# Tim has eight pairs of trousers, 15 shirts, six ties, and four jackets, of which two pairs of…

Tim has eight pairs of trousers, 15 shirts, six ties, and four jackets, of which two pairs of…
Tim has eight pairs of trousers, 15 shirts, six ties, and four jackets, of which two pairs of trousers, five shirts, two ties, and two jackets are green. Suppose that one morning Tim selects his outfit completely at random. Then, the total number of possible ways that he could select his outfit is 8 15 6 4 D 2880. These are the sample points of Tim’s experiment. Since selection is completely at random, we assume that the sample points are equally likely. There are 2 5 2 2 D 40 possible ways that he could choose a completely green outfit, so P .Tim is dressed completely in green on a particular day/ D 40=2880 D :014. Notice that in this example there were far too many sample points to actually list them. Nevertheless, we could calculate the required probability by simple counting. Counting methods are thus extremely useful in calculating probabilities when sample points are equally likely, and we will see more sophisticated examples later. Apr 06 2022 11:33 AM

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