Determine the probability that at least one of the seven eggs selected has money inside

Question

An Easter basket of 25 plastic eggs has ten eggs that have money inside. If 7 eggs are selected at random from the basket, determine the probability that atleast one of the seven eggs selected has money inside

Answer 1

So here's how the problem breaks down.  There are 25 eggs with 10 having money.  The problem asks for the probability of taking out AT LEAST one that does have money if you took out 7 eggs.  So the way to figure out the probability of that happening is if you add all the possible outcomes of taking out at least one egg.  There are 7 cases in which you take out at least one egg with money.

 

1. You take out only one egg with money: (10/25)*(15/24)*(14/23)*(13/22)*(12/21)*(11/20)*(10/19)

 

The first case in the above combination is the case where you take out one with money and the rest of the cases are the cases where the eggs do not have money.  The numbers decrease because you are not replacing the eggs in the basket after you take them out. So as you keep taking out the eggs, the denominator decreases which represents how many eggs there are left. Same for the numerator.  This decreases since you keep taking out of one of the two types.  

 

If you want to simplify this equation, you can write it as so: ( (10! / 9!) *(15!/9!)) / (25!/18!)

 

But then you have to take into account the other cases as well where you get more than one egg with money in it. This means there are six more cases.

 

2. You take out two eggs with money: (10/25)*(9/24)*(15/23)*(14/22)*(13/21)*(12/20)*(11/19)

 

By now you might see the pattern here. You continue the equations until you reach the final one where you take out 7 eggs with money in them.  The simplified equations would be as follows for each case:

 

1. ( (10! / 9!) * (15!/9!)) / (25!/18!)

 

2. ( (10! / 8!) * (15!/10!)) / (25!/18!)

 

3. ( (10! / 7!) * (15!/11!)) / (25!/18!)

 

4. ( (10! / 6!) * (15!/12!)) / (25!/18!)

 

5. ( (10! / 5!) * (15!/13!)) / (25!/18!)

 

6. ( (10! / 4!) * (15!/14!)) / (25!/18!)

 

7. ( (10! / 3!) * (15!/15!)) / (25!/18!) or ( (10! / 8!) * (1) ) / (25!/18!)

 

If you take the probabilities of each of these 7 cases and then add them together then you get the whole probability of pulling at least one egg with money in it. Sorry it's so complicated

Answer 2

Kaylee:

 

I think I have this one.  Of the 25 eggs, 10 have money inside.  So if you select 7 eggs, there would be the probability of 7/10 that at least one contained money.

 

See how that compares with other answers.

 

Larry