{"id":1802,"date":"2019-08-23T13:21:48","date_gmt":"2019-08-23T17:21:48","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/simplestats\/?post_type=chapter&p=1802"},"modified":"2019-10-05T18:33:14","modified_gmt":"2019-10-05T22:33:14","slug":"5-2-2-simple-probability-calculations","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/simplestats\/chapter\/5-2-2-simple-probability-calculations\/","title":{"raw":"5.2.2 Simple Probability Calculations","rendered":"5.2.2 Simple Probability Calculations"},"content":{"raw":"[latexpage]\r\n\r\nThis section is a brief side quest which shows you how to calculate combinations of probabilities. For example, back to die rolling, what is the probability of throwing a two or<\/em> a four?\r\n\r\n \r\n\r\nI'm certain you already know the answer. In this case the \"outcomes of interest\" are two instead of one, so the probability is two out of six possible outcomes:\r\n\r\n \r\n\r\n$$p(\\textrm{two or four})=\\frac{\\textrm{number of outcomes we are interested in}}{\\textrm{number of all outcomes}}=\\frac{2}{6}=\\frac{1}{3}=0.333$$\r\n\r\n \r\n\r\nOr I could have just as easily simply added the two outcomes' individual probabilities:\r\n\r\n \r\n\r\n$$p(\\textrm{two or four})=\\frac{\\textrm{number of two's}}{\\textrm{all outcomes}} + \\frac{\\textrm{number of four's}}{\\textrm{all outcomes}}=\\frac{1}{6}+\\frac{1}{6}=\\frac{2}{6}=0.333$$\r\n\r\n \r\n\r\nAnd this is it: to combine the probabilities of two outcomes which cannot happen at the same time (a.k.a. disjoint events<\/em><\/strong>[footnote]You can recognize dijoint event by the usage of \"or\": it's one or<\/em> the other (or<\/em> a third one, etc.). When flipping one coin, you can either get heads or<\/em> tails; when you roll one die, you can get only one<\/em> of its sides at a time. Hence, we add<\/em> their probabilities.[\/footnote]), you simply have to add them together.<\/strong> (Recall we already used this when we started with the probability of getting heads or<\/em> tails being 1; it's simply the probability of getting heads (0.5) added to the probability of getting tails (0.5)).\r\n\r\n \r\n
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Do It! 5.4 Adding Probabilities<\/em><\/p>\r\n\r\n<\/header>\r\n

\r\n\r\n \r\n\r\nSince we already imagined a bowl with ten consecutively numbered balls inside, let's save ourselves the effort of imagining a new one and reuse it again. What is the probability of randomly selecting the #5 ball or<\/em> the #7 ball or<\/em> the #9 ball out of the ten numbered balls in our bawl?\r\n\r\n<\/div>\r\n(Answer: 0.3)<\/sub>\r\n\r\n \r\n\r\n<\/div>\r\n \r\n\r\nOn the other hand, combining probabilities of events that can<\/em> happen at the same time<\/em>, or that happen one after another<\/em> in time (both a.k.a. independent<\/em> events<\/em><\/strong>[footnote]Events are called independent when the outcome of one doesn't affect the outcome of the other whatsoever. (Contrast this with getting heads in a coin toss, which precludes getting tails; same with throwing any number on a die as it precludes the other numbers from being thrown.)[\/footnote])<\/strong>\u00a0is a tad more complicated and requires multiplication.<\/strong>\r\n\r\n \r\n\r\nFor example, the probability of throwing double two's when rolling two dice (or throwing a two with one die and then immediately throwing again another two) is:\r\n\r\n \r\n\r\n$$p(\\textrm{double two's})=\\frac{\\textrm{number of two's (1st die)}}{\\textrm{all outcomes (1st die)}}\\times\\frac{\\textrm{number of two's (2nd die)}}{\\textrm{all outcomes (2nd die)}}=$$\r\n\r\n$=\\frac{1}{6}\\times\\frac{1}{6}=\\frac{1}{36}=0.028$\r\n\r\n \r\n\r\nOr, if we flip a coin three times (or three coins at the same time), the probability of getting three tails is the probability of getting tails once out of one coin flip (i.e., 0.5) multiplied by the same probability and then multiplied by the same probability again (or simply 0.53<\/sup>):\r\n\r\n \r\n\r\n$$p(\\textrm{three tails})=\\frac{1}{2}\\times\\frac{1}{2}\\times\\frac{1}{2}=\\frac{1}{8}=0.125$$\r\n\r\n \r\n\r\nThus the probability of flipping three tails in a row (or three tails with three coins at the same time) is 1.25 percent.\r\n\r\n \r\n
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Do it! 5.5 Multiplying Probabilities<\/em><\/p>\r\n\r\n<\/header>\r\n

\r\n\r\n \r\n\r\nUsing the same imaginary bowl with ten consecutively numbered balls inside as in the previous exercise, what is the probability of randomly selecting first the #3 ball, then the #4 ball, and then the #5 ball, if you return the selected balls immediately back in the bowl before selecting the next one?<\/em>\r\n\r\n<\/div>\r\n(Answer: 0.001)<\/sub>\r\n\r\n \r\n\r\n<\/div>\r\n \r\n\r\nNow take the time to note the italicized condition at the end of the question in the exercise you just did. It's important enough to necessitate its own scary-red warning,\r\n\r\n \r\n
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Watch Out!! #10<\/strong><\/span>... for Replacement When Working with Probabilities<\/em><\/p>\r\n\r\n<\/header>\r\n

\r\n\r\n \r\n\r\nWhat would have happened had I not specified that in the calculation in Do It! 5.5<\/em> you should consider the selected balls being returned right after their random selection? Why, you would have tempered with the number of all possible outcomes, of course.\r\n\r\n<\/div>\r\nAfter all, after randomly selecting the first ball, unless you imagine returning it back in the bowl<\/em>, there will be only (10-1=) 9 balls left from which to make the second selection. Then after removing the second ball, and again not returning it back in the bowl<\/em>, you'd have left only (9-1=) 8 imaginary balls from which to select your third ball. Then, unlike the $\\frac{1}{10}\\times\\frac{1}{10}\\times\\frac{1}{10}$ you should have used above, the calculation now becomes:\r\n\r\n \r\n\r\n$$p(\\textrm{\"3\", \"4\", \"5\" balls in a row})=\\frac{1}{10}\\times\\frac{1}{9}\\times\\frac{1}{8}=0.0013$$\r\n\r\n \r\n\r\nThe difference between this result and the one in the exercise seems small but that's only because we're working with small numbers. It's still important to understand how random selection with replacement<\/em> differs from random selection without replacement<\/em> and to use the correct calculations.\r\n\r\n \r\n\r\n<\/div>\r\n \r\n\r\nBefore we move on using probabilities with actual data, you could use a bit more practice.\r\n\r\n \r\n
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Do It! 5.6 Adding and Multiplying Probabilities, With and Without Replacement<\/em><\/p>\r\n\r\n<\/header>\r\n

\r\n\r\n \r\n\r\nImagine you and four of your friends (let's call them Adam, Bhav, Chen, and Dila) are in a class of 25 students. Assume that it's the first time your class meets and your professor doesn't know any of you; she only has the class roster in front of her so any name she calls, she calls from the roster at random. Answer the following questions:\r\n