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# Filling out frequency table for independent events

## Video transcript

Voiceover:One rainy Saturday morning, Adam woke up to hear his mom complaining about the house being dirty. "Mom is always grouchy when it rains," Adam's brother said to him. So Adam decided to figure out if this statement was actually true. For the next year, he charted every time it rained and every time his mom was grouchy. What he found was very interesting. Rainy days and his mom being grouchy were entirely independent events. Some of his data are
shown in the table below. Fill in the missing values
from the frequency table. Let's see, we have raining
days and not raining days and the total days that
he kept the data for. And then he tabulated on or let's say, the raining days whether his mom was grouchy or not grouchy. And on a not raining day whether his mom was grouchy or not grouchy. And there's a total of 35 days it rained, 330 days that it didn't rain. And then 73 times his mom was grouchy and 292 times his mom was not grouchy. So the first thing is how
do we figure this out? We have these 4 boxes here. It's not clear that we can just... we have enough information to fill it out just with this table. But we have to remember what they told us. They told us that his mom being grouchy and it raining were
entirely independent events. Another way of saying that
is the probability of his... Let me do this in color that
you're more likely to see. Another way of saying that... So independent events, that means that the probability... My pen is acting up a little bit. Probability that mom is grouchy. So let me write that. Mom... My pen is really... Mom is grouchy given it is raining. It shouldn't really matter
whether it's raining. It should just be the same
thing as the probability of mom being grouchy in general. So what does that tell us? Well we can figure out the probability that mom is grouchy in general. She's grouchy 73 out of 365 days. So the probability that
mom is grouchy in general is going to be 73 divided by 365. And so [are these] just
based on the data we have. That's the best estimate
that mom is grouchy. The probability that mom is grouchy. It's the percentage of days
that she's been grouchy. So that is .2. So based on the data, the best estimate of the probability of mom being grouchy is .2 or 20%. And so we should have the probability of mom being grouchy
given that it's raining should be 20% as well. So this number... So given that it's raining, we should also have 20% of the time, mom is grouchy because these are independent events. It shouldn't matter whether
it's raining or not. This should be 20%... She should be grouchy 20%
of time that it's raining and she should be grouchy 20% of the time that it's not raining. That's what would be
consistent with the data saying that these were
entirely independent events. So what is 20% of 35? Well 20% is 1/5th. 1/5th of 35 is 7. And once again, all I did is I said 20% of 35 is 7. And if that's 7 then 35 minus 7. That's gonna be 28 right over there. And then if this is 7, then 73 minus 7 is going to be 66. And 330... I guess there's a couple
of way we could do it. We could take... Actually we could just take 292 minus 28 is going to be... Let's see 292 minus 8 would be 284. Minus another 20, 264. And do the numbers all add up? Yes. 66 plus 264 is 330. So the key realization here is what he's saying he found
was very interesting. Rainy days and his mom being grouchy were entirely independent events. That means that the probability of his mom being grouchy... It shouldn't matter whether
it's raining or not. It should just be... It should be the same probability of whether it's raining or not. And our best estimate of the probability of his mom being grouchy is on the total days, is 20%. And so if the data's backing up that it's independent events then the best way to
fill this out would be the probability of his mom being grouchy on a rainy day or not rainy
day should be the same. And that's what we filled
out right over here.