S1: Discrete Random Variables Problem: Watch

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cherc2005
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#1
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This is from Edexcel S1 Paper Jan 2003, Q5 part (e):

(see attached image below - Question & Answer)

I have looked at the mark-scheme for it and know the answer but I simply dont know how to arrive at the answers.
Could someone please explain this part of the question to me and how to work it out.
Thanks.
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Jonny W
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P(X = 0) = 1/2
P(X = 1) = 1/4
P(X = 3) = 1/4

There are nine possible outcomes for (X1, X2):

(0, 0) _____ probability 1/4 ______ value of (X1 + X2) = 0
(0, 1) _____ probability 1/8 ______ value of (X1 + X2) = 1
(0, 3) _____ probability 1/8 ______ value of (X1 + X2) = 3
(1, 0) _____ probability 1/8 ______ value of (X1 + X2) = 1
(1, 1) _____ probability 1/16 _____ value of (X1 + X2) = 2
(1, 3) _____ probability 1/16 _____ value of (X1 + X2) = 4
(3, 0) _____ probability 1/8 ______ value of (X1 + X2) = 3
(3, 1) _____ probability 1/16 _____ value of (X1 + X2) = 4
(3, 3) _____ probability 1/16 _____ value of (X1 + X2) = 6

So

P(X1 + X2 = 0) = 1/4 (1st line of table)
P(X1 + X2 = 1) = 1/8 + 1/8 = 1/4 (2nd and 4th lines of table)
P(X1 + X2 = 2) = 1/16 (5th line of table)
P(X1 + X2 = 3) = 1/8 + 1/8 = 1/4 (3rd and 7th lines of table)
P(X1 + X2 = 4) = 1/16 + 1/16 = 1/8 (6th and 8th lines of table)
P(X1 + X2 = 6) = 1/16 (9th line of table)
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cherc2005
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Is this the same thing as one of those diagram/charts i.e. when you display all possible outcomes in a box-type chart?

Thanks For the help, i get it now!
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Jonny W
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(Original post by cherc2005)
Is this the same thing as one of those diagram/charts i.e. when you display all possible outcomes in a box-type chart?
Yes, you can put the probabilities in a 4-by-4 table:



\begin{array}{c|ccc}

x_2 \backslash x_1 & 0 & 1 & 2 & 3 \\\hline

0 & 1/4 & 1/8 & 0 & 1/8 \\

1 & 1/8 & 1/16 & 0 & 1/16 \\

2 & 0 & 0 & 0 & 0 \\

3 & 1/8 & 1/16 & 0 & 1/16

\end{array}

then add up the diagonals, starting in the top-left corner:

1/4
1/8 + 1/8 = 1/4
0 + 1/16 + 0 = 1/16
1/8 + 0 + 0 + 1/8 = 1/4
1/16 + 0 + 1/16 = 1/8
0 + 0 = 0
1/16
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