Suppose 20% of babies are born early, 50% are born on time, and 30% are born late. A nurse uses a random-number table to find the experimental probability that of 5 births , at least 1 baby will be born early. The digits 0 and 1 represents babies born early. The digits 2, 3, 4, 5, and 6 represent babies born on time. The digits 7, 8, 9 represent babie s born late.

Question

23059 - 78234 - 01785 - 12359 - 26789
14568 - 24579 - 13579 - 01239 - 24589
03489 - 12456 - 01458 - 23567 - 01238
01235 - 34567 - 23478 - 12546 - 23589

Find the experimental probability that of 5 babies born, at least 1 will be born early.
(1 point)

A 1/5
B 2/5
C 3/5
D 4/5

Answer 1

so add up the points for success, and divide by the total # points

Answer 2

@oobleck What??

Answer 3

10. Suppose 20% of babies born are born early, 50% are born on time, and 30% are born late. A nurse uses a random-numbertable to find the experimental probability that of 5 births, at least 1 baby will be born early. The digits O and 1 represent babies born early. The digits 2, 3, 4, 5, and 6 represent babies born on time. The digits 7, 8, and 9 represent babies born late.

Answer 4

To find the experimental probability that at least 1 baby will be born early, we need to find the probability that none of the 5 babies are born early, and then subtract that from 1.

The probability that a baby is not born early is 80%, or 0.8. So the probability that none of the 5 babies are born early is:

0.8 * 0.8 * 0.8 * 0.8 * 0.8 = 0.32768

To find the probability that at least 1 baby will be born early, we subtract this from 1:

1 - 0.32768 = 0.67232

So the experimental probability that of 5 births, at least 1 baby will be born early, is about 0.672 or 67.2%.