Find the probability with normal distribution I am doing some revision for an exam I have tomorrow and I can't work out how an answer is achieved.

The weight $X$ kg of a bag of cement can be modelled by a normal
  distribution with mean $50$ kg and standard deviation $2$ kg.
  $P(x > 53) = 0.0668$
Find the probability, $a$, that two weigh more than $53$ kg and one weighs less than $53$ kg.

So I did: $a = 0.0688^2 \times (1-0.0688)$
However the answer is $a = 3 \times 0.0688^2 \times (1-0.0688)$
I can't understand where the multiplication by $3$ comes from, although three are three bags, surely that has already been taken into account by multiplying $3$ probablities?
(The exam is tomorrow so I would appreciate answers before then, but I would still like to know the answer irrespective of the exam)
 A: I assume that we have bought $3$ bags, and want the probability that $2$ of them weigh more than $53$ kg and one weighs less.
So we have an experiment in which the probability of success is $0.0688$. The experiment is repeated independently $3$ times and we want the probability of exactly $2$ successes. This is a binomial distribution situation, with $n=3$ and $p=0.0688$. The probability that the number of successes is $2$ is equal to 
$$\binom{3}{2}p^2(1-p).$$
Another way of thinking about it is that we buy a bag, then buy another, then another. We can end up with $2$ bags at $\gt 53$, and one bag at $\le 53$, in $3$ different orders: SSF, SFS, and FSS (S stands for success, F for failure).  The probability of SSF is $(p)(p)(1-p)$. The probability of SFS is $(p)(1-p)(p)$. And the probability of FSS is $(1-p)(p)(p)$. Each of these is $p^2(1-p)$. Add up.
A: Ok, if I've understood correct, the question is given three bags $X_1,X_2,X_3$ to find the probability that two of them weight over $53$ kg and the other one weights less than $53$ kg. 
You should consider three possibilities: that the latter bag refers to $X_1$ or to $X_2$ or to $X_3$. You have found the probability only of the one possible case, for example that $X_1,X_2$ weight over $53$ kg and $X_3$ weights less than $53$ kg.
A: For each i.i.d. $X_k$, consider a Bernoulli random variable $U_k = I(X_k > 53)$. Then $U_k$ are also iids, and $\mathbb{P}(U_k = 1) = \mathbb{P}(X_k > 53)$. Ultimately, you are interested in $Y = U_1 + U_2 + U_3$, which follow binomial distribution with $n=3$ and $p = \mathbb{P}(X_k > 53)$.
Thus, the answer is 
$$
  \mathbb{P}(Y=2) = \binom{3}{2} p^2 \left(1-p\right)  = 3 p^2 \left(1-p\right)   
$$
