IE 300 / GE 331 Discussion Problem Set 2
•
(Independence) 2110
(a)
Let A denote the event that a player defeats all four opponents in a game. Since the
results from opponents are independent, we have
P(A) = (0.8)
4
= 0.4096
(b)
Let B denote the event that a player defeats at least two opponents in a game. B occurs
if and only if the player defeats the first two opponents. Therefore,
P(B) = (0.8)(0.8) = 0.64
(c)
Let C denote the event that a player defeats all four opponents at least once when the
game is played three times. C’ is then the event that the player does not defeat all four
opponents in all of the three games. The player does not defeat all four opponents in
one game is A’. So
P(
C′
) = P(
A′
)
3
= (1
−
0.4096)
3
= 0.2058
P(C) = 1
−
P(
C′
) = 0.7942
•
(Total probability rule) 2146
Let A denote the event that a transaction can be completed in less than 100 milliseconds.
Let B denote the event that a transaction is changing an item. B’ is the event that a
transaction is adding an item. Then according to the total probability rule,
𝑃𝑃
(
𝐴𝐴
) =
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 Spring '09
 NegarKayavash
 Conditional Probability, Probability, Probability theory, total probability rule

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