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Mathematics 18 Online
OpenStudy (anonymous):

P(e1)=P(e2)=0.15 P(e3)=0.4 P(e4)=2*P(e5) What is P(e5) and P(e4)?

OpenStudy (campbell_st):

the sum of the probabilities is always 1 so at the moment if there are 5 events in the sample space then P(e1) + P(e2) + P(e3) = 0.7 then P(e4) + P(e5) = 0.3 (1) and you know P(e4) = 2*P(e5) (2) so making a substitution of (2) into equation (1) you have 2*P(e5) + P(e5) = 0.3 or 3P(e5) = 0.3 you should be able to solve for P(e5) and also find P(e4) by putting your solution into (2)

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