Let $P(H) = p$ be the probability of one head. In many scenarios, this probability is assumed to be $p = \frac{1}{2}$ for an unbiased coin. In this instance, $P(H) = 3P(T)$ so that $p = 3(1-p) \implies 4p = 3$ or $p = \frac{3}{4}$.
You are interested in the event that out of three coin tosses, at least 2 of them are Heads, or equivalently, at most one of them is tails. So you are interested in finding the likelihood of zero tails, or one tails.
The probability of zero tails would be the case where you only received heads. Since each coin toss is independent, you can multiply these three tosses together:
$P(H)P(H)P(H) = p^3$ or in your case, $(\frac{3}{4})^3 = \frac{27}{64}$.
Now we must consider the case where one of your coin flips is a tails. Since you have three flips, you have three independent opportunities for tails. The likelihood of two heads and one tails is $3(p^2)(1-p)$. The reason for the 3 coefficient is the fact that there are three possible events which include two heads and one tails: $HHT, HTH, THH$. In your case (where the coin is 3 times more likely to have heads): $3(\frac{3}{4})^2(\frac{1}{4}) = \frac{27}{64}$.
Adding those events together you get $p^3 + 3(p^2)(1-p) = \frac{54}{64}$. Note that the 3 coefficient is a result of the binomial expansion as used by @Back2Basic
"Show that the probability of a tie (we get the same number of heads) is the same as getting exactly 4 heads and 4 tails on 8 coin flips."
Let $H,T$ denote the number of heads and tails repectively flipped
by you in $4$ flips .
Let $H',T'$ denote the number of heads and tails respectively flipped
by your friend in $4$ flips.
Then the probability of a tie is $\Pr\left(H=H'\right)$.
But $T'$has the same distribution as $H'$ and there is independence, so we observe:
$\Pr\left(H=H'\right)=\Pr\left(H=T'\right)=\Pr\left(H=4-H'\right)=\Pr\left(H+H'=4\right)$
The last probability can be recognized as the probability of $4$
heads by $8$ flips (the flips of you and your friend taken together).
"Use this answer to calculate the probability of someone winning (getting more heads than the other person)."
Now we have:
- $\Pr\left(\text{tie}\right)=\Pr\left(H+H'=4\right)=2^{-8}\binom{8}{4}$.
- $1=\Pr\left(\text{you win}\right)+\Pr\left(\text{tie}\right)+\Pr\left(\text{friend wins}\right)$
- $\Pr\left(\text{you win}\right)=\Pr\left(\text{friend wins}\right)$
leading to: $$\Pr\left(\text{you win}\right)=\frac{1}{2}\left(1-2^{-8}\binom{8}{4}\right)$$
"Also, If I toss the coin 5 times, while my friend only tosses hers 4 times, calculate the probability that I will get strictly more heads than my friend."
If you toss $5$ times then think of it as a match1 as described above that is followed by an extra toss of you, and call the whole thing match2.
Now apply that:
$$\Pr\left(\text{you win match2}\right)=$$$$\Pr\left(\text{you win match1}\right)+\Pr\left(\text{match1 ends in a tie}\wedge\text{extra toss is a head}\right)=\Pr\left(\text{you win match1}\right)+\Pr\left(\text{match1 ends in a tie}\right)\Pr\left(\text{extra toss is a head}\right)=$$$$\frac{1}{2}\left(1-2^{-8}\binom{8}{4}\right)+2^{-8}\binom{8}{4}\times\frac12=\frac12$$
There is another (more elegant) route to this result.
Let $H,T$ denote the number of heads and tails repectively flipped
by you in $5$ flips .
Let $H',T'$ denote the number of heads and tails respectively flipped
by your friend in $4$ flips.
The probability of winning for you is $\Pr(H>H')$ and just as above we find:
$\Pr\left(H>H'\right)=\Pr\left(H>T'\right)=\Pr\left(H>4-H'\right)=\Pr\left(H+H'>4\right)$
The RHS is the probability that by $9$ flips there are more heads than tails. Symmetry then tells us that this equals $\frac12$.
Best Answer
The easiest way to this is by defining appropriate random variables:
$Y_1$ being the number of heads the first person has throw, and $Y_2$ being the number of heads the second person has thrown. Then by the information:
$Y_1\sim Bin(3,\frac{2}{3})$ and $Y_2\sim \text{Geom}(\frac{1}{3})$
I think there is an implicit assumption that $Y_1$ and $Y_2$ are independent. Then the desired probability is:
$$ \mathbb{P}(Y_1\geq Y_2)= \mathbb{P}(Y_1=1,Y_2=0)+ \mathbb{P}(Y_1=2,Y_2=0)+ \mathbb{P}(Y_1=2,Y_2=1) + \mathbb{P}(Y_1=3,Y_2=0) +\mathbb{P}(Y_1=3,Y_2=1)+ \mathbb{P}(Y_1=3,Y_2=2)$$
Also when dealing with probability the average number is usually the expectation, and they are probably referring to the expectation:
$$ \mathbb{E}[Y_1+Y_2] $$