# Three people go to the market in the morning with the following probabilities: 1/2, 2/3, and 3/4. What is the likelihood that two go to the market on a given morning?

Nov 6, 2015

We'll have to work out the probabilities for all three combinations of two, where one stays home . Let's call the people $A , B , C$

#### Explanation:

We'll define the probabilities by the letter $P$
(the $\neg$ sign means "not", $\wedge$ means "and")
$P \left(A\right) = \frac{1}{2} \to P \left(\neg A\right) = \frac{1}{2}$
$P \left(B\right) = \frac{2}{3} \to P \left(\neg B\right) = \frac{1}{3}$
$P \left(C\right) = \frac{3}{4} \to P \left(\neg C\right) = \frac{1}{4}$

Then we get (remember "and" means multiply)
$P \left(A \wedge B \wedge \neg C\right) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4} = \frac{2}{24}$
$P \left(A \wedge \neg B \wedge C\right) = \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{4} = \frac{3}{24}$
$P \left(\neg A \wedge B \wedge C\right) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{6}{24}$

Since these combinations are mutually exclusive we add :
$P \left(\text{total}\right) = \frac{2}{24} + \frac{3}{24} + \frac{6}{24} = \frac{11}{24}$

Note :
If the question had been "at least two" we would have to add:
$P \left(A \wedge B \wedge C\right) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{6}{24}$ for a total of $\frac{17}{24}$

$P \left(A \wedge B\right) + P \left(A \wedge C\right) + P \left(B \wedge C\right)$ without the NOT's, then $P \left(A \wedge B \wedge C\right)$ would have been included three times. In fact $P \left(\text{total}\right)$ would then be greater than 1 (a capital sin).