Team A has the probability 0.3 of winning every time they play (there are no ties). Let X denote the # of times Team A wins in 5 games.
Find the distribution of X.
For this problem since I have to use the Bernoulli Trials which states that $p(k) = ({}^n\mathrm C_k)\cdot p^k\cdot q^{n-k}$, for 0, would my answer be $p(0) = ({}^5\mathrm C_0)(0.3)^0(0.7)^{5-0}=0.16807$?
On
First, because it is dangerous to use formulae without understanding where they come from, a brief refresher. Each game can win in either a win ($W$) or a loss ($L$), with $p = P(W) = 0.3$ and $q = P(L) = 1-P(W) = 0.7$. For any string of five $W$s and $L$s, we can compute the probability that the team won/lost those games using the independence of the games. For example, suppose that the team won the first two games, then lost the last three. What is the probability of that event?
We want to compute $P(WWLLL)$. By independence, this is given by $$ P(WWLLL) = P(W) P(W) P(L) P(L) P(L) = p^2 q^3 = (0.3)^2 (0.7)^3 \approx 0.031.$$ In general, if we fix a string with two $W$s and three $L$s, the probability of that particular sequence of wins and losses will be about $0.031$, which means that we need to count up the number of ways of forming strings with two $W$s and three $L$s. To do this, we can imagine having 5 slots in which to place $W$s, then choosing 2 of them. There are 5 choices for the first slot, and 4 choices for the second slot, so there are $P_{5,2} = 5\cdot 4 = 20$ possible choices if we distinguish different orders---that is, if we keep track of (for example) winning the first game against Team A, and the second game against Team B.
Since we could switch the two wins around, we divide by $2!$ (the number of possible rearrangements) to kill of the redundancy. Hence there are $$ \frac{P_{5,2}}{2} = \frac{5!}{3!2!} = \binom{5}{2} $$ different ways of winning two games and losing three games. Therefore \begin{align} P(\text{2 wins}) &= (\text{number of ways to win two games}) \cdot P(\text{any particular sequence containing 2 $W$s}) \\ &= \binom{5}{2} (0.3)^2 (0.7)^3. \end{align} This same reasoning can be used to argue that the probability of winning $k$ games out of five is given by $$ P(\text{$k$ wins}) = \binom{5}{k} (0.3)^k (0.7)^{5-k}, $$ where $k$ is a number between 0 and 5, inclusive. Thus, using this formula and a calculator, we have \begin{align} P(0) &= \frac{5!}{(5-0)! 0!} (0.3)^0 (0.7)^5 \approx 0.168 \\ P(1) &= \frac{5!}{(5-1)! 1!} (0.3)^1 (0.7)^4 \approx 0.360 \\ P(2) &= \frac{5!}{(5-2)! 2!} (0.3)^2 (0.7)^3 \approx 0.309 \\ P(3) &= \frac{5!}{(5-3)! 3!} (0.3)^3 (0.7)^2 \approx 0.132 \\ P(4) &= \frac{5!}{(5-4)! 4!} (0.3)^4 (0.7)^1 \approx 0.028 \\ P(5) &= \frac{5!}{(5-5)! 5!} (0.3)^5 (0.7)^0 \approx 0.002. \end{align}
Note also that the same logic gives the binomial distribution. If we play $n$ identical independent games, and the probability of winning each game is $p$ (with $q = 1-p$ being the probability of losing), then we have \begin{align} P(\text{$k$ wins}) &= (\text{number of ways to get $k$ wins}) \cdot P(\text{any particular sequence with $k$ wins}) \\ &= \binom{n}{k} p^k q^{n-k} \\ &= \frac{n!}{(n-k)!k!} p^k q^{n-k}, \end{align} which is the binomial distribution.
Yes, indeed. $X$ follow a binomial distribution of trial amount $5$ and Bernouli success rate $0.3$.
$$p_{\small X}(k)~=~\mathsf P(X\,{=}\,k) ~=~ {}^5\mathrm C_k ~(0.3)^k(0.7)^{5-k}~\big[k\in\{0,1,2,3,4,5\}\big]$$
Everything else is just subtitution of each value of $k$, and indeed $p_{\small X}(0)=0.7^5=0.16807$