{"id":1784,"date":"2019-08-22T22:15:31","date_gmt":"2019-08-23T02:15:31","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/simplestats\/?post_type=chapter&p=1784"},"modified":"2019-10-05T18:25:48","modified_gmt":"2019-10-05T22:25:48","slug":"5-2-1-calculating-probabilities","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/simplestats\/chapter\/5-2-1-calculating-probabilities\/","title":{"raw":"5.2.1 Working with Probabilities","rendered":"5.2.1 Working with Probabilities"},"content":{"raw":"[latexpage]\r\n\r\nWe express probabilities as proportions<\/strong> (and we also denote them with\u00a0p,\u00a0<\/em>just like we do proportions[footnote]If you need a reminder, the relevant part is in Section 2.3.1, here: https:\/\/pressbooks.bccampus.ca\/simplestats\/chapter\/2-3-1-adding-percentages<\/a>\/[\/footnote]), as this is indeed what they are:\r\n\r\n \r\n\r\n$$p=\\frac{\\textrm{number of specific outcomes we are interested in}}{\\textrm{number of all possible outcomes}}$$\r\n\r\n \r\n\r\nOr, the probability of a specific outcome is the proportion of the number of such outcomes out of the number of all possible outcomes.\r\n\r\n \r\n\r\nThus the probability of getting heads in a coin toss is:\r\n\r\n \r\n\r\n$$p(\\textrm{heads})=\\frac{\\textrm{number of heads sides of a coin}}{\\textrm{number of all sides of a coin}}=\\frac{1}{2}=0.5$$\r\n\r\n \r\n\r\nThe same of course applies to tails:\r\n\r\n \r\n\r\n$$p(\\textrm{tails})=\\frac{\\textrm{number of tails sides of a coin}}{\\textrm{number of all sides of a coin}}=\\frac{1}{2}=0.5$$\r\n\r\n \r\n\r\nHeads and tails together exhaust all possible outcomes, so the probability that a coin will fall on any of its two sides is:\r\n\r\n \r\n\r\n$$p(\\textrm{heads or tails})=\\frac{2}{2}=\\frac{1}{2}+\\frac{1}{2}=0.5+0.5=1$$\r\n\r\n \r\n\r\nNow how about we extend our example to something that has more that two outcomes? With six sides, a conventional die will serve us perfectly.\r\n\r\n \r\n\r\nFollowing the same logic as with the coin, the probability to throw, say, a five is:\r\n\r\n \r\n\r\n$$p(\\textrm{five})=\\frac{\\textrm{number of \"five\" sides of a die}}{\\textrm{number of all sides of a die}}=\\frac{1}{6}=0.167$$\r\n\r\n \r\n\r\nThe same goes for throwing a one, a two, a three, a four, or a six:\r\n\r\n \r\n\r\n$$p(\\textrm{one})=p(\\textrm{two})=p(\\textrm{three})=p(\\textrm{four})=p(\\textrm{five})=p(\\textrm{six})=\\frac{1}{6}=0.167$$\r\n\r\n \r\n\r\nOr, imagine you have a bowl with ten balls inside (i.e., the balls have numbers from 1 to 10). The probability of selecting each one out (without looking!) is, you guessed it, 1 out of 10, as each number appears only once and there are ten possible outcomes:\r\n\r\n \r\n\r\n$$p(1)=p(2)=p(3)=\\ldots=p(10)=\\frac{1}{10}=0.1$$\r\n\r\n \r\n\r\nWhile this principle applies to N<\/em> of any size -- so we can increase the number of outcomes as much as we want -- note the key prerequisite for the calculations to work: the outcomes must happen randomly.<\/strong> A coin toss and a die throw are classical examples of random chance. But when picking balls out of a bowl we have to make sure we don't look or we might (consciously or subconsciously) choose<\/em> one. Choosing a ball with a specific number introduces bias and thus invalidates randomness -- i.e., it invalidates the principle of the outcomes having the same probability. Without this principle we cannot calculate anything: the only way to know the probability of an outcome is, in a sense, to divide the total probability, as it were, (i.e., 1) by the number of all possible outcomes, giving us equal probability for each. We know the probability of an outcome only<\/em> if<\/em> we know how many outcomes are possible in total and they all have the same probability.\u00a0<\/strong>(Chapter 6 has more on the topic as it's devoted to the topic of how random selection works.)\r\n\r\n ","rendered":"

We express probabilities as proportions<\/strong> (and we also denote them with\u00a0p,\u00a0<\/em>just like we do proportions[1]<\/sup><\/a>), as this is indeed what they are:<\/p>\n

 <\/p>\n

  <\/span>   <\/span>\"\[p=\frac{\textrm{number of specific outcomes we are interested in}}{\textrm{number of all possible outcomes}}\]\"<\/p>\n

