Question

not really math but a & b are stumping me. Obvs I'm new to stats:
Suppose that a balanced coin is independently tossed two times. Define the following events, and their probabilities:
Event A - head appears on the first toss

Event B - head appears on the second toss

Event C - both tosses yield the same outcome

a.Are A and B independent?
b.Are A and C independent?
c.Are all of the three events independent?

Answer 1

2 events \(A\) and \(B\) are independent if the following equality holds:
\(P(A \cap B)=P(A)\cdot P(B)\)

So you need to find each probability P(A), P(B), P(C), as well as the corresponding intersections of events to determine if they're independent.

Answer 2

or you can think of it like this

two events are independent if one probability doesn't affect the other (and vice versa)

Answer 3

so if event A happening changes event B, or vice versa, then you don't have independent events

Answer 4

It will be very helpful if you list out your sample space (what are all the possible outcomes you can get from the two tosses?) And then write out what each event A, B, C can represent

Answer 5

I understand that a is independent but not sure how to prove. This is what I got, but I'm sure my notation is wrong:
S={HH,HT,TH,TT}
Event A - 1/2 probability
Event B - 1/2 probability
Event C - 1/2 probability

P(A∩B) = P(A)P(B)
so
P(A∩B) =(1/2)(1/2)
HH =1/4

Answer 6

yup looks good for a) :) and you're probabilities P(A), P(B), P(C) are good

Answer 7

Just to add a note for the last one. To show independence of 3 events, you must show all of these are true:
\(P(A \cap B)=P(A)P(B)\)
\(P(A \cap C)=P(A)P(C)\)
\(P(B \cap C)=P(B)P(C)\)
\(P(A\cap B \cap C)=P(A)P(B)P(C)\)

Answer 8

thanks I'll give a try

Answer 9

i'm not sure for question be, i think they are dependent but the equation that I get tells me they are independent:
P(A∩C)=P(A)P(C)
P(A∩C)=(1/2)(1/2)
HH =1/4

Answer 10

I think your right.

Answer 11

I think my math is right but I have come to the conclusion it is actually independent

Answer 12

Yes that looks right. Th

Answer 13

That would imply independence though, which is in agreement with your math,

Answer 14

the final one i got dependent, because the math told me
P(A∩B ∩C) = P(A)P(B)P(C)
1/4 =(1/2)(1/2)(1/2)
=1/8

Last one shows that all three together are not independent but are actually dependent when considering all three events together

Answer 15

yep looks good :)

Answer 16

THANKS!