Questions: The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC). A local hospital randomly selects five births and lets the random variable X count th number not resulting in a defect. Assume the births are independent. What is the probability that exactly two births do not result in defects? 0.0003 0.1328 0.0008 0.0082

Solution Steps

We are tasked with finding the probability that exactly 2 out of 5 randomly selected births do not result in a birth defect. The probability of a birth not resulting in a defect is given as \( p = \frac{32}{33} \), while the probability of a birth resulting in a defect is \( q = \frac{1}{33} \).

The probability of exactly \( x \) successes (births not resulting in defects) in \( n \) trials (total births) can be calculated using the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

Where:

• \( n = 5 \) (total births)
• \( x = 2 \) (births not resulting in defects)
• \( p = \frac{32}{33} \) (probability of success)
• \( q = \frac{1}{33} \) (probability of failure)

Substituting the values into the formula:

\[ P(X = 2) = \binom{5}{2} \cdot \left(\frac{32}{33}\right)^2 \cdot \left(\frac{1}{33}\right)^{5-2} \]

Calculating \( \binom{5}{2} \):

\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \cdot 4}{2 \cdot 1} = 10 \]

Now substituting back into the probability formula:

\[ P(X = 2) = 10 \cdot \left(\frac{32}{33}\right)^2 \cdot \left(\frac{1}{33}\right)^3 \]

Calculating \( \left(\frac{32}{33}\right)^2 \) and \( \left(\frac{1}{33}\right)^3 \):

\[ \left(\frac{32}{33}\right)^2 = \frac{1024}{1089} \] \[ \left(\frac{1}{33}\right)^3 = \frac{1}{35937} \]

Thus,

\[ P(X = 2) = 10 \cdot \frac{1024}{1089} \cdot \frac{1}{35937} \]

Calculating the final probability gives:

\[ P(X = 2) \approx 0.0003 \]

Final Answer