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Assume that there are two urns. The first urn contains 4 red balls, 3 blue balls, and 3 white balls. The second urn contains 2 red balls, 4 blue balls, and 4 white balls. You randomly select an urn and take two balls from the urn. The probability that you pick the first urn is 40%. What is the probability that
(a) the two balls are red?
(b) the second ball is blue?
(c) the second ball is blue given that the first ball is red?

My attempt:

U1 = first urn, U2, = second urn, R,B,W = red, blue, white respectively

(a) P(R,R) = P(first ball R) * P(second ball R)
$P$(first ball R) = $P(R|U1)P(U1) + P(R|U2)P(U2)$ = $(\frac{4}{10})(\frac{4}{10}) + (\frac{2}{10})(\frac{6}{10}) = \frac{28}{100}$

For the second ball, we know that a red ball has already been drawn, so we need to take into account two possibilities: first ball was drawn from urn 1 and first ball was drawn from urn 2

if first ball was from U1: $P$(second ball R) = ($\frac{3}{9}$)($\frac{4}{10}$) + ($\frac{2}{10}$)($\frac{6}{10}$) $= 0.253333$
if first ball was from U2: $P$(second ball R) = ($\frac{4}{10}$)($\frac{4}{10}$) + ($\frac{1}{9}$)($\frac{6}{10}$) $= 0.22667$
Therefore, $P$(second ball R) = $0.253333 + 0.22667 = 0.48$

Final answer: $P(R,R) = (0.28)(0.48) = 0.1344$


(b) $P$(second ball B)$ = P(R,B) + P(W,B) + P(B,B)$

$P(R,B) = P$(first R) * $P$(second B)
We know that P(first R) = 0.28. For the second ball B, we split it into two cases: when the first ball drawn from the first urn, and drawn from the second urn.
if first ball from U1: $P$(second ball B) = ($\frac{3}{9}$)($\frac{4}{10}$) + ($\frac{4}{10}$)($\frac{6}{10}$)
if first ball from U2: $P$(second ball B) = ($\frac{3}{10}$)($\frac{4}{10}$) + ($\frac{4}{9}$)($\frac{6}{10}$)
$P$(second ball B) $= 0.733333$ so $P(R,B) = (0.28)(0.733333) = 0.205333333$

$P(W,B) = P$(first W) * $P$(second B)
$P$(first ball W) = $P(W|U1)P(U1) + P(W|U2)P(U2)$ = $(\frac{3}{10})(\frac{4}{10}) + (\frac{4}{10})(\frac{6}{10}) = \frac{36}{100}$
For the second ball B, we split it into two cases: when the first ball drawn from the first urn, and drawn from the second urn.
if first ball from U1: $P$(second ball B) = ($\frac{3}{9}$)($\frac{4}{10}$) + ($\frac{4}{10}$)($\frac{6}{10}$)
if first ball from U2: $P$(second ball B) = ($\frac{3}{10}$)($\frac{4}{10}$) + ($\frac{4}{9}$)($\frac{6}{10}$)
$P$(second ball B) $= 0.733333$ so $P(W,B) = (0.36)(0.733333) = 0.264$

$P(B,B) = P$(first B) * $P$(second B)
$P$(first ball B) = $P(B|U1)P(U1) + P(B|U2)P(U2)$ = $(\frac{3}{10})(\frac{4}{10}) + (\frac{4}{10})(\frac{6}{10}) = \frac{36}{100}$
For the second ball B, we split it into two cases: when the first ball drawn from the first urn, and drawn from the second urn.
if first ball from U1: $P$(second ball B) = ($\frac{2}{9}$)($\frac{4}{10}$) + ($\frac{4}{10}$)($\frac{6}{10}$)
if first ball from U2: $P$(second ball B) = ($\frac{3}{10}$)($\frac{4}{10}$) + ($\frac{3}{9}$)($\frac{6}{10}$)
$P$(second ball B) $= 0.0.6488888$ so $P(W,B) = (0.36)(0.0.648888) = 0.2336$

Therefore, $P$(second ball B) $= 0.7029333333$

(c) $P(R,B) = 0.205333333$ from part (b)

Could someone check and see if I did this correctly? I feel like I may have overcounted things since there are so many scenarios. Thank you

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  • $\begingroup$ Please add the self-study tag. $\endgroup$ – Xi'an Apr 3 at 7:48
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a) It's healthier if you condition on the chosen urn: $$\begin{align}P(RR)&=P(RR|U_1)P(U_1)+P(RR|U_2)P(U_2)\\ &= \frac{4}{10}.\frac{3}{9}.\frac{4}{10}+\frac{2}{10}.\frac{1}{9}.\frac{6}{10}=\frac{1}{15}\end{align}$$ Your answer differs from this one, because you still consider the choice of urns in the second draw. I haven't calculated (b) and (c), however since they follow a similar approach, I don't think you got the right answer for them. Apply the above logic to (b) and (c). Also, in (c), it asks for $P(B_2|R_1)$; you seem to calculate a joint probability; not a conditional one.

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  • $\begingroup$ thank you. For part (b) I calculated P(RB) + P(WB) + P(BB) using the same logic as that you gave. For part (c), would I do P(B2|R1) = P(RB) / P(R) and use P(RB) from (b) and calculate P(R) as 0.28, which I did in my original answer? $\endgroup$ – IrCa Apr 4 at 1:16
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    $\begingroup$ @IrCa Yes, you can do that, your P(R) calculation is correct. $\endgroup$ – gunes Apr 5 at 10:59

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