Urn I and urn II each have two red chips and two white chips. Two chips are drawn simultaneously from each urn. Let X1 be he number of red chips in the first sample and X2 be the number of red chips in the second sample. Find the pdf of X1 + X2.

solution: 0: 1/36, 1: 2/9, 2: 1/2, 3: 2/9, 4: 1/36.

can someone help me understand this problem

  • $\begingroup$ Please proof-read the second sentence of your question. If two white chips are drawn simultaneously from each urn, there can be no red chips in either sample, can there? $\endgroup$ – Dilip Sarwate Apr 22 '13 at 3:00

The draw from each urn is two red with probability $\frac 16$, because to get two red you need to star with a red (probability $\frac 12$) and then get the other red (probability $\frac 13)$. Similarly, you get two white with probability $\frac 16$, so you get one of each with probability $\frac 23$. Now to get four reds you need to get two reds out of each urn. If you want two of each you have a number of ways to get there-add them up.

  • $\begingroup$ this is the way its done in the book oi38.tinypic.com/6dza08.jpg can someone please help me understand it done this way $\endgroup$ – notamathwiz Apr 22 '13 at 5:25
  • $\begingroup$ which part did you get stuck? $\endgroup$ – Halil Duru Apr 22 '13 at 5:29
  • $\begingroup$ the $p_x(m)=$ part $\endgroup$ – notamathwiz Apr 22 '13 at 5:33
  • $\begingroup$ @notamathwiz: now to get one red chip, you need to draw two whites out of one urn and one white/one red out of the other. There are 2 ways to pick which urn gives the red, so the total chance is $2 \cdot \frac 16 \cdot \frac 23=\frac 29$ Getting two reds can happen three different ways-can you list them? $\endgroup$ – Ross Millikan Apr 22 '13 at 13:28

$R1$ $R2$ $W1$ $ W2 $ --Urn I

$R3$ $ R4 $ $W3 $ $ W4$ --Urn II

you draw $4$ chips - $2$ from first urn and $2$ from second..How many combinations are there?

$C(4,2)*C(4,2)=36$ different combinations--

Well , among them only one have $4$ reds.

Only one have $4$ whites or $0$ reds.

1 red and 3 white will be 8 cases:

$R1$ $ W1$ $ W3 $ $W4 $ ||$ R2 ... || R3... $ || R4...

$R1$ $ W2$ $ W3 $ $W4 $ || $ R2... || R3... $ || R4...

By symmetry, 3 red and 1 white will be 8 cases too..

And the remaining 18 combinations will be 2 red and 2 white.

Hence , we get your pdf of red chips in the total sample.

  • $\begingroup$ umm is there a computational way to do this problem? something that is more concise than writing out all possibilities? $\endgroup$ – notamathwiz Apr 22 '13 at 5:08
  • $\begingroup$ the computational way is similar --see Milikan's answer--by the way you don't need to write all cases -- counting is easier with the product rule.. $\endgroup$ – Halil Duru Apr 22 '13 at 5:19

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