Answer:
a) To estimate the probability of a randomly selected pregnant woman delivering twins, we divide the number of twin births by the total number of births:
\[ P(\text{Twin birth}) = \frac{125,336}{3,858,214 + 125,336 + 4,233 + 266 + 47} \]
\[ P(\text{Twin birth}) \approx \frac{125,336}{3,988,096} \]
\[ P(\text{Twin birth}) \approx 0.031 \]
So, the estimated probability of a randomly selected pregnant woman delivering twins is approximately 0.031.
b) To estimate the probability of a randomly selected pregnant woman delivering quadruplets, we divide the number of quadruplet births by the total number of births:
\[ P(\text{Quadruplet birth}) = \frac{266}{3,858,214 + 125,336 + 4,233 + 266 + 47} \]
\[ P(\text{Quadruplet birth}) \approx \frac{266}{3,988,096} \]
\[ P(\text{Quadruplet birth}) \approx 0.0000667 \]
So, the estimated probability of a randomly selected pregnant woman delivering quadruplets is approximately 0.000067.
c) To estimate the probability of a randomly selected pregnant woman delivering more than a single child, we sum the number of multiple births (twins, triplets, quadruplets, quintuplets, or higher) and divide it by the total number of births:
\[ P(\text{More than single birth}) = \frac{125,336 + 4,233 + 266 + 47}{3,858,214 + 125,336 + 4,233 + 266 + 47} \]
\[ P(\text{More than single birth}) \approx \frac{129,882}{3,988,096} \]
\[ P(\text{More than single birth}) \approx 0.0326 \]
So, the estimated probability of a randomly selected pregnant woman delivering more than a single child is approximately 0.033.
