In a basket there are 2 chocolate bunnies for 70 CZK each, 3 chocolate eggs for 50 CZK each, and 5 candies for 30 CZK each. We randomly select three items from the basket. What is the probability that we will get sweets worth 130 CZK?
Charles' solution:
(1) We can get three sweets worth 130 CZK in only two ways:
a. one bunny and two candies,
b. two eggs and one candy.
(2) We are selecting 3 items from all 10 sweets in the basket. The total number of such selections is given by the binomial coefficient $K={10\choose3}=120$.
(3) The number of selections that contain one bunny and two candies is $K_1=2\cdot{5\choose2}=20$.
(4) The number of selections that contain two eggs and one candy is $K_2={3\choose2}\cdot5=15$.
(5) The total number of favourable selections is therefore $K_1\cdot K_2$. The probability that three randomly selected sweets will cost 130 CZK is $P=\frac{K_1\cdot K_2}{K}=\frac{300}{120}=2.5$.
Did Charlse make a mistake in his reasoning? If so, determine what the correct result should look like and where the error is.
The mistake is in step (5). The total number of favourable selections is the sum of $K_1+K_2$. The probability that three randomly selected sweets will be worth 130 CZK is $P=\frac{K_1+K_2}{K}=\frac{35}{120}\cong 0.2917$.
The mistake is in step (2). The number of different selections is smaller because the same sweets repeat in the basket. The number of different triples of items is $K=9$, of which two triples meet the given criteria (1a,1b). The resulting probability is $P=\frac29\cong0.2222$.
The mistake is in step (3). Charles calculated the number of selections $K_1$, in which there is one bunny and two candies, incorerectly. $K_1=1\cdot {7\choose2}=21$. The resulting probability is $P=2.6250$.
The mistake is in step (5). The product of $K_1\cdot K_2$ needs to be divided by the product of the repeated candies. The resulting probability is $P=\frac{K_1\cdot K_2}{2\cdot 3\cdot5}\cdot\frac{1}{120}\cong0.0417$.
The solution is correct.