Problem 16 - A First Course in Probability

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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').

Step 2: 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').

Step 3: 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 ∩ B') using the formula: \(P(A \cap B') = P(A) - P(A \cap B)\) We can find P(A ∩ 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 ∩ B'): \(P(A \cap B') = P(A) - P(A \cap B) = 0.98 - 0.144 = 0.836\)

Step 4: 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: \(P(A | B') = \frac{P(A \cap B')}{P(B')} = \frac{0.836}{0.85} ≈ 0.9835\) So, the probability that the baby survives if the woman does not have a C section is approximately 98.35%.