A survey finds that 40% of students ride in a car to school. Eri would like to estimate the probability that if 3 students were randomly selected, only 1 rides in a car.
To simulate this probability, she lets the numbers 0, 1, 2, and 3 represent a student who rides in a car to school and then 4, 5, 6, 7, 8, and 9 represent a student who does not ride in a car to school. She then has a computer randomly select 3 numbers. She repeats this process for 20 trials.
The results of these trials are shown in this list.
468, 380, 120, 220, 553,945, 935, 607, 473, 490,074, 981, 692, 518, 408,954, 943, 389, 594, 569
Based on this simulation, what is the estimated probability that only 1 of 3 randomly selected students rides in a car to school?
Enter your answer, as a decimal, in the box.
A survey finds that 40% of students ride in a car to school. Eri would like to estimate the probability that if 3 students were randomly selected, only 1 rides in a car.
To simulate this probability, she lets the numbers 0, 1, 2, and 3 represent a student who rides in a car to school and then 4, 5, 6, 7, 8, and 9 represent a student who does not ride in a car to school. She then has a computer randomly select 3 numbers. She repeats this process for 20 trials.
The results of these trials are shown in this list.
468, 380, 120, 220, 553,945, 935, 607, 473, 490,074, 981, 692, 518, 408,954, 943, 389, 594, 569
Based on this simulation, what is the estimated probability that only 1 of 3 randomly selected students rides in a car to school?
Enter your answer, as a decimal, in the box.
To find the estimated probability that only 1 of 3 randomly selected students rides in a car to school, we count the number of times the simulation results in exactly one number between 0 and 3.
From the list, we can see that the numbers 381, 220, 490, 408, 389, and 569 have only one number between 0 and 3. These are the results that meet the given condition.
Therefore, out of 20 trials in the simulation, 6 trials resulted in only 1 student riding in a car to school.
The estimated probability is 6/20, or 0.3. Answer: \boxed{0.3}.