a soccer team\'s shirts have arrived in a big box, and people just start grabbin
ID: 3261565 • Letter: A
Question
a soccer team's shirts have arrived in a big box, and people just start grabbing them, looking for the right size. the box contains 6 medium, 10 large, and 11 extra-large shirts. A player wants a medium for himself and one for his sister. Find the probability of each event described. Complete parts a-d below.
a) The first two the player grabs are the wrong sizes. The probability that the first two shirts the player grabs are the wrong size is (Round to three decimal places as needed.) b) The first medium shirt the player finds is the third one he checks The probability that the first medium shirt the player finds is the third one he checks is The probabilit that the first medium shirt the player finds is the third one he checks is Round to three decimal places as needed.) c) The first four shirts the player picks are all extra-large. The probability that the first four shirts the player picks are all extra-large (Round to three decimal places as needed) d) At least one of the first four shirts the player checks is a medium. The probability that at least one of the first four shirts the player checks is a medium is Round to three decimal places as needed.)Explanation / Answer
Ans:
1) The first shirt is 21/27, because there are 20 shirts, 16 the wrong size. The second shirt you choose is 20/26, because you have already removed one of the wrong-sized shirts, leaving 19, 15 of which are the wrong size.
Thus, 21/27 x 20/26=0.5983
2) You are choosing 3 shirts. The first two can't be medium so
21/27 and 20/26 again. The third shirt is medium, so 6 medium shirts over 25 left, 6/25.
Thus, 21/27 x 20/26 x 6/25=0.1436
3) Same principle leads to
11/27 x 10/26 x 9/25 x 8/24=0.0188
So basically, the top number is the number of shirts that are left that fit what you're looking for, and the bottom number is the total number of shirts left, the number of possible choices.
4) This question means that you can have 1 medium, 2 mediums, 3 mediums, or 4 mediums because it says at least one. So you have to calculate the probabilities of each and add them together. So follow the same steps four times, and add.
When 1 is medium=6/27*21/26*20/25*19/24=0.1137
When 2 are medium=6/27*5/26*21/25*20/24=0.0299
When 3 are medium=6/27*5/26*4/25*21/24=0.0060
When 4 are medium=6/27*5/26*4/25*3/24=0.0008
Probability that atleast 1 is medium=0.1137+0.0299+0.006+0.0008=0.1504
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