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
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.
To find the experimental probability that at least 1 baby will be born early out of 5 births, we need to analyze the digits in the random-number table.
Out of the 20 digits in each row, the digits 0 and 1 represent babies born early. Looking at each row, we can count the number of times these digits appear:
- Row 1: 1 (1 baby born early)
- Row 2: 0 (no babies born early)
- Row 3: 1 (1 baby born early)
- Row 4: 1 (1 baby born early)
Out of the 4 rows, 3 rows have at least 1 baby born early. Therefore, the experimental probability is 3 out of 4.
The experimental probability that of 5 babies born, at least 1 baby will be born early is 3/4.
So, the correct answer is D. start fraction 4 over 5 end fraction.