups. If you have a group of 53 men, you can use binomial probability to calculate various probabilities related to the number of color-blind men.
In our exercise, we will determine:
  • The probability that exactly 5 men are color-blind.
  • The probability that no more than 5 men are color-blind.
  • The probability that none are color-blind.
  • The probability that at least 1 man is color-blind.
Binomial Coefficient
The binomial coefficient, often represented as \(\binom{n}{k}\), is a key part of the binomial probability formula. It calculates the number of ways to choose \(k\) successes from \(n\) trials.
For example, to find the probability that exactly 5 out of 53 men are color-blind, we need the binomial coefficient \( \binom{53}{5} \). You can calculate it as:
\[ \binom{53}{5} = \frac{53!}{5!(53-5)!} \] This computation shows how many ways we can choose 5 color-blind men from 53. Once you have this value, you multiply it by the probability of success raised to the power of \(k\) (number of successes) and the probability of failure raised to the power of \(n-k\) (number of failures).
Learning how to use the binomial coefficient effectively allows you to tackle many different types of probability problems involving fixed numbers of trials and binary outcomes (success/failure).
Cumulative Probability
Cumulative probability sums up the probabilities of multiple outcomes. For instance, if we want to find the probability that no more than 5 men are color-blind in our group of 53, we calculate the cumulative probability for 0, 1, 2, 3, 4, and 5 color-blind men.
Mathematically, this is represented as:
\[ P(X \leq 5) = \sum_{k=0}^{5} \binom{53}{k} (0.042)^k (1-0.042)^{53-k} \] Here, we need to add up the probabilities for each case from 0 to 5. This result helps us to understand the broader range of possible outcomes rather than focusing on a single specific outcome.
Cumulative probabilities are particularly useful when calculating probabilities for ranges of outcomes, such as finding the likelihood that 'at most' or 'at least' a certain number of events will occur. By adding the probabilities for all these individual cases, we obtain a more comprehensive understanding of the scenario.

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Most popular questions from this chapter

Assume that \(n\) is a positive integer. Use mathematical induction to prove each statement S by following these steps. See Example \(I\). (a) Verify the statement for \(n=1\) (b) Write the statement for \(n=k\) (c) Write the statement for \(n=k+1\) (d) Assume the statement is true for \(n=k\). Use algebra to change the statement in part (b) to the statement in part (c). (e) Write a conclusion based on Steps (a)-(d). $$2+4+8+\dots+2^{n}=2^{n+1}-2$$

Assume that \(n\) is a positive integer. Use mathematical induction to prove each statement S by following these steps. See Example \(I\). (a) Verify the statement for \(n=1\) (b) Write the statement for \(n=k\) (c) Write the statement for \(n=k+1\) (d) Assume the statement is true for \(n=k\). Use algebra to change the statement in part (b) to the statement in part (c). (e) Write a conclusion based on Steps (a)-(d). $$5+10+15+\dots+5 n=\frac{5 n(n+1)}{2}$$

The table gives the results of a 2008 survey of Americans aged \(18-24\) in which the respondents were asked, "During the past 30 days, for about how many days have you felt that you did not get enough sleep?"$$\begin{array}{|l|c|c|c|c|}\hline \text { Number of Days } & 0 & 1-13 & 14-29 & 30 \\ \hline \text { Percent (as a decimal) } & 0.23 & 0.45 & 0.20 & 0.12 \\\\\hline\end{array}$$ Using the percents as probabilities, find the probability that, out of 10 respondents in the \(18-24\) age group selected at random, the following were true. Fewer than 2 did not get enough sleep on 14 or more days.

The recursively defined sequence \(a_{1}=k\) $$ a_{n}=\frac{1}{2}\left(a_{n-1}+\frac{k}{a_{n-1}}\right), \quad \text { if } n>1$$ can be used to compute \(\sqrt{k}\) for any positive number \(k\). This sequence was known to Sumerian mathematicians 4000 years ago, and it is still used today. Use this sequence to approximate the given square root by finding \(a_{6} .\) Compare your result with the actual value. (Source: Heinz-Otto, P., Chaos and Fractals, Springer-Verlag.) (a) \(\sqrt{2}\) (b) \(\sqrt{11}\)

The management of a firm wishes to survey the opinions of its workers, classified as follows for the purpose of an interview:\(30 \%\) have worked for the company 5 or more years, \(28 \%\) are female,\(65 \%\) contribute to a voluntary retirement plan, and \(50 \%\) of the female workers contribute to the retirement plan. Find each probability if a worker is selected at random. (a) A male worker is selected. (b) A worker is selected who has worked for the company less than 5 yr. (c) A worker is selected who contributes to the retirement plan or is female.

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