SOLUTION: 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.
Choices:
a) 54/243
b) 45/253
Algebra.Com
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.
Choices:
a) 54/243
b) 45/253
c) 65/566
d) 25/324
Answer by greenestamps(13203) (Show Source): 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.
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.
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).
But the actual calculations -- based on my interpretation of the problem -- don't show that answer.
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.
Based on that interpretation, here are the calculations....
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.
Case (1)...
If we call the 3 available toys A, B, and C, then we can get either
AAABC or ABBBC or ABCCC (3 ways)
For each of those 3 ways, the number of different orders to get those toys is
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.
Case (2)...
For this case, we can get either
AABBC or AABCC or ABBCC (3 ways)
For each of those 3 ways, the number of different orders to get those toys is
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.
The total number of ways of getting at least 1 of each of the 3 available toys is then 60+90 = 150.
The total number of possible ways of getting 1 of the 3 toys in each of 5 meals is 3^5 = 243.
So the probability of getting at least 1 of each toy in 5 meals is 150/243 = 50/81.
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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