 <\/p>\n

Or, the probability of a specific outcome is the proportion of the number of such outcomes out of the number of all possible outcomes.<\/p>\n

 <\/p>\n

Thus the probability of getting heads in a coin toss is:<\/p>\n

 <\/p>\n

  <\/span>   <\/span>\"\[p(\textrm{heads})=\frac{\textrm{number of heads sides of a coin}}{\textrm{number of all sides of a coin}}=\frac{1}{2}=0.5\]\"<\/p>\n

 <\/p>\n

The same of course applies to tails:<\/p>\n

 <\/p>\n

  <\/span>   <\/span>\"\[p(\textrm{tails})=\frac{\textrm{number of tails sides of a coin}}{\textrm{number of all sides of a coin}}=\frac{1}{2}=0.5\]\"<\/p>\n

 <\/p>\n

Heads and tails together exhaust all possible outcomes, so the probability that a coin will fall on any of its two sides is:<\/p>\n

 <\/p>\n

  <\/span>   <\/span>\"\[p(\textrm{heads or tails})=\frac{2}{2}=\frac{1}{2}+\frac{1}{2}=0.5+0.5=1\]\"<\/p>\n

 <\/p>\n

Now how about we extend our example to something that has more that two outcomes? With six sides, a conventional die will serve us perfectly.<\/p>\n

 <\/p>\n

Following the same logic as with the coin, the probability to throw, say, a five is:<\/p>\n

 <\/p>\n

  <\/span>   <\/span>\"\[p(\textrm{five})=\frac{\textrm{number of "five" sides of a die}}{\textrm{number of all sides of a die}}=\frac{1}{6}=0.167\]\"<\/p>\n

 <\/p>\n

The same goes for throwing a one, a two, a three, a four, or a six:<\/p>\n

 <\/p>\n

  <\/span>   <\/span>\"\[p(\textrm{one})=p(\textrm{two})=p(\textrm{three})=p(\textrm{four})=p(\textrm{five})=p(\textrm{six})=\frac{1}{6}=0.167\]\"<\/p>\n

 <\/p>\n

Or, imagine you have a bowl with ten balls inside (i.e., the balls have numbers from 1 to 10). The probability of selecting each one out (without looking!) is, you guessed it, 1 out of 10, as each number appears only once and there are ten possible outcomes:<\/p>\n

 <\/p>\n

  <\/span>   <\/span>\"\[p(1)=p(2)=p(3)=\ldots=p(10)=\frac{1}{10}=0.1\]\"<\/p>\n

 <\/p>\n

While this principle applies to N<\/em> of any size — so we can increase the number of outcomes as much as we want — note the key prerequisite for the calculations to work: the outcomes must happen randomly.<\/strong> A coin toss and a die throw are classical examples of random chance. But when picking balls out of a bowl we have to make sure we don’t look or we might (consciously or subconsciously) choose<\/em> one. Choosing a ball with a specific number introduces bias and thus invalidates randomness — i.e., it invalidates the principle of the outcomes having the same probability. Without this principle we cannot calculate anything: the only way to know the probability of an outcome is, in a sense, to divide the total probability, as it were, (i.e., 1) by the number of all possible outcomes, giving us equal probability for each. We know the probability of an outcome only<\/em> if<\/em> we know how many outcomes are possible in total and they all have the same probability.\u00a0<\/strong>(Chapter 6 has more on the topic as it’s devoted to the topic of how random selection works.)<\/p>\n

 <\/p>\n

  1. If you need a reminder, the relevant part is in Section 2.3.1, here: https:\/\/pressbooks.bccampus.ca\/simplestats\/chapter\/2-3-1-adding-percentages<\/a>\/ ↵<\/a><\/li><\/ol><\/div>","protected":false},"author":533,"menu_order":6,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1784","chapter","type-chapter","status-publish","hentry"],"part":28,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/simplestats\/wp-json\/pressbooks\/v2\/chapters\/1784","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/simplestats\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/simplestats\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/simplestats\/wp-json\/wp\/v2\/users\/533"}],"version-history":[{"count":11,"href":"https:\/\/pressbooks.bccampus.ca\/simplestats\/wp-json\/pressbooks\/v2\/chapters\/1784\/revisions"}],"predecessor-version":[{"id":1886,"href":"https:\/\/pressbooks.bccampus.ca\/simplestats\/wp-json\/pressbooks\/v2\/chapters\/1784\/revisions\/1886"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/simplestats\/wp-json\/pressbooks\/v2\/parts\/28"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/simplestats\/wp-json\/pressbooks\/v2\/chapters\/1784\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/simplestats\/wp-json\/wp\/v2\/media?parent=1784"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/simplestats\/wp-json\/pressbooks\/v2\/chapter-type?post=1784"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/simplestats\/wp-json\/wp\/v2\/contributor?post=1784"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/simplestats\/wp-json\/wp\/v2\/license?post=1784"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}