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Unformatted text preview: 1.2. DISCRETE PROBABILITY DISTRIBUTIONS 25 First toss Second toss Third toss Outcome H H H H H H T T T T T T (Start) ω ω ω ω ω ω ω ω 1 2 3 4 5 6 7 8 H T Figure 1.8: Tree diagram for three tosses of a coin. Let A be the event “the first outcome is a head,” and B the event “the second outcome is a tail.” By looking at the paths in Figure 1.8, we see that P ( A ) = P ( B ) = 1 2 . Moreover, A ∩ B = { ω 3 , ω 4 } , and so P ( A ∩ B ) = 1 / 4 . Using Theorem 1.4, we obtain P ( A ∪ B ) = P ( A ) + P ( B ) P ( A ∩ B ) = 1 2 + 1 2 1 4 = 3 4 . Since A ∪ B is the 6element set, A ∪ B = { HHH,HHT,HTH,HTT,TTH,TTT } , we see that we obtain the same result by direct enumeration. In our coin tossing examples and in the die rolling example, we have assigned an equal probability to each possible outcome of the experiment. Corresponding to this method of assigning probabilities, we have the following definitions....
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This note was uploaded on 09/15/2009 for the course SCF scf taught by Professor Scf during the Spring '09 term at Indian Institute Of Management, Ahmedabad.
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