- #1

vparam

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- Homework Statement
- Suppose that in the 2050 election, polls show Party A has a 60% chance of winning the senate and a 50% chance of winning the house, and that if Party A wins the house, it has an 80% chance of winning the senate. Let X be the random variable whose value is the the number of congressional houses won by Party A. (a) Find the expected value of X.

- Relevant Equations
- ##p(S|H) = \frac{p(S \cap H)}{p(H)}##

##E(X)=\sum_{s \in S}{p(s)X(s)}##

I submitted this solution, and it was marked incorrect. Could I get some feedback on where I went wrong?

Let S represent the event that Party A wins the senate and H represent the event that Party A wins the house.

There are 4 cases: winning the senate and house (##S \cap H##), winning just the senate (##S \cap \overline{H}##), winning just the house (##H \cap \overline{S}##), and winning neither (##\overline{S} \cap \overline{H}##).

I computed the probabilities for each compound event as follows:

$$p(S \cap H) = p(S|H)p(H)= (.8)(.5) = .4$$ $$p(S \cap \overline{H}) = p(S)p(\overline{H}) = (.6)(.5) = .3$$ $$p(H \cap \overline{S})=(p(S) - p(S \cap H))p(H) = (.6 - .4)(.5) = .1$$ $$p(\overline{S} \cap \overline{H}) = p(\overline{S})p(\overline{H}) = (.4)(.5) = .2$$

This means that the expected value is:

$$E(X) = 2(.4) + 1(.3) + 1(.1) + 0(.2) = 1.2$$

Let S represent the event that Party A wins the senate and H represent the event that Party A wins the house.

There are 4 cases: winning the senate and house (##S \cap H##), winning just the senate (##S \cap \overline{H}##), winning just the house (##H \cap \overline{S}##), and winning neither (##\overline{S} \cap \overline{H}##).

I computed the probabilities for each compound event as follows:

$$p(S \cap H) = p(S|H)p(H)= (.8)(.5) = .4$$ $$p(S \cap \overline{H}) = p(S)p(\overline{H}) = (.6)(.5) = .3$$ $$p(H \cap \overline{S})=(p(S) - p(S \cap H))p(H) = (.6 - .4)(.5) = .1$$ $$p(\overline{S} \cap \overline{H}) = p(\overline{S})p(\overline{H}) = (.4)(.5) = .2$$

This means that the expected value is:

$$E(X) = 2(.4) + 1(.3) + 1(.1) + 0(.2) = 1.2$$