# Probability question involving binomial theorom

I'm having difficulty solving this question :

In a multiple choice test there are 5 questions each with three possible answers. For each question a student chooses an answer at random. Find the probability that she gets

B) three or fewer.

This is how i solved A:

N: 5.
P: .33.
Q: .67

$P(X=3)=p(3)= 10(.33^3)(.67^2) = 0.1613$

10 is the number of possibilities to get 3 questions right from 5 questions. I wad wondering if this method is correct and also i have no idea how to apply my current formula to section B.

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Your method for (A) is correct, though it would be better not to round $\frac13$ so early: the precise answer of $\frac{40}{243}$ is closer to $0.1646$.
• calculate the probabilities for exactly 4 or 5 correct and subtract these probabilities from $1$.
@Ahoura Ghotbi: not really - the second is fast enough in this case: $1 - 5 \times \frac{2}{243} - 1 \times \frac{1}{243}$ –  Henry Apr 4 '12 at 23:55
No, there isn't. Well, I am lying somewhat. Various pieces of software will give you the answer. But by hand you can't really do much better. For larger problems, one can save some time by using $\binom{n}{k}p^kq^{n-k}$ to compute $\binom{n}{k+1}p^{k+1}q^{n-k-1}$. –  André Nicolas Apr 4 '12 at 23:58