Question

Eric surveys students at his school and finds that 80% have a pet. He wants to estimate the probability that, if he randomly selected 4 students, more than 2 would have a pet.

To estimate this probability, he lets the numbers 1, 2, 3, and 4 represent a student who has a pet and 5 represent a student who does not have a pet. He then has a computer randomly select 4 numbers and repeats this 20 times.

The results of these trials are shown in this list.

5533, 2245, 1555, 3341, 4252,5335, 5321, 4155, 2131, 3414,1532, 4251, 2523, 3311, 2352,2332, 5451, 3344, 1121, 5243

Based on this simulation, what is the estimated probability that more than 2 of 4 randomly selected students would have a pet?

Enter your answer, as a decimal, in the box.

Answer 1

0.75

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Explanation:

You'll look at each four digit group separately.

Since we want more than 2 people having a pet, this means the group of four must have 3 pet owners or all 4 are pet owners.

The digit 5 represents "no pet". If we see it show up twice or more, then we must cross the group off the list. Something like 5533 has the first two people that don't own pets, while the last two people do own pets. In this case, we don't have a "more than 2 pet owners" situation happening. So we must cross it off the list.

Here is the list of groups to cross off

- 5533
- 1555
- 5335
- 4155
- 5451

There are five such groups we have eliminated. That leaves 20-5 = 15 groups where there are more than 2 pet owners.

We have 15 groups we want out of 20 total. It leads to the probability of 15/20 = 0.75