relation between occurances in statistics (coin flip)

Question

we toss a coin 50 times and define three occurances:

a) in the first 4 tossings we got heads.

b)in the last 4 tossings we got heads.

c)totally, we got 20 times head(out of 50 tossings)

1) are A and b disjoint to each other?

2)are a, b ,c independent to each other?

3)calculate $P(a|b \cup c) $

my attempt:

1)a and b are not disjoint to themselves, because we defined those occurances in relation to tossing a coin 50 times. that means that there is a chance that we will obtain hads in the frst 4 tossings and in the last 4 tossings.

2)if the occurances were independent of eachother, we'll get that $p(a|b) = p(a)$.if we got 4 heads in the beginning, doesn't mean that we will get 4 heads in the end, thus p(a|b)=p(a). however, regarding p(a|c) or p(b|c) i'm not sure, and would appreciate an explanation regarding it, my intuition s that $P(a \cup c) = p(a) + p(c) - p(a \cap c)$ .

3)according to 2, i'll calculate $p(a|b\cup c)$ as $p((a|b)\cup c)$. $p(a) = \frac{1}{2^4}$ $p(c)=\frac{2}{5}$, so $p((a|b)\cup c)$ = $p(a \cup c) = \frac{1}{32} + \frac{2}{5} -p(a \cap c) $

please correct me if i've done a mistake so i can learn how to do it correctly. thank you very much!

Answer 1

You are correct about the first question, and you give the correct reason. The event are not disjoint since you can have both of them occurring (both a and b can be true at the same time).

On the second question you are correct that a and b are independent to each other. I think your explanation is sufficient: the outcome of the first four throws does not affect the outcome of the last 4 throws.

You are also good to be uncertain/suspicious about the other case: Is c independent to a (or b). It's not so easy to tell now, because event c is about all the tossings, so it is plausible to think that it might affect the first four (or the last four). It would be good to know more about your intuition on this. In the question you just write the formula for the probability of the union of 2 events (it's just the formula, you do not say how this relates to your problem).

I will not give you a straight answer (yet), I will just ask you some extra questions to help you build your intuition about this part.

1. Let's assume that event c was : we got 3 total heads out of the 50 tossings. How do you answer whether c and a are independent now?

2. Let's assume that event c was : we got 4 total heads out of the 50 tossings. How do you answer whether c and a are independent now?

After you've answered these questions try the original one again.

Finally, about the 3rd question. It's not clear to me why you are asked to calculate this probability but let's say it's part of the exercise.

First of all be careful with your calculations. $\frac{1}{2^4} p(c) = \frac{2}{5}$ is clearly wrong, as it implies that p(c) is greater than 1.

Also please break down your steps further so it is easier for someone to follow your reasoning. How do you do your very first step? I am not sure exactly what you mean, but be careful: a|b is not an event. We use this notation to describe conditional events and we only use the notation to ask about the probability of this conditional event. It does not make much sense to ask about the union of a conditional event with an event. In other words ((a|b) ∪ c) does not make much sense. Especially since I see you later equating a|b to a (because their probabilities are the same).

Maybe it would be better if you started with the conditional probability definition.