• Oct 21st 2007, 07:46 AM
Infiniti
"Of three urns, the first contains 2 white and 4 black cards, the second contains 8 white and four black cards, and the third contains 1 white and 3 black card. If one card is selected from each urn, find the probability that exactly 2 white cards are drawn"
• Oct 21st 2007, 08:03 AM
red_dog
Use Poisson's rule.
Let $\displaystyle A_1$: the card drawn from the first urn is white.
$\displaystyle A_2$: the card drawn from the second urn is white.
$\displaystyle A_3$: the card drawn from the third urn is white.
We have:
$\displaystyle p_1=p(A_1)=\frac{1}{2}, \ q_1=\frac{1}{2}$
$\displaystyle p_2=p(A_2)=\frac{2}{3}, \ q_2=\frac{1}{3}$
$\displaystyle p_3=p(A_3)=\frac{1}{4}, \ q_3=\frac{3}{4}$

So, the probability that exactly two cards are white is equal to the coefficient of $\displaystyle x^2$ in the expression
$\displaystyle (p_1x+q_1)(p_2x+q_2)(p_3x+q_3)$
That means $\displaystyle p_1p_2q_3+p_1q_2p_3+q_1p_2p_3$.
Now, plug in $\displaystyle p_1,p_2,p_3,q_1,q_2,q_3$.