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.
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
so add up the points for success, and divide by the total # points
@oobleck What??
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.
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%.
To find the experimental probability that at least 1 baby will be born early out of 5 births, we need to count the number of outcomes where at least 1 baby is born early and divide it by the total number of possible outcomes.
Looking at the given random-number table, we can see that the digits 0 and 1 represent babies born early. Therefore, any outcome that includes a 0 or 1 represents a birth where at least 1 baby is born early.
Analyzing each row of the random-number table, we can identify the following outcomes where at least 1 baby is born early:
- Row 1: 23059
- Row 2: 14568
- Row 3: 03489
- Row 4: 01235
So there are 4 outcomes where at least 1 baby is born early out of the total 5 outcomes.
Therefore, the experimental probability that at least 1 baby will be born early out of 5 births is 4/5.
So the correct answer is option D) 4/5.