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Chapter 3: Problem 16

Ninety-eight percent of all babies survive delivery. However, 15 percent of all births involve Cesarean (C) sections, and when a C section is performed, the baby survives 96 percent of the time. If a randomly chosen pregnant woman does not have a C section, what is the probability that her baby survives?

Short Answer

Expert verified
The probability that the baby survives if the woman does not have a C section is approximately 98.35%.

Step by step solution

01

Identify the events

We need to identify the events we are interested in. Let A be the event that the baby survives and B be the event that the woman has a C section. We are looking for the probability of the baby surviving given that there is no C section, or P(A | B').
02

Find the probabilities of events

We have been given: - P(A) = 0.98 (The probability of a baby surviving) - P(B) = 0.15 (The probability of a C section) - P(A | B) = 0.96 (The probability of a baby surviving given there's a C section) Now, we have to find the probability of a baby surviving given that there is no C section, or P(A | B').
03

Apply the conditional probability formula

The conditional probability formula is: \(P(A | B') = \frac{P(A \cap B')}{P(B')}\) First, we need to find P(B'), which is the probability of not having a C section: P(B') = 1 - P(B) = 1 - 0.15 = 0.85 Next, we can find P(A 鈭� B') using the formula: \(P(A \cap B') = P(A) - P(A \cap B)\) We can find P(A 鈭� B) using the given conditional probability P(A | B): \(P(A \cap B) = P(A | B) * P(B) = 0.96 * 0.15 = 0.144\) Now we can find P(A 鈭� B'): \(P(A \cap B') = P(A) - P(A \cap B) = 0.98 - 0.144 = 0.836\)
04

Calculate the probability of surviving given no C section

We can now put all these values to calculate the probability of surviving given no C section: \(P(A | B') = \frac{P(A \cap B')}{P(B')} = \frac{0.836}{0.85} 鈮� 0.9835\) So, the probability that the baby survives if the woman does not have a C section is approximately 98.35%.

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Key Concepts

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

Events in Probability
In probability, an event is essentially a set of outcomes to which we assign a probability. Events in probability can either be simple or compound. A simple event involves one possible outcome, while a compound event involves a combination of outcomes.
For example, in the 'baby survival' scenario, several events are at play:
  • Event A: The baby survives.
  • Event B: A Cesarean (C) section occurs.
  • Event B': A natural birth occurs (no C section).
Understanding these events is crucial, as it forms the basis for computing probabilities and determining their relationships and dependencies with respect to one another. Identifying the correct events helps in applying the right probability techniques to solve problems.
Probability Formula
To solve probability problems, it is essential to make use of probability formulas which include conditional probability, multiplication rules, and the addition rule.
The Conditional Probability Formula, for instance, is used to find the likelihood of event A happening given that event B has happened. It is expressed as:\[P(A | B) = \frac{P(A \cap B)}{P(B)}\]
In simpler terms, we calculate the intersection of events A and B (both events happening together) and divide it by the probability of event B occurring. Applying this to our exercise, we aim to find P(A | B'), the probability of the baby surviving when there is no C section. This requires understanding and calculating the individual probabilities before using the formula.
Bayes' Theorem
Bayes' Theorem is a concept in probability that helps us update our beliefs about the probability of a hypothesis based on new evidence. Although not explicitly required in the exercise, Bayes' Theorem provides a framework for understanding conditional probabilities.
It states:\[P(B | A) = \frac{P(A | B) \cdot P(B)}{P(A)}\]
This formula allows us to find the probability of event B given that event A has happened, using the probability of A given B, the probability of B, and the probability of A. While solving the given exercise does not demand Bayes' Theorem directly, the comprehension of how probabilities update with new information enriches our understanding of conditional probability.
Probability Distribution
Probability distribution provides a comprehensive picture of all possible outcomes of a random variable and the likelihood of each outcome. It can be discrete, with specific values, or continuous for a range of values.
In real-world scenarios like baby deliveries, we often deal with discrete distributions since we have defined outcomes, such as survival versus non-survival. Calculating distributions involves:
  • Listing all possible outcomes.
  • Assigning and summing probabilities to ensure they add up to 1.
  • Utilizing formulas and theorems to find specific probabilities.
Understanding probability distributions helps us visualize and handle uncertainties effectively, as seen in evaluating baby survival rates under different childbirth methods.

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Most popular questions from this chapter

A town council of 7 members contains a steering committee of size \(3 .\) New ideas for legislation go first to the steering committee and then on to the council as a whole if at least 2 of the 3 committee members approve the legislation. Once at the full council, the legislation requires a majority vote (of at least 4 ) to pass. Consider a new piece of legislation, and suppose that each town council member will approve it, independently, with probability \(p .\) What is the probability that a given steering committee member's vote is decisive in the sense that if that person's vote were reversed, then the final fate of the legislation would be reversed? What is the corresponding probability for a given council member not on the steering committee?

\(A\) and \(B\) alternate rolling a pair of dice, stopping either when \(A\) rolls the sum 9 or when \(B\) rolls the sum 6\. Assuming that \(A\) rolls first, find the probability that the final roll is made by \(A .\)

The color of a person's eyes is determined by a single pair of genes. If they are both blue-eyed genes, then the person will have blue eyes; if they are both brown-eyed genes, then the person will have brown eyes; and if one of them is a blue-eyed gene and the other a brown-eyed gene, then the person will have brown eyes. (Because of the latter fact, we say that the brown-eyed gene is dominant over the blue-eyed one.) A newborn child independently receives one eye gene from each of its parents, and the gene it receives from a parent is equally likely to be either of the two eye genes of that parent. Suppose that Smith and both of his parents have brown eyes, but Smith's sister has blue eyes. (a) What is the probability that Smith possesses a blue eyed gene? (b) Suppose that Smith's wife has blue eyes. What is the probability that their first child will have blue eyes? (c) If their first child has brown eyes, what is the probability that their next child will also have brown eyes?

3.25. The following method was proposed to estimate the number of people over the age of 50 who reside in a town of known population 100,000: "As you walk along the streets, keep a running count of the percentage of people you encounter who are over \(50 .\) Do this for a few days; then multiply the percentage you obtain by 100,000 to obtain the estimate." Comment on this method. Hint: Let \(p\) denote the proportion of people in the town who are over \(50 .\) Furthermore, let \(\alpha_{1}\) denote the proportion of time that a person under the age of 50 spends in the streets, and let \(\alpha_{2}\) be the corresponding value for those over \(50 .\) What quantity does the method suggested estimate? When is the estimate approximately equal to \(p ?\)

In any given year, a male automobile policyholder will make a claim with probability \(p_{m}\) and a female policyholder will make a claim with probability \(p_{f},\) where \(p_{f} \neq p_{m} .\) The fraction of the policyholders that are male is \(\alpha, 0<\alpha<1 .\) A policyholder is randomly chosen. If \(A_{i}\) denotes the event that this policyholder will make a claim in year \(i,\) show that $$ P\left(A_{2} | A_{1}\right)>P\left(A_{1}\right) $$ Give an intuitive explanation of why the preceding inequality is true.
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