# Ice-cream problem: banana ice-cream and caramel syrup

#### Desert Fox

##### New member
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We have 5 types of ice-cream: vanilla, banana, coconut, chocolate, forest fruits. And we have 4 types of syrup: strawberry, raspberry, blueberry, caramel. A person buys a portion of ice-cream (we are considering random choice here!) . The portion contains three different types of ice-cream and two different types of syrup.
What is the probability (expressed in percents) of a portion containing banana ice-cream and caramel syrup?

I am in big struggle to find the solution. I need a clear, stand-by-step solution.... and a remark of the required formulas for the solution.

Here it is my attempt, please, correct me if i am wrong:
For ice-cream: he can choose 3 different types out of 5. So, there is 3/5 probability of containing banana ice-cream.
For syrup: he can choose 2 different types out of 4. So, there is 2/4 probability of containing caramel syrup.
Hence, the probability of a portion containing banana ice-cream and caramel syrup is: 3/5 multiplied by 2/4 = 6/20, which is exactly 30%.

But in my textbook (where I have the answer, but I don't have the solution) the given answer is: 15%

Where I am wrong??? Please, help....
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#### Subhotosh Khan

##### Super Moderator
Staff member
We have 5 types of ice-cream: vanilla, banana, coconut, chocolate, forest fruits. And we have 4 types of syrup: strawberry, raspberry, blueberry, caramel. A person buys a portion of ice-cream (we are considering random choice here!) . The portion contains three different types of ice-cream and two different types of syrup.
What is the probability (expressed in percents) of a portion containing banana ice-cream and caramel syrup?

I am in big struggle to find the solution. I need a clear, stand-by-step solution.... and a remark of the required formulas for the solution.

Here it is my attempt, please, correct me if i am wrong:
For ice-cream: he can choose 3 different types out of 5. So, there is 3/5 probability of containing banana ice-cream.
For syrup: he can choose 2 different types out of 4. So, there is 2/4 probability of containing caramel syrup.
Hence, the probability of a portion containing banana ice-cream and caramel syrup is: 3/5 multiplied by 2/4 = 6/20, which is exactly 30%.

But in my textbook (where I have the answer, but I don't have the solution) the given answer is: 15%

Where I am wrong??? Please, help....​
I don't quite understand - what does a portion mean in this context?

#### Toasted

##### New member
I don't quite understand - what does a portion mean in this context?
I think in this case, it just means 3 different ice-creams and 2 different syrups, no more no less, and no doubles of ice-cream or syrups. Portion is just what they are calling that amount of food

#### stapel

##### Super Moderator
Staff member
I think "portion" is intended to mean "serving", with each serving containing three scoops.

#### mmm4444bot

##### Super Moderator
Staff member
Are you reading the question similar to having all possible combinations available, and a machine dispenses one at random?

If so, then I agree that there's a 30% chance the machine will dispense something containing both banana and caramel flavors. :cool:

#### Desert Fox

##### New member
I don't quite understand - what does a portion mean in this context?
It means 3 different ice-creams and 2 different syrups. In the context every portion consists of three different ice-creams and two different syrups.
Ice-creams and syrups are chosen at random....

#### Jomo

##### Elite Member
We have 5 types of ice-cream: vanilla, banana, coconut, chocolate, forest fruits. And we have 4 types of syrup: strawberry, raspberry, blueberry, caramel. A person buys a portion of ice-cream (we are considering random choice here!) . The portion contains three different types of ice-cream and two different types of syrup.
What is the probability (expressed in percents) of a portion containing banana ice-cream and caramel syrup?

I am in big struggle to find the solution. I need a clear, stand-by-step solution.... and a remark of the required formulas for the solution.

Here it is my attempt, please, correct me if i am wrong:
For ice-cream: he can choose 3 different types out of 5. So, there is 3/5 probability of containing banana ice-cream.​

It is not 3/5!

In how many ways can you choose 3 ice creams from 5? Answer 5C3
In how many ways can the 3 chosen ice creams include banana is cream? Answer comes from choosing 2 ice creams from 4 (not including banana ice cream). To these 2 non banana ice cream we WILL add banana ice cream so in the end we do have 3 ice cream flavors including banana. This can be done in 4C2 ways.

So the probability that one chooses 3 ice creams including banana ice creams is (#of favorable outcomes)/(# of total outcomes) = 4C2/5C3

Whenever you try to compute a (simple) probability and you get the wrong result make sure you are computing
(#of favorable outcomes)/(# of total outcomes)

If you think the answer is 3/5, then you should be able to list the 5 total outcomes and the 3 favorable outcomes!!!!!!!!

You would get 3/5 under this situation. You have a 5 sided die, numbered 1-5, where each side is the same size and you are asked to find the probability of rolling the die and getting an odd number. Since there are 3 odd numbers (1, 3, 5) and 5 possible results (1, 2, 3, 4 or 5) then the result is 3/5. NOTE that you chose one number (you tossed the die once) NOT three like when choosing 3 different ice creams!

#### Jomo

##### Elite Member
It means 3 different ice-creams and 2 different syrups. In the context every portion consists of three different ice-creams and two different syrups.
Ice-creams and syrups are chosen at random....
Why not list all possible combinations of 3 ice creams and 2 syrups (call this number D). Then count the ones that have banana ice cream and caramel syrup. (Call this number N). Then your answer will simply be N/D which you should convert to %.