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.