document.write( "Question 1041891: A restaurant has 3 kinds of toys and is giving one toy for each
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Algebra.Com's Answer #657056 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "The first answer I gave you was incorrect.  I regret this.\r\n" );
document.write( "Please ignore it and accept my apology.  Here is the correct \r\n" );
document.write( "solution: \r\n" );
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document.write( "We will first find the probability of the complement event,\r\n" );
document.write( "where there are only 1 or 2 types of toys.  Then we will\r\n" );
document.write( "subtract from 1 to find the desired probability. \r\n" );
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document.write( "Let the 5 purchases be purchases 1,2,3,4, and 5, say in\r\n" );
document.write( "order of the earliest purchase to the latest purschase.\r\n" );
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document.write( "Case 1. 3 purchases getting one kind of toy and 2 purchases\r\n" );
document.write( "        getting another kind of toy.\r\n" );
document.write( "Case 2. 4 purchases getting one kind of toy and 1 purchase\r\n" );
document.write( "        getting another kind of toy.\r\n" );
document.write( "Case 3. 5 purchases getting the same kind of toy all 5 times.\r\n" );
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document.write( "Now we calculate the number in each of the above 3 cases.\r\n" );
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document.write( "Case 1. 3 purchases getting one kind of toy and 2 purchases\r\n" );
document.write( "        getting a second kind of toy.\r\n" );
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document.write( "Choose the 3 purchases for getting one kind of toy.\r\n" );
document.write( "That's 5 purchases choose 3 or 5C3 ways.\r\n" );
document.write( "Choose the remaining 2 purchases for getting the  \r\n" );
document.write( "second kind of toy.\r\n" );
document.write( "That's 2 purchases choose 2 or 2C2 ways.\r\n" );
document.write( "Choose the type toy for the three of one kind 3C1 ways\r\n" );
document.write( "Choose the type toy for the two of another kind in 2C1 ways   \r\n" );
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document.write( "For case 1, 5C3×2C2×3C1×2C1 = 10×1×3×2 = 60 ways\r\n" );
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document.write( "Case 2. 4 purchases getting one kind of toy and 1 purchase\r\n" );
document.write( "        getting another kind of toy\r\n" );
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document.write( "Choose the 4 purchases for getting one kind of toy.\r\n" );
document.write( "That's 5 purchases choose 4 or 5C4 ways.\r\n" );
document.write( "Choose the remaining 1 purchase for getting the  \r\n" );
document.write( "other kind of toy.\r\n" );
document.write( "That's 1 purchase choose 1 or 1C1 ways.\r\n" );
document.write( "Choose the kind of toy for the four purchases getting the \r\n" );
document.write( "same kind of toy.\r\n" );
document.write( "That's 3 toy types choose 1 or 3C1\r\n" );
document.write( "Choose the toy type for the two purchases getting the \r\n" );
document.write( "other kind of toy.\r\n" );
document.write( "That's 2 toy types choose 1 or 2C1 \r\n" );
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document.write( "For case 2, 5C4×1C1×3C1×2C1 = 5×1×3×2 = 30 ways \r\n" );
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document.write( "Case 3. 5 purchases getting the same kind of one toy \r\n" );
document.write( "in each purchase.       \r\n" );
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document.write( "That's 3 toy types choose 1, or 3C1\r\n" );
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document.write( "For case 3, 3C1 = 3\r\n" );
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document.write( "Total for all three cases: 60+30+3 = 93\r\n" );
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document.write( "So the numerator for the probability of the complement\r\n" );
document.write( "event is 93\r\n" );
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document.write( "Now we calculate the denominator for the probability.\r\n" );
document.write( "That's all the ways the toy types can come up.\r\n" );
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document.write( "Choose the toy type to get in the 1st purchase in 3 ways.\r\n" );
document.write( "Choose the toy type to get in the 2nd purchase also in 3 ways.\r\n" );
document.write( "Choose the toy type to get in the 3rd purchase also in 3 ways.\r\n" );
document.write( "Choose the toy type to get in the 4th purchase also in 3 ways.\r\n" );
document.write( "Choose the toy type to get in the 5th purchase also in 3 ways.\r\n" );
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document.write( "That's 3×3×3×3×3 = 35 = 243 ways.\r\n" );
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document.write( "So the probability of the complement event is 93/243 = 31/81.\r\n" );
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document.write( "Therefore the probability of the desired event is \r\n" );
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document.write( "1 - 31/81 = 81/81 - 31/81 = 50/81 \r\n" );
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document.write( "THAT's the correct answer.  Sorry I got it wrong the first time.\r\n" );
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document.write( "Edwin
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