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

Question

Mathematics

Maeve Nixon

8 May, 02:06

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

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Answers (2)

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Aliza Stout · Answer 1

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.

Oscar Weber · Answer 2

(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.