last_question_-_tutorial_3

last_question_-_tutorial_3 - We argued that E (T54 ) = E...

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Unformatted text preview: We argued that E (T54 ) = E (T5 ) + E T54|5 (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 T54|5 = E E T54|5 X We also showed in the tutorial class that 1 1 + E (T54 ) E T54|5 X = i = 1 + E (T54|5 ) if i = 4 if i = 1, 2, 3, 6 if i = 5 (2) (3) so E (T54|5 ) = E E T54|5 X = 1 × 1/6 + [1 + E (T54 )] × 4/6 + 1 + E (T54|5 ) × 1/6 (4) Solving for E (T54|5 ) we get E (T54|5 ) = 64 + E (T54 ) 55 Now substituting E (T54|5 ) 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.

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