Probability that the stapler is the one not working

conditional probabilityprobabilityprobability theoryproblem solving

I had three staplers in my drawer. I knew that one of them always worked, that one of them
never worked, and that the third one worked with probability $1/2$. But I did not know which was
which because they all looked the same. I wanted to establish which stapler was the one that
never worked. I took one stapler at random and tried it and it worked. Then I took another
stapler at random and tried it twice and both times it did not work. What is the probability
that the second stapler that I tried, which twice failed to work, is the one which never works?

Is the probability 1/2? since if the first stapler is:

1)Stapler which works(which probability of being taken out is 1/3)–> there's a 1/2 probability that the second stapler is the not working one. Then the probability of this case is P= (1/3)x(1/2)= 1/6

2)Stapler which works with 1/2(which probability of being taken out is 1/3)–> there's a 1 probability that the second stapler is not working. Then the probability of this case is P= (1/3)x1= 1/3

Hence by adding the 2 probabilities I yield P(stapler that's not working)= P(1st case)+ P(2nd case)= (1/6)+(1/3)= 1/2

Any recommended formulas maybe that you recommend I use and hints if this is wrong?

Best Answer

We can apply Bayes' theorem to solve this problem. We first need to find the scenarios in which we arrive at the outcome you observed. Either:

  1. The first stapler always works and the second stapler never works;
  2. The first stapler always works and the second stapler works half of the time, and this second stapler did not work twice;
  3. The first stapler works half of the time, and it worked, and the second stapler never works.

We then have to divide the probability of either the first or the third scenario happening, by the probability of one of these scenarios happening. We have:

$$P(1) = \frac{1}{3} \frac{1}{2}$$ $$P(2) = \frac{1}{3} \frac{1}{2} \left(\frac{1}{2}\right)^2$$ $$P(3) = \frac{1}{3} \frac{1}{2} \frac{1}{2}$$

We thus find:

$$P = \frac{P(1) + P(3)}{P(1) + P(2) + P(3)} = \frac{\frac{1}{6} + \frac{1}{12}}{\frac{1}{6} + \frac{1}{24} + \frac{1}{12}} = \frac{4 + 2}{4 + 1 + 2} = \frac{6}{7} \approx 0.857$$

Related Question