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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)

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  • $\begingroup$ In tex mode, use \times instead of * i.e. $\times$ instead of $*$ $\endgroup$
    – user17762
    May 17, 2012 at 22:15

3 Answers 3

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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.

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  • $\begingroup$ Thank you for your answer, I'm not sure why the order matters though? $\endgroup$
    – Jonathan.
    May 17, 2012 at 21:16
  • $\begingroup$ Think in more familiar terms. You have coins that land heads with probability $0.07$, and tails with probability $0.93$. What is the probability that if you toss $3$ of these coins, you will get a total of $2$ heads and $1$ tail? The standard binomial model gives $\binom{3}{2}(0.07)(0.93)$. Or else equivalently imagine tossing a funny coin $3$ times (so order matters). We will have a total of $2$ heads and $1$ tail in $3$ different ways, HHT, HTH, THH. Calculate probability of each, add up. $\endgroup$ May 17, 2012 at 21:52
  • $\begingroup$ Tanks that makes more sense. But with the coins I've always wondered why it isn't HHT, HTH, THH, HHT, THH? $\endgroup$
    – Jonathan.
    May 17, 2012 at 22:04
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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.

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  • $\begingroup$ Why exactly does the order matter? Thanks for your answer. $\endgroup$
    – Jonathan.
    May 17, 2012 at 21:34
  • $\begingroup$ @Jonathan. The order doesn't matter, that's the point. $\endgroup$
    – SBF
    May 18, 2012 at 11:16
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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 $$ \mathbb{P}(Y=2) = \binom{3}{2} p^2 \left(1-p\right) = 3 p^2 \left(1-p\right) $$

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