document.write( "Question 1100394: In a restaurant, waiters offer 1 toy per meal. If there are 3 assorted toys available, find the probability of getting 3 assorted toys in 5 meals.\r
\n" ); document.write( "\n" ); document.write( "Choices:
\n" ); document.write( "a) 54/243
\n" ); document.write( "b) 45/253
\n" ); document.write( "c) 65/566
\n" ); document.write( "d) 25/324
\n" ); document.write( "

Algebra.Com's Answer #714973 by greenestamps(13203)\"\" \"About 
You can put this solution on YOUR website!

I'm wondering if you have shown the problem and the answer choices correctly, or if I am misinterpreting the problem.

\n" ); document.write( "The answer has to be a fraction with a denominator that is a power of 3, or is a simplified fraction that is equivalent to one with a denominator that is a power of 3.

\n" ); document.write( "Based on that, the answer, without doing any calculations, should be a, 54/243. Note, however, that that is not a good answer choice, since it is a fraction that is not in simplified form. (54/243 = 2/9).

\n" ); document.write( "But the actual calculations -- based on my interpretation of the problem -- don't show that answer.

\n" ); document.write( "My interpretation of the problem is we are to find the probability that, in 5 random meals, we get at least one of the 3 available toys.

\n" ); document.write( "Based on that interpretation, here are the calculations....

\n" ); document.write( "There are two ways to get at least 1 of each toy in 5 meals: (1) we get 3 of 1 toy and 1 of each of the others; or (2) we get 2 each of 2 toys and 1 of the other.

\n" ); document.write( "Case (1)...
\n" ); document.write( "If we call the 3 available toys A, B, and C, then we can get either
\n" ); document.write( "AAABC or ABBBC or ABCCC (3 ways)
\n" ); document.write( "For each of those 3 ways, the number of different orders to get those toys is \"%285%21%29%2F%28%283%21%29%281%21%29%281%21%29%29+=+120%2F6+=+20\"
\n" ); document.write( "So the total number of ways of getting 3 of 1 toy and 1 each of the other 2 toys in 5 meals is 3*20 = 60.

\n" ); document.write( "Case (2)...
\n" ); document.write( "For this case, we can get either
\n" ); document.write( "AABBC or AABCC or ABBCC (3 ways)
\n" ); document.write( "For each of those 3 ways, the number of different orders to get those toys is \"%285%21%29%2F%28%282%21%29%282%21%29%281%21%29%29+=+120%2F4+=+30\"
\n" ); document.write( "So the total number of ways of getting 2 each of 2 toys and 1 of the other in 5 meals is 3*30 = 90.

\n" ); document.write( "The total number of ways of getting at least 1 of each of the 3 available toys is then 60+90 = 150.

\n" ); document.write( "The total number of possible ways of getting 1 of the 3 toys in each of 5 meals is 3^5 = 243.

\n" ); document.write( "So the probability of getting at least 1 of each toy in 5 meals is 150/243 = 50/81.

\n" ); document.write( "I will be interested to see if you get solutions from any of the other tutors, with the problem interpreted so that 54/243 is the correct answer....
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