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StephenElston
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Fixed equation typo in lab 04-04
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Module04/04-04-Probability.ipynb

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"\n",
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"There are two possible outcomes for each experiment; so the total number of possible combinations of five passengers being searched or not searched is 2<sup>5</sup> or 32. So within those 32 sets of possible result combinations, how many have three searches? We can use the <sub>n</sub>C<sub>k</sub> formula to calculate this:\n",
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"\n",
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"\\begin{equation}_{5}C_{3} = \\frac{5!}{3!(5 - 3)!} = \\frac{120}{6\\times 4} = \\frac{120}{24} = 5\\end{equation}\n",
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"\\begin{equation}_{5}C_{3} = \\frac{5!}{3!(5 - 3)!} = \\frac{120}{6\\times 2} = \\frac{120}{24} = 5\\end{equation}\n",
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"\n",
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"So 5 out of our 32 combinations had 3 searches and 2 non-searches.\n",
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"\n",

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