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Unformatted text preview: We argued that
E (T54 ) = E (T5 ) + E T545 (1) and also showed that E (T5 ) = 6. To ﬁnd the second term in the equation
above, we deﬁned X to be the outcome of the experiment (roll) after the ﬁrst
5 has been appeared and argued that
E T545 = E E T545 X
We also showed in the tutorial class that 1
1 + E (T54 )
E T545 X = i = 1 + E (T545 ) if i = 4
if i = 1, 2, 3, 6
if i = 5 (2) (3) so
E (T545 ) = E E T545 X
= 1 × 1/6 + [1 + E (T54 )] × 4/6 + 1 + E (T545 ) × 1/6 (4) Solving for E (T545 ) we get
E (T545 ) = 64
+ E (T54 )
55 Now substituting E (T545 ) from the equation above in equation (1)
E (T54 ) = E (T5 ) + 64
+ E (T54 )
55 (5) so
E (T54 ) = 6 + 64
+ E (T54 )
55 solving for E (T54 ) we get
E (T54 ) = 36. 1 (6) ...
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This note was uploaded on 03/17/2012 for the course STAT 333 taught by Professor Chisholm during the Winter '08 term at Waterloo.
 Winter '08
 Chisholm
 Probability

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