What Is the Probability of Choosing Banana Ice-Cream and Caramel Syrup?

[DesertFox]

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%. My answer is: 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...

 

[phinds]

Unless you have left something out of the problem statement, I'd say the book is wrong and you are right.

 

[DesertFox]

Unless you have left something out of the problem statement, I'd say the book is wrong and you are right.

A new idea came in my mind: the probability of choosing banana ice-cream "increases" after every single choice. What do I mean? I make three successive choices of ice-cream. The first time, I choose one type out of 5 different types. The second time, I choose one type out of 4 different types. The third time, I choose one type out of 3 different types of ice-cream.
So, maybe the 3/5 probability of containing banana is not correct?
The same will be valid for the probability of caramel syrup...

I hope somebody will help me...

 

[phinds]

A new idea came in my mind: the probability of choosing banana ice-cream "increases" after every single choice. What do I mean? I make three successive choices of ice-cream. The first time, I choose one type out of 5 different types. The second time, I choose one type out of 4 different types. The third time, I choose one type out of 3 different types of ice-cream.
So, maybe the 3/5 probability of containing banana is not correct?
The same will be valid for the probability of caramel syrup...

I hope somebody will help me...

No, that doesn't work. Another way to got at it is to figure out the probability of NOT GETTING that combination. That turns out to be 70%.

A simple way to do "all combinations" without actually DOING "all" combinations is this:

if you were to draw each of the 5 ice creams types without replacement and indicate whether or not the banana is the one drawn at a particular one, and then write the possible draws down, you get

00001 (banana is drawn last)
00010
00100
01000
10000 (banana is drawn first)

BUT ... you