My gut reaction, thinking about when to add probabilities, is that it’s done less often than multiplying. However, there is one obvious type of situation in which you add:<\/p>\n
Example 1<\/strong><\/p>\n Each ticket bought has 1\/1000 probability of winning a new PC. Josh buys five tickets. What is his probability of winning?<\/p>\n Solution:<\/p>\n In this case, Josh’s probability of winning is the sum of each of his tickets’ winning chances:<\/p>\n P(Josh wins)=1\/1000+1\/1000+1\/1000+1\/1000+1\/1000=5\/1000=1\/200<\/p>\n Note that if one ticket wins, the others can’t. Such a premise – where if event X happens, then event Y cannot – is called mutual exclusivity<\/em>. In turn, we refer to X and Y as mutually exclusive<\/em> events. Normally, when probabilities are added, they belong to mutually exclusive events, either of which will achieve the same outcome.<\/p>\n Generally, probabilities are multiplied under the following conditions:<\/p>\n Example 2<\/strong><\/p>\n At a fishing hole, trout is caught with 55% probability; whitefish, 32%. Assuming you catch two fish, what is the probability that the first is a trout, then the second, a whitefish?<\/p>\n Solution:<\/p>\n We assume there are so many fish present that catching one does not affect the probability of which kind you’ll catch next.<\/p>\n P(TW)=P(T)xP(W)=(0.55)(0.32)=0.176<\/p>\n Next we look at a problem whose solution needs both operations:<\/p>\n Example 3<\/strong><\/p>\n A husband and wife have a joint chequing account. Each cheque needs only one signature; either can sign. The cheques are all written in either black or blue. The wife prefers blue ink; 80% of the time she signs in blue. The husband prefers black; 75% of the time, he signs in black. The wife is more interested in managing the account: in total, 85% of the cheques are written by her.<\/p>\n What is the probability that a cheque from the couple is signed in black?<\/p>\n Solution:<\/p>\n Since only one person signs the cheque, we can add the probabilities of either doing so:<\/p>\n P(Black)=P(wife wrote it, signed in black)+P(husband wrote it, signed in black)<\/p>\n Recall that the wife writes 85% of the cheques, and signs in black only 20% of the time:<\/p>\n P(wife wrote the cheque and signed it in black)=0.85×0.20<\/p>\n If the wife writes 85% of the cheques, the husband must write the other 15% of them. He signs in black 75% of the time:<\/p>\n P(husband wrote the cheque and signed it in black)=0.15×0.75<\/p>\n Therefore,<\/p>\n P(Black)=P(wife wrote it, signed in black)+P(husband wrote it, signed in black)<\/p>\n becomes<\/p>\n P(Black)=0.85×0.20 + 0.15×0.75=0.17 + 0.1125=0.2825<\/p>\n Apparently, the probability a given cheque will be in black is 0.2825, or 28.25%.<\/p>\n I’ll be talking much more about probability in coming posts:)<\/p>\n Source:<\/p>\n Ross, Sheldon. A First Course in Probability<\/u>. New York: Macmillan, 1988.<\/p>\n\n