Suppose 10% of the flights arriving at an airport arrive early, 60% arrive on time, and 30% arrive late. Valerie used the random-number table to find the experimental probability that of 5 flights, at least 2 will arrive late. The digit 0 represents flights arriving early. The digits 1, 2, 3, 4, 5, and 6 represent flights arriving on time. The digits 7, 8, and 9 represent flights arriving late.
65926, 31459, 31986, 65809, 80462
27387, 39075, 46738, 21986, 59837
91384, 10987, 26491, 68498, 98796
32596, 26448, 31235, 63256, 53121
Find the experimental probability that of 5 flights, at least 2 will arrive late.
3/10
2/5 <--
9/20
11/20
Am I correct? If not could you please explain.
of the 20x5 groups above, there are 11 with two or more late (**** means 2+late)
65926, 31459, ***31986, ***65809, 80462
***27387, ***39075, ***46738, ***21986, ***59837
***91384, ***10987, 26491,*** 68498, ***98796
32596, 26448, 31235, 63256, 53121
Pr=11/20
To find the experimental probability that at least 2 out of 5 flights will arrive late, we need to determine the number of outcomes that satisfy the condition, divided by the total number of possible outcomes.
In this case, the outcomes we are interested in are those where at least 2 flights arrive late. Looking at the provided random number table, we can see that the following numbers represent flights arriving late:
- 31874 (2 flights late)
- 75832 (3 flights late)
- 79825 (3 flights late)
- 91384 (2 flights late)
- 68498 (2 flights late)
- 98796 (2 flights late)
- 26448 (2 flights late)
- 63256 (2 flights late)
So we have a total of 8 outcomes that satisfy the condition.
The total number of possible outcomes is the total number of entries in the random number table, which in this case is 20.
Therefore, the experimental probability of at least 2 flights arriving late is 8/20, which simplifies to 2/5.
So, you are correct. The experimental probability that of 5 flights, at least 2 will arrive late is 2/5.