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xailer
04-18-2006, 12:32 PM
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
04-18-2006, 02:33 PM
1. Does Bayes' Formula help?

\;\;\;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
04-18-2006, 05:29 PM
\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).