#### xailer

##### New member
1. P(A|B) = 7/10, P(B|A) = 7/15, P(A U B) = 3/5

Find P(A n B) and P(A' n B)

2. We put 4 coins X and 2 coins Y randomly in a row. What is probability that one of X coins will be at the beginning and also end of the row? Result should be 3/14

Number of outcomes is 6!

Number of outcomes where X coins are at the beginning and the end of the row is (4! / 2! * 4!)

So result should be (4! / 2! * 4!) / 6! But it is not.

thank you

#### Gene

##### Senior Member
1. Does Bayes' Formula help?

$$\displaystyle \;\;\;P(A|B)\;=\;\frac{P(A\,\cap\,B)}{P(B)}$$

2. Since all Xs & Ys are the same, there are 6!/(4!*2!) outcomes.
I'm confused by your "should be"s. Does the book say 3/14 is the answer? That's not what I get.

#### pka

##### Elite Member
$$\displaystyle \begin{array}{l} P(A|B) = \frac{7}{{10}}\quad \Rightarrow \quad P(A \cap B) = \frac{7}{{10}}P(B) \\ P(B|A) = \frac{7}{{15}}\quad \Rightarrow \quad P(A \cap B) = \frac{7}{{15}}P(A) \\ P(A \cup B) = P(A) + P(B) - P(A \cap B) \\ \left\{ \begin{array}{l} \frac{8}{{15}}P(A) + P(B) = \frac{3}{5} \\ P(A) + \frac{3}{{10}}P(B) = \frac{3}{5} \\ \end{array} \right. \\ \end{array}$$
Solve for P(A) & P(B).