1. ## probabilities

13 . show that:
i. P(A∩B∩C)=P(A|B∩C)P(B|C)P(C)
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14.two boxes contain the number of the ball as indicated. a box is chosen randomly and then two balls are
drawn One after the other, from that box
, without Replacement .what is the ....................W B W B
probability that both Of the drawn balls ...............6 4 5 5
have the same color?
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15. 14.two boxes contain the number of the ball as indicated.aball is drawn from box 1 and put in box 2 .a ball
Is then drawn randomly from box 2 .what
Is the probability that the ball drawn is ...............................W B W B
White ? .................................................. .....................m1 m2 m1 m2

2. Originally Posted by compufatwa
13 . show that:
i. P(A∩B∩C)=P(A|B∩C)P(B|C)P(C)
$P(B|C) = \frac{P(B\cap C)}{P(C)}$
$P(A|B\cap C) = \frac{P(A\cap B\cap C)}{P(B\cap C)}$
Thus,
$P(A|B\cap C)P(B|C)P(C) = \frac{P(A\cap B\cap C)}{P(B\cap C)} \cdot \frac{P(B\cap C)}{P(C)} \cdot P(C) = P(A\cap B\cap C)$

I do not understand 14 and 15.

3. In #14, let’s name one box X containing six white balls and four black balls.
Then box Y contains five of each color.
$P(X) = P(Y) = \frac{1}{2}$ and let $W_1 W_2$ denote the first and second balls drawn are white.

Here is what the question asks: $P\left( {W_1 W_2 \cup B_1 B_2 } \right)$.

Because they come from either X or Y we have
$\begin{array}{l} P\left( {W_1 W_2 \cup B_1 B_2 } \right) = P\left( {W_1 W_2 X) + P(B_1 B_2 X} \right) + P\left( {W_1 W_2 Y) + P(B_1 B_2 Y} \right) \\
= P\left( {W_1 W_2 |X)P(X) + P(B_1 B_2 |X} \right)P(X) + P\left( {W_1 W_2 |Y)P(Y) + P(B_1 B_2 |Y} \right)P(Y) \\ \end{array}$

But since each box is equally likely the becomes
$P\left( {W_1 W_2 \cup B_1 B_2 } \right) = \frac{1}{2}\left[ {\frac{{6 \cdot 5}}{{10 \cdot 9}} + \frac{{4 \cdot 3}}{{10 \cdot 9}} + \frac{{5 \cdot 4}}{{10 \cdot 9}} + \frac{{5 \cdot 4}}{{10 \cdot 9}}} \right]$

You can do #15 the same way.