You toss three coins. Is the event that there is at most one tails independent
from the event that there is both a heads and tails present?
I am new to probability and I have no idea what I am doing. This is what I have so far:
(Defining my probability space):
Let $S$ be a string length 3 on alphabet {$H,T$}.
Let $A$ be an event that there is at most one tail.
Let $B$ be an event that there is both tail and head present.
After this I have no idea how to solve it. Any help would be appreciated. Thank you.
Best Answer
So we have to calculate the probabilities of both, and see what we get.
The probability that there is at most one tail is the probability that there are $3$ heads $=(\frac{1}{2})^3 = \frac{1}{8}$ plus the possibility of $1$ tail, which can come on any of the three flips, ($=\frac{1}{8}*3 = \frac{3}{8}$), which is $1/2$.
The probability that both a heads and a tails is present is the opposite of either all turning up heads or all turning up tails, which comes out to be $\frac{6}{8}=\frac{3}{4}$.
Now we have to calculate the probability of both the above happening i.e. both a head and tail turn up, and there is at most one tail. This is the same as saying that exactly one tail turns up. The probability of that happening is $\frac{3}{8}$.
Independence of two events $A$ and $B$ happens when $P(A)P(B)=P(A \cap B)$. In this case, $\frac{1}{2} \cdot\frac{3}{4}$ is equal to $\frac{3}{8}$, whence the events are independent. Please ask if any doubts.