Question

in a school, 40% of the students have brown eyes. find the experimental probability that in a group of 4 students, at least one of them has brown eyes. the problem has been simulated by generating random numbers. the digits 0 - 9 were used. let numbers \0\, \1\, \2\, and \3\ represent the 40% of students with brown eyes. a sample of 20 random numbers is shown. experimental probability = ?%

Explanation:

Step1: Count the total number of simulations

There are \(20\) groups of random - number simulations.

Step2: Count the number of groups with no brown - eyed students (i.e., groups with no \(0\), \(1\), \(2\), \(3\))

Looking at the groups:

• For the group \(7918\): has \(1\) (so not the group we are looking for)
• For the group \(7910\): has \(0\) (so not the group we are looking for)
• For the group \(2546\): has \(2\) (so not the group we are looking for)
• For the group \(1390\): has \(1\) and \(3\) and \(0\) (so not the group we are looking for)
• For the group \(6075\): has \(0\) (so not the group we are looking for)
• For the group \(1230\): has \(1\), \(2\), \(3\), \(0\) (so not the group we are looking for)
• For the group \(2386\): has \(2\) and \(3\) (so not the group we are looking for)
• For the group \(0793\): has \(0\) and \(3\) (so not the group we are looking for)
• For the group \(7359\): has \(3\) (so not the group we are looking for)
• For the group \(3048\): has \(3\) and \(0\) (so not the group we are looking for)
• For the group \(2816\): has \(2\) and \(1\) (so not the group we are looking for)
• For the group \(6147\): has \(1\) (so not the group we are looking for)
• For the group \(5978\): no \(0\), \(1\), \(2\), \(3\)
• For the group \(5621\): has \(2\) and \(1\) (so not the group we are looking for)
• For the group \(9732\): has \(3\) and \(2\) (so not the group we are looking for)
• For the group \(9436\): has \(3\) (so not the group we are looking for)
• For the group \(3806\): has \(3\) and \(0\) (so not the group we are looking for)
• For the group \(5971\): has \(1\) (so not the group we are looking for)
• For the group \(6173\): has \(1\) and \(3\) (so not the group we are looking for)
• For the group \(1430\): has \(1\), \(4\), \(3\), \(0\) (so not the group we are looking for)

The number of groups with no brown - eyed students (i.e., no \(0\), \(1\), \(2\), \(3\)) is \(1\) (the group \(5978\))

Step3: Calculate the number of groups with at least one brown - eyed student

The number of groups with at least one brown - eyed student \(n=20 - 1=19\)

Step4: Calculate the experimental probability

The experimental probability \(P=\frac{n}{N}\times100\%\), where \(n = 19\) (number of favorable outcomes) and \(N = 20\) (total number of outcomes)

\(P=\frac{19}{20}\times100\%=95\%\)

Answer:

\(95\%\)