A study found that 3939% of the assisted reproductive technology (art) cycles resulted in pregnancies. twenty-fourfour 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 a randomly selected art cycle to result in a pregnancy and produce a multiple birth? explain

(a) Let A be an ART cycle pregnancy and;B be the event that an ART cycle resulted in a multiple birth.
P(A and B) = P(A).P(B│A)
Look for P(A) and P(B│A).
P(A) = 39/100 = 0.39
P(B│A) =24/100 = 0.24
The probability that a randomly selected ART cycle resulted in a pregnancy and produced a multiple birth is P(A and B) is:
P(A and B) = P(A). P(B│A)
= 0.39 * 0.24
= 0.0936

(b) Let B ’ - the complement of B. This means that this is the probability that a randomly selected ART cycle that resulted in a pregnancy did not make a multiple birth is P(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) = 0.24.
P(B’│A) = 1 – P(B│A)
= 1 – 0.24
= 0.76
(c) Knowing that P(A and B) = 0.0936. An event that happens with a probability of 0.05 or less is normally considered unusual. Use this information to know whether it would be unusual for a randomly selected ART cycle to result in a pregnancy and produce a multiple birth.

MsEHolt MsEHolt · Answer 2 · 2018-07-05T04:51:41+02:00

The correct answers are:

0.0936;
0.2964; and
no.

Explanation:

39% of ART cycles result in pregnancy. Of those, 24% result in multiple births. To find the probability that a random ART cycle results in a pregnancy with multiple births, we want to find 24% of 39%; this means we multiply these:
0.24(0.39) = 0.0936.

To find the probability that it will not result in multiple births, we subtract from 100%; since 24% of the pregnancies are multiple births, 100-24 = 76% will not be multiple births.

Now we find 76% of 39%; we multiply again: 0.76(0.39) = 0.2964.

In statistics, a result is considered unusual if the probability is 0.05 or less. Our probability was 0.0936, which is greater than 0.05; thus it is not unusual.