\begin{array}{l}{\text { Suppose a coin is tossed three times independently, with probability of land- }} \\ {\text { ing heads } 0 \leq p \leq 1 \text { and a complement probability } 1-p \text { of landing tails. }} \\ {\text { Consider the events } A=" \text { at most one tails" and } B=\text { "all tosses are the same' }} \\ {\text { (a) Find the probability of event } A \text { . }} \\ {\text { (b) Find the probability of event } B \text { . }} \\ {\text { (c) Find the probability of event } A \cap B \text { . }} \\ {\text { (d) For which values of } p \text { are events } A \text { and } B \text { independent and why? }}\end{array}
I did not read the question carefully, and just assumed $P(Heads) = P(Tails) = \frac{1}{2}$ , but then question d didn't make any sense. So, now I have no idea how to go about this. I've tried googling for biased coins, or coins with probability p, but most of the answers seem to involve something called binomial distribution, which I have not covered.
Does someone know of another way of solving it?
UPDATE:
$$P(B) = p^3 + (1-p)^3$$
$$P(A \cap B) = p^3$$
$$P(A \cap B) = P(A)P(B) \iff p = \frac{1}{2} , p=0 \text{ or } p=1$$
Best Answer
Guide:
There are only $8$ possible outcomes, $HHH, HHT, \ldots, TTT$. Find the probability of each outcome.
For example for part one, the computation is just $$P(HHH)+P(HHT)+P(HTH)+P(THH)=p^3 + 3p^2(1-p)$$
After you solve for the first three parts, solve for $p$ in $P(A \cap B)=P(A)P(B)$.