Question 1171674
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<table border = "1" cellpadding = "5"><tr><td></td><td>Less than 15 minutes</td><td>15 - 29 minutes</td><td>30 - 44 minutes</td><td>45 - 59 minutes</td><td>60 or more minutes</td><td>Total</td></tr><tr><td>Private vehicle</td><td>636</td><td>908</td><td>590</td><td>257</td><td>256</td><td>2647</td></tr><tr><td>Public transportation</td><td>9</td><td>54</td><td>96</td><td>62</td><td>108</td><td>329</td></tr><tr><td>Other means</td><td>115</td><td>70</td><td>23</td><td>7</td><td>7</td><td>222</td></tr><tr><td>Total</td><td>760</td><td>1032</td><td>709</td><td>326</td><td>371</td><td>3198</td></tr></table>


Problem 1)
Find the probability that either the commuter used public transportation and/or the commuter had a commute of 60 or more minutes. (2 marks)


Add up all the values in the "public transportation" row and all of the values in the "60 or more minutes" column:

9+54+96+62+108+256+7 = 592

note: don't add in the totals as they already account for the values listed above.


As a shortcut we could add the totals for the row and column: 329+371 = 700
Then we subtract off 108 because it was counted twice: 700-108 = 592


Either way, there are 592 people who either took public transportation, their commute time was 60 minutes or more, or both events occurred. 


Divide 592 over the grand total in the lower right corner and we get (592)/(3,198) = 0.185 approximately



Answer: Approximately 0.185

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Problem 2) Given that the commuter used public transportation, find the probability that the commuter had a commute of 60 or more minutes. (2 marks)


Focus solely on the "public transportation" row. There are 108 people whose commute time was 60 or more minutes. There are 329 people total who took public transportation.


The probability of a commuter having a comute time of 60 or more minutes, given they took public transportation, is 108/329 = 0.328



Answer: Approximately 0.328

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