Answer:
The probability that a birth was to a teen mother if we know that the pregnancy was unintended is 0.1942.
Step-by-step explanation:
Denote the events as follows:
T = birth was to a teen mother
Y = birth was to a young-adult mother
A = birth was to an adult mother
U = a birth was unintended
I = a birth was intended.
Given:
P (T) = 0.09
P (Y) = 0.24
P (A) = 0.67
P (I | T) = 0.23
P (U | T) = 0.77
P (I | Y) = 0.50
P (U | Y) = 1 - P (I | Y) = 0.50
P (I | A) = 0.75
P (U | A) = 1 - P (I | A) = 0.75 = 0.25
Consider the tree diagram below.
According to the Bayes' theorem,
[tex]P(B|A)=\frac{P(A|B)P(B)}{P(A|B)P(B)+P(A|B^{c})P(B^{c})}[/tex]
Using the Bayes' theorem compute the probability that a birth was to a teen mother if we know that the pregnancy was unintended as follows:
[tex]P(T|U)=\frac{P(U|T)P(T)}{P(U|T)P(T)+P(U|Y)P(Y)+P(U|A)P(A)} \\=\frac{(0.77\times0.09)}{(0.77\times0.09)+(0.50\times0.24)+(0.25\times0.67)} \\=0.19423\\\approx0.1942[/tex]
The probability that a birth was to a teen mother if we know that the pregnancy was unintended is 0.1942.
