Question

Suppose 20% of babies born 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 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.

23059 78234 01785 12359 26789
14568 24579 13579 01239 24589
03489 12456 01458 23567 01238
01235 34567 23478 13546 23589

Find the experimental probability that of 5 babies born, at least 1 will be born early.
A. one-fifth
B. two-fifths
C. three-fifths
D. start fraction 4 over 5 end fraction

Answer 1

To use the random-number table, we can look at each digit separately and see what it represents. For example, in the first row, the first digit is 2, which represents a baby born on time. The second digit is 3, which also represents a baby born on time. Continuing in this way, we can generate five "births": on time, on time, early, on time, late.

To find the probability of at least 1 baby being born early, we need to find the probability of the complement (no babies born early) and subtract it from 1. The probability of no babies born early is (0.8)^5 = 0.32768, since each digit has an 80% chance of not being a 0 or 1. Therefore, the probability of at least 1 baby being born early is 1 - 0.32768 = 0.67232, which is approximately 0.67 or two-thirds.

The answer is not given in the choices, but the closest is B) two-fifths, which is not correct.