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)
Best Answer
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.