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I am trying to solve the following problem:

A worker has the probability of $1/2$ of getting an electrical contract, if he gets the electrical part, the probability of getting the hydraulic contract is $3/4$. If he doesn't get the electrical contract, the probability of getting the hydraulic contract is $1/3$.

  • What is the probability of him getting both contracts?

  • What is the probability of him getting exactly one contract?

  • What is the probability of him getting no contract?

I'm confused by this. I tried to set up the sample space and answer it but this left me very confused, I don't think it works this way. So the concept that seemed more suitable was conditional probability so I wrote:

$$P(H|E) + P(H|\overline{E}) + P(\overline{H}|E) + P(\overline{H}|\overline{E})=1 $$

We know that $P(H|E)=3/4$ and $P(H|\overline{E}) =1/3$ but how do we determine the remaining terms? I thought that $P(\overline{H}|E)=1-P(H|E)$ and $P(\overline{H}|\overline{E})=1-P(H|\overline{E})$ but this also doesn't seems to work. Also, when we sum up the equations, we obtain the former equation but it gets

$$P(H|E) + P(H|\overline{E}) + P(\overline{H}|E) + P(\overline{H}|\overline{E})=2 $$

Which is nonsense. How should I answer this?

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    $\begingroup$ As an aside, the identity you seem to have been reaching for was $\Pr(H\cap E)+\Pr(H\cap \bar{E})+\Pr(\bar{H}\cap E)+\Pr(\bar{H}\cap \bar{E})=1$ which follows from Law of total probability and/or by partitioning the sample space. Where conditional probability comes into play here is in remembering $\Pr(A\cap B) = \Pr(A)\Pr(B\mid A)$. You are told in the problem that $\Pr(E)=1/2$, that $\Pr(H\mid E) = 3/4$ and that $\Pr(H\mid \bar{E})=1/3$ which combined with these identities and more that you should know is enough to answer all the parts to the problem. $\endgroup$ – JMoravitz Sep 22 at 16:39
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Hint: I agree with you that from the text it can be read off that $P(H|E)=\frac34, P(H|\overline E)=\frac13, P(E)=\frac12$

Then we we can use the law of total probability $P(H)=P(H|E)\cdot P(E)+P(H|\overline E)\cdot P(\overline E)$. With $P(\overline E)=1-P(E)$ we get

$$P(H)=\frac34\cdot \frac12+\frac13\cdot \frac12 =\frac46=\frac23\Rightarrow P(\overline H)=\frac13$$

Now you can use a table to maintain the overview. With the help of the Bayes`theorem it is easy to calculate the values for the intersections to complete the table. For instance, the probability of getting both contracts is $P(H\cap E)=P(H|E)\cdot P(E)=\frac34\cdot \frac12$.

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Hint. Draw the binary tree with four leaves corresponding to the four possible ways to get (or not) the two contracts.

Start with the electrical from the root, then two branches from each descendant.

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If you want to use conditional probability it is often in the form Prob(A|B)Prob(B) or Prob(A|B_i)Prob(B_i). The fact that you have a sum greater than 2, as you note, is a good indication you are doing something wrong.

Often when we have events in sequence we can find the probabilities by multiplying the probability the first event happens times the probability the second one does. To get both contracts, for example, it would be 1/2 * 3/4.

What would be the probability of getting neither?

As the probabilities must sum to 1, you get for free now the probability of getting exactly one contract. Even better, however, is to compute that independently and see if it matches what we have here.

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