4.8: Exercises

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Q	A
At a swim meet, the competitors in the 100-m freestyle are Ben, Chad, Grover, and Tim. These four swimmers make up our sample space \(\Omega\) for the winner of this heat. Is Chad \(\in \Omega\)?	Yes.
Is Tim an outcome?	Yes.
Is Ben an event?	No, since outcomes are elements of the sample space, while events are subsets of the sample space.
Is { Chad, Grover } an event?	Yes.
Is { Ben } an event?	Yes.
Suppose I told you that Pr({Ben})=.1, Pr({Chad})=.2, Pr({Grover})=.3, and Pr({Tim})=.3. Would you believe me?	Better not. This is not a valid probability measure, since the sum of the probabilities of all the outcomes, Pr(\(\Omega\)), is not equal to 1.
Suppose I told you that Pr({Ben, Chad})=.3, and Pr({Ben, Tim})=.4, and Pr({Grover})=.4. Could you tell me the probability that Ben wins the heat?	Yes. If Pr({Ben, Chad})=.3 and Pr({Grover})=.4, that leaves .3 probability left over for Tim. And if Pr({Ben, Tim})=.4, this implies that Pr({Ben})=.1.
And what’s the probability that someone besides Chad wins?	Pr(\(\overline{\{\text{Chad}\}}\)) = \(1 -\)Pr({Chad}), so we just need to figure out the probability that Chad wins, and take one minus that. Clearly if Pr({Ben, Chad})=.3 (as we were told), and Pr({Ben})=.1 (as we computed), then Pr({Chad})=.2, and the probability of a non-Chad winner is .8.
Okay, so we have the probabilities of our four swimmers Ben, Chad, Grover, and Tim each winning the heat at .1, .2, .4, and .3, respectively. Now suppose Ben, Chad, and Grover are UMW athletes, Tim is from Marymount, Ben and Tim are juniors, and Chad and Grover are sophomores. We’ll define \(U\)={Ben,Chad,Grover}, \(M\)={Tim}, \(J\)={Ben,Tim}, and \(S\)={Chad,Grover}. What’s Pr(\(U\))?	.7.