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Chapter 4: Problem 24

Sickle cell anemia. Sickle cell anemia is a genetic blood disorder where red blood cells lose their flexibility and assume an abnormal, rigid, "sickle" shape, which results in a risk of various complications. If both parents are carriers of the disease, then a child has a \(25 \%\) chance of having the disease, \(50 \%\) chance of being a carrier, and \(25 \%\) chance of neither having the disease nor being a carrier. If two parents who are carriers of the disease have 3 children, what is the probability that (a) two will have the disease? (b) none will have the disease? (c) at least one will neither have the disease nor be a carrier? (d) the first child with the disease will the be \(3^{r d}\) child?

Short Answer

Expert verified
(a) 0.140625, (b) 0.421875, (c) 0.578125, (d) 0.140625

Step by step solution

01

Define Variables and Probability

For this exercise, we will use the binomial probability formula. Define \( p = 0.25 \) for a child having the disease (since the probability of a child having the disease is 25%). Thus, \( q = 1 - p = 0.75 \) (the probability of not having the disease). Each child is an independent event.
02

Probability of Exactly Two Children Having the Disease

To find the probability that exactly 2 out of 3 children will have the disease, use the binomial probability formula:\[P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\]Here, \( n = 3 \), \( k = 2 \), and \( p = 0.25 \):\[P(X=2) = \binom{3}{2} (0.25)^2 (0.75)^1 = 3 \times 0.0625 \times 0.75 = 0.140625\]
03

Probability of None Having the Disease

To find the probability that none of the children will have the disease, use the same formula with \( k = 0 \):\[P(X=0) = \binom{3}{0} (0.25)^0 (0.75)^3 = 1 \times 1 \times 0.421875 = 0.421875\]
04

Probability of at Least One Child Not Having Disease or Being a Carrier

The probability of a child neither having the disease nor being a carrier is given as \( 0.25 \). The probability that at least one child fits this category is the complement of all children being either diseased or carriers. Calculate the probability of 3 children not fitting and subtract from 1:\[P(at \ least \ one \ not \ having \ disease \ or \ being \ carrier) = 1 - P(all \ not \ 0.25) = 1 - (0.75)^3 = 1 - 0.421875 = 0.578125\]
05

Probability for the First Child with Disease Being Third Child

The situation requires that the first two children do not have the disease, and the third child is the first with the disease. This is:\[P(first \ and \ second \ without \ disease, \ third \ with \ disease) = (0.75) (0.75) (0.25) = 0.140625\]

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Key Concepts

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

Genetic Disorder
Sickle cell anemia is a genetic disorder affecting the shape of red blood cells, causing them to become rigid and sickle-shaped. This can lead to a variety of health complications due to the lack of flexibility in the blood cells.
Genetic disorders like sickle cell anemia are passed down from parents to offspring, following specific inheritance patterns. In genetics, these patterns help predict the likelihood of a child inheriting a disease.
Both parents being carriers means they have one sickle cell gene. Each pregnancy has a new roll of the dice, independent of each other. The chance remains the same for every child with 25% for having the disease, 50% for being a carrier, and 25% for neither.
It's crucial to understand these probabilities to grasp the genetic transmission of disorders. This knowledge aids in predicting risks and understanding outcomes.
Independent Events
When considering genetic outcomes like those for sickle cell anemia, each child's genetic makeup is considered an independent event. In probability, an event is said to be independent if the outcome of any event is not affected by previous events.
In this context, whether one child inherits the disease or not does not impact the probability of the next child inheriting it.
  • Suppose for child one, the outcome is that they have the disease.
  • For the next child, the probability remains unchanged.
  • It's still 25% for having the disease, regardless of the first child's outcome.
Understanding this concept is fundamental when calculating probabilities for multiple children as it shows why we treat each as a separate probabalistic experiment, reinforcing the continuity of probabilities across independent events.
Probability Formula
The binomial probability formula is essential in calculating the likelihood of specific outcomes. When a scenario involves a fixed number of independent trials, like determining the likelihood of children having a genetic disorder, this formula proves invaluable.
The formula is expressed as \[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \] where:
  • \( n \) is the number of trials (e.g., number of children).
  • \( k \) is the number of successful outcomes you are interested in (e.g., children with the disease).
  • \( p \) is the probability of success on a single trial (e.g., probability one child will have the disease).
  • \( 1-p \) is the probability of failure.
This formula lets you calculate the probability of getting exactly \( k \) successes out of \( n \) trials, such as exactly two of the three children having the disease. Knowing how to use this formula is a powerful tool for anyone working with probabilities.
Complement Rule
In probability, the complement rule is used to find the likelihood of the opposite occurring when it's easier to calculate the probability of an event not happening.
The complement rule is expressed as \[ P( ext{Not A}) = 1 - P(A) \]Applying this to genetics, especially for sickle cell anemia, it provides a way to determine events like at least one child neither having the disease nor being a carrier.
Suppose we want to calculate the probability of at least one child out of three fitting this description:
  • First, calculate the probability of none fitting the description (all having the disease or being carriers).
  • Subtract this probability from 1 to find the complement probability.
  • In our example, this becomes: \( 1 - (0.75)^3 = 0.578125 \).
The complement rule is a simple yet powerful concept that simplifies probability calculations, particularly in complex scenarios.

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Most popular questions from this chapter

Exercise 4.36 states that the distribution of speeds of cars traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. The speed limit on this stretch of the I-5 is 70 miles/hour. (a) A highway patrol officer is hidden on the side of the freeway. What is the probability that 5 cars pass and none are speeding? Assume that the speeds of the cars are independent of each other. (b) On average, how many cars would the highway patrol officer expect to watch until the first car that is speeding? What is the standard deviation of the number of cars he would expect to watch?

CAPM. The Capital Asset Pricing Model (CAPM) is a financial model that assumes returns on a portfolio are normally distributed. Suppose a portfolio has an average annual return of \(14.7 \%\) (i.e. an average gain of \(14.7 \%\) ) with a standard deviation of \(33 \%\). A return of \(0 \%\) means the value of the portfolio doesn't change, a negative return means that the portfolio loses money, and a positive return means that the portfolio gains money. (a) What percent of years does this portfolio lose money, i.e. have a return less than \(0 \% ?\) (b) What is the cutoff for the highest \(15 \%\) of annual returns with this portfolio?

Arachnophobia. A Gallup Poll found that \(7 \%\) of teenagers (ages 13 to 17 ) suffer from arachnophobia and are extremely afraid of spiders. At a summer camp there are 10 teenagers sleeping in each tent. Assume that these 10 teenagers are independent of each other. \({ }^{33}\) (a) Calculate the probability that at least one of them suffers from arachnophobia. (b) Calculate the probability that exactly 2 of them suffer from arachnophobia. (c) Calculate the probability that at most 1 of them suffers from arachnophobia. (d) If the camp counselor wants to make sure no more than 1 teenager in each tent is afraid of spiders, does it seem reasonable for him to randomly assign teenagers to tents?

Pew Research reported that the typical response rate to their surveys is only \(9 \%\). If for a particular survey 15,000 households are contacted, what is the probability that at least 1,500 will agree to respond? \(^{42}\)

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