PLEASE HELP!
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 the tableBased 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.
5533 2245 1555 3341
4252 5335 5321 4155
2131 3414 1532 4251
2523 3311 2352 2332
5451 3344 1121 5243

The probability of an event occurring is the number of times that it occurred when the experiment was conducted as a fraction of the total number of times the experiment was conducted.

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

Learn more about probability here:

brainly.com/question/3733849

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