A study found that 40% of the assisted reproductive technology (ART) cycles resulted in pregnancies. Twenty-three percent of the ART pregnancies resulted in multiple births.
(a) Find the probability that a random selected ART cycle resulted in a pregnancy and produced a multiple birth.
(b) Find the probability that a randomly selected ART cycle that resulted in a pregnancy did not produce a multiple birth.
(c) Would it be unusual for randomly selected ART cycle to result in a pregnancy and product a multiple birth? Explain.
Solution
(a) Let A be the event that an ART cycle pregnancy and B be the event that an ART cycle resulted in a multiple birth. Note that the probability that two events A and B will occur in sequence is as follows.
P(A and B) = P(A).P(B│A)
Determine P(A) and P(B│A).
P(A) = 
= 0.4
P(B│A) = 
= 0.23
The probability that a randomly selected ART cycle resulted in a pregnancy and produced a multiple birth is P(A ns B). Substitute the value for P(A) and P(B│A) into the formula and simplify to find P(A and B).
P(A and B) = P(A). P(B│A)
= 0.4 * 0.23
=0.092
Thus, the probability that a randomly selected ART cycle resulted in a pregnancy and produced a multiple birth is 0.092.
(b) Let B ’ be the complement of B. Hence, the probability that a randomly selected ART cycle that resulted in a pregnancy did not product a multiple birth is P(B’│A). Notice that, in the sample space of ART cycle resulting in pregnancies, the event {B’│A} is the set of all outcomes that are not included in the event {B│A}. Therefore, the event {B’│A} is the complement of the event {B│A}.
Determine P(B’│A) using the formula the complement of an event, P(E’) = 1 – P(E), where E is an event and E’ is its complement. Recall that P(B│A) = 0.23.
P(B’│A) = 1 – P(B│A)
= 1 – 0.23
= 0.77
Therefore, the probability that a randomly selected ART cycle that resulted in a pregnancy did not produce a multiple birth is 0.77.
(c) An event that occurs with a probability of 0.05 or less is typically considered unusual. Use this information to determine whether it would be unusual for a randomly selected ART cycle to result in a pregnancy and produce a multiple birth. Recall that P(A and B) = 0.092.