$X$ has distribution $B (30, 0.6)$. Find $P(X \geq 16)$.
I know how to find $2$ or $3$ numbers where you use combinations and simply add probabilities for each variable. But this value includes $14$ variables and I don't think it wants me to do them all separately.
$Y$ has distribution $B(4, 0.7)$. Three values are chosen at random. Find the probability that their total is $10$.
I don't know the method for this type of question.
On
You have everything to solve the problem. What you need is
$$ P( X \geq 16) = \sum_{k=16}^{n} \binom{n}{k} p^k (1-p)^k $$ which is trivial given you have both $p$ and $n$.
EDIT for the second problem, think about this: in how many ways can you get $10$ from $0,1,2,3,4$ with three samples with replacement? It's either one $2$ and two $4$ or two $3$ and one $4$. Obviously, the order doesn't matter. Can you find the probability now?
For first question, are you expected to approximate the probability using normal distribution? The distribution of $X$ is approximately normal, with the same mean and variance as the $B$, i.e. $$X'\sim N[30\times.6, 30\times.6\times(1-.6)].$$ Then find (or look up) the probability $P(X'\ge15.5)$.
More: When $n$ is large enough and $p$ is not close to $0$ nor $1$, binomial distribution can be approximated by a normal distribution. (For the criteria of "enough" see Wikipedia page)
If $X\sim B(n,p)$, then the mean of $X$ is $np$ and the variance of $X$ is $np(1-p)$. This pair of value can be supplied to form an approximate normal distribution:
$$X'\sim N\left(\mu=np, \sigma^2 = np(1-p)\right)$$
Then, with a slight continuity correction (binomial distribution is discrete but normal distribution is continuous), the required probability can be found as
$$P(X\ge16)\approx P(X'\ge15.5)=P\left(Z\ge\frac{15.5-np}{\sqrt{np(1-p)}}\right)$$
This is usually found by looking up $Z$-table or numerical integration.
If $Y_1+Y_2+Y_3=10$, then the possible values of the $Y_i$'s are $$(2,4,4),(3,3,4),(3,4,3),(4,2,4),(4,3,3),(4,4,2)$$ Find the probability of each case, and sum.