Coin toss using additivity property

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nicholaskong100

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additivity property.jpg

Some questions:

1) How do I know if P(A) is at least 3 heads or if P(A's complement) is at least 3 heads instead? Does it matter?

2) At least 3 heads would mean the other 97 toss can be heads or tails. But the book says the union of A and A complement are disjoint meaning A and A complement must have different outcomes. Would that mean the 97 toss cannot be heads, the 97 must be all tails?
 
At least thee heads is not exactly none, not exactly one, nor not exactly two.
The probability of exactly no heads is [imath]\dbinom{100}{0}\cdot 2^{-100}[/imath]
The probability of exactly one head is [imath]\dbinom{100}{1}\cdot 2^{-100}[/imath]
The probability of exactly two heads is [imath]\dbinom{100}{2}\cdot 2^{-100}[/imath]
 
nicholaskong100 said:
View attachment 28558

Some questions:

1) How do I know if P(A) is at least 3 heads or if P(A's complement) is at least 3 heads instead? Does it matter?

2) At least 3 heads would mean the other 97 toss can be heads or tails. But the book says the union of A and A complement are disjoint meaning A and A complement must have different outcomes. Would that mean the 97 toss cannot be heads, the 97 must be all tails?
Click to expand...
If you define A as the event "at least 3 of the 100 tosses are heads", then Ac is "fewer than 3 are heads". (But how you define an event is up to you!)

The idea is that it's easier to find the probability of "fewer than 3 of the 100 tosses are heads", than of "at least 3 are heads". That's because the former means "0, or 1, or 2 are heads", while the latter means "3, or 4, or 5, ... or 100 are heads".

You show a few common misunderstandings due to not having a lot of experience with these ideas. The event A isn't about some particular 3 coins, so that you can't talk about "the other 97".

Also, though a lot less important, it is A, not P(A), that is "at least 3 of the 100 tosses are heads"; you need to distinguish between events and their probabilities in order to think clearly about these things.
 
Dr.Peterson said:
If you define A as the event "at least 3 of the 100 tosses are heads", then Ac is "fewer than 3 are heads". (But how you define an event is up to you!)

The idea is that it's easier to find the probability of "fewer than 3 of the 100 tosses are heads", than of "at least 3 are heads". That's because the former means "0, or 1, or 2 are heads", while the latter means "3, or 4, or 5, ... or 100 are heads".

You show a few common misunderstandings due to not having a lot of experience with these ideas. The event A isn't about some particular 3 coins, so that you can't talk about "the other 97".

Also, though a lot less important, it is A, not P(A), that is "at least 3 of the 100 tosses are heads"; you need to distinguish between events and their probabilities in order to think clearly about these things.
Click to expand...
additivity.png

Oops, my book had it written in terms of probability. I took a screenshot of how it is written. So is the additivity property suppose to be written as A + A's complement = 1 instead?
 
pka said:
At least thee heads is not exactly none, not exactly one, nor not exactly two.
The probability of exactly no heads is [imath]\dbinom{100}{0}\cdot 2^{-100}[/imath]
The probability of exactly one head is [imath]\dbinom{100}{1}\cdot 2^{-100}[/imath]
The probability of exactly two heads is [imath]\dbinom{100}{2}\cdot 2^{-100}[/imath]
Click to expand...
subset_formula.jpgThanks, I understand it better now. It makes sense to use the binomial coefficient to compute the different subsets of the complement of the event. Less work to do.
 
nicholaskong100 said:
View attachment 28561

Oops, my book had it written in terms of probability. I took a screenshot of how it is written. So is the additivity property suppose to be written as A + A's complement = 1 instead?
Click to expand...
No, the book is right. You were technically wrong here:
1628988290589.png
nicholaskong100 said:
View attachment 28562Thanks, I understand it better now. It makes sense to use the binomial coefficient to compute the different subsets of the complement of the event. Less work to do.
Click to expand...
Not quite. First find the probability of 0, 1, or 2 (which includes multiplying by (1/2)^100), then subtract from 1. You did things in the wrong order, and got a negative probability, which is impossible.
 
Once again, I suggest that you first think about a simpler problem.

Suppose that you flip a coin three times.

There are 8 possibilities

HHH
HHT
HTH
HTT
THH
THT
TTH
TTT

If the coin is fair, they are all equally likely.

What is the probability that at least one is heads?

HHH
HHT
HTT
HTH
THH
THT
TTH

7 out of 8 or 7/8

What is the probability that none are heads?

Obviously TTT, or 1 out of 1/8.

Add them up because they are mutually exclusive

(7/8) + (1/8) = 8/8 = 1.

That makes sense. Either at least one is heads or ay most none is heads. That is exhaustive and mutually exclusive.

So let's extend the logic.

Out of a hundred., at least three heads and at most two is exhaustive and mutually exclusive.

P(at most two) + P(at least three) = 1.

Thus P(at least three) = 1 - P(at most 2).

Can we break the probability of at most 2 into the probability of mutually exclusive outcomes?

Sure. P(at most two) = P(exactly none) + P(exactly one) + P(exactly two).
 
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