I am a bit confused over a problem that is as follows:
A coin is tossed three times. Consider the following events
A: Heads on the first toss.
B: Tails on the second.
C: Heads on the third toss.
D: All three outcomes the same (HHH or TTT). E: Exactly one head turns up.
(a) Which of the following pairs of these events are independent?
(1) A, B
(2) A, D
(3) A, E
(4) D, E
(b) Which of the following triples of these events are independent?
(1) A, B, C
(2) A, B, D
(3) C, D, E
I can well understand the triples cases and why all of them are dependent but I am a bit confused about the pairs of events. The answer says that (A,B) and (A,D) are independent events. So I am having two troubles: 1) I can't seem to show that (A,B) are independent and 2) I seem to get that (A,E) are also independent even though I feel intuitively that it shouldn't.
Can someone help me with this ? What is the best method to approach such questions since I am just checking whether the probability of 1 event occurring is affecting the probability of the other event.
Thanks
Best Answer
In cases like this, it's not hard to simply enumerate the entire sample space. Here we have exactly $8$ possible events, all equally probable:
$$\{HHH,\;HHT,\;HTH,\;THH,\;HTT,\;THT,\;TTH,\;TTT\}$$
Again, each of these events has probability $\frac 18$.
Now, let's look at $A$: half of the events start with $H$ so $P(A)=\frac 12$. Specifically the $A$-events are $$\{HHH,\;HHT,\;HTH,\;HTT\}$$
Similarly: the $B$-events are $$\{HTH,\;HTT,\;TTH,\;TTT\}$$ so, again $P(B)=\frac 12$
We note that $\{HTH,\;HTT\}$ are all the events which are simultaneously $A$ and $B$ events.
As to the conditional probabilities, we see that exactly two $A$-events are also $B$-events so $P(B|A)=\frac 12=P(B)$. Similarly, exactly two $B$-events are also $A$-events so $P(A|B)=\frac 12=P(A)$. As the conditional probabilities match the unconditional probabilities the events are independent.
Note: you don't need to check both of $P(A|B)$ and $P(B|A)$. If one of them shows independence, so will the other. Still, when you are just starting out it's good to write everything out.
As to $A$ and $E$ we need to list the $E$ events: $$\{HTT,\;THT,\;TTH\}$$ and we see that $HTT$ is the only event which is both an $E$ and an $A$ event. Thus $P(A|E)=\frac 13\neq P(A)$ so the events are dependent.