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Deck one contains k white and j red cards. Deck two contains r white and p red cards. A card is chosen at random from deck one and put into deck two, then a card is chosen at random from deck two and put into deck one, and finally a card is selected from deck one. What is the probability that the final selected card is white?

This is what i have so far

let $A$ = card chosen from deck one is white

$B$ = 2nd card chosen from deck two is white

$C$ = 3rd card chosen from deck one is white

$P(B) = P(B|A)P(A) + P(B|A^c)P(A^c) = (r+1)/(r+p+1) * k/(j+k) + r/(r+p+1) * j/(j+k)$

$P(C) = P(C|BA) + P(C|BA^c) + P(C|B^cA) + P(C|B^cA^c)$

?? takes into account 4 scenarios: white out+white in, red out+white in, white out+red in, red out + red in

I am very lost now :)

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  • $\begingroup$ Just consider the possible paths to drawing a white card. There are only $4$ such paths, according to which color card is moved at each stage. $\endgroup$
    – lulu
    Commented Sep 6, 2022 at 21:31

1 Answer 1

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You're almost correct. $P(B)=P(B|A)P(A)+P(B|A^c)P(A^c)=\frac{r+1}{r+p+1}\frac{k}{k+j}+\frac{r}{r+p+1}\frac{j}{k+j}=\frac{jr+kr+k}{(k+j)(r+p+1)}$

Next,

$P(C)=P(C|BA)P(BA)+P(C|BA^c)P(BA^c)+P(C|B^cA)P(B^cA)+P(C|B^cA^c)P(B^cA^c)$

$P(BA)=P(A)P(B|A)=\frac{r+1}{r+p+1}\frac{k}{k+j}$

$P(BA^c)=P(A^c)P(B|A^c)=\frac{r}{r+p+1}\frac{j}{k+j}$

$P(B^cA)=P(A)P(B^c|A)=\frac{p}{r+p+1}\frac{k}{k+j}$

$P(B^cA^c)=P(A^c)P(B^c|A^c)=\frac{p+1}{r+p+1}\frac{j}{k+j}$

Finally,

$P(C)=\frac{k}{k+j}\frac{r+1}{r+p+1}\frac{k}{k+j}+\frac{k+1}{k+j}\frac{r}{r+p+1}\frac{j}{k+j}+\frac{k-1}{k+j}\frac{p}{r+p+1}\frac{k}{k+j}+\frac{k}{k+j}\frac{p+1}{r+p+1}\frac{j}{k+j}=\frac{jkp+jkr+jk+jr+k^2p+k^2r+k^2-kp}{(k+j)^2(r+p+1)}$

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