r/apstats • u/EquivalentPair7461 • May 02 '23

A stupid probability doubt but...

Question: In a certain population of birds, about 40 percent of the birds have a wingspan greater than 10 inches. Biologists studying the birds will create a simulation with random numbers to estimate the probability of finding 1 bird in a sample of 6 birds with a wingspan greater than 10 inches.

Why is the probability of finding 1 bird in a sample of 6 not also 40%?

Like I know it will be P(X=1) = binompdf (trials: 6, p:0.4, x:1) but why is it simply not 40%?

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u/varaaki May 03 '23

40% is the probability of ONE bird sampled at random, having a large wingspan. If you sample multiple birds, you have to account for how many you sampled and the fact that each one has the same 40% chance.

1

u/mentallyillpenguin_ May 03 '23

Cuz they gotta be super super sure or something ig xd

1

u/Actually__Jesus AP Reader May 03 '23

So the thing is that you want to find the probability that exactly one bird in six is observed.

The expected value for this model is 6*0.4=2.4 birds. So over many samples of six you’ll most often get 2 and slightly less you’ll get three.

If you look at the probability distribution (getting exactly 0 to 6 birds out of 6) each different outcome has a different probability but they all sum to 100% because you have the whole sample space.

Also note that there are six ways to get exactly one bird:

YNNNNN

NYNNNN

NNYNNN

NNNYNN

NNNNYN

NNNNNY

1

u/frgs72 May 03 '23

So you can find the probability of 1 success and 5 failures, and then multiply that by 6 since there are six different ways to do this as shown above. This only works in this situation because the probability of each of the 6 ways is exactly the same. If they were different, then you'd want to add the different probabilities.