In one series of Kinder Surprise chocolate eggs, sold at airports, each egg contains a toy Airbus plane, in one of five liveries. If the toys are distributed randomly among the eggs, what is the expected number of eggs you must buy to get a complete set?

Bolaji Atanda and Mohaned Al-Bassam got completely correct solutions.

This is Bolaji’s solution:

The amount of tries needed to meet a objective *[he means the expected (average) number of tries needed]* when the probability is x/n is n/x.

So if there was a probability of 1/5 you will need 5 tries to meet an objective. Just like how if you had a probability of 5/5 you would need 1

try.

So on the first result there was a 5/5 chance of getting a unique livery then it becomes 4/5 then 3/5 then 2/5 then 1/5.

So on the first go it will take 1 try to get a unique one.

On the 2nd go there is a probability of 4/5 so it will take 5/4 tries

On the 3rd go there is a probability of 3/5 so it will take 5/3 tries to get a unique one.

On the 4th go there is a probability of 2/5 so it will take 5/2 tries to get a unique one

On the 5th go there is a probability of 1/5 so it will take 5/1 tries to get a unique one.

So to know the amount of tries needed I would just have to add up 5/5 + 5/4 + 5/3 + 5/2 + 5/1

This amounts to 11.416

(Notice that the answer is a *fraction* although it is a mean average of *whole numbers*.)

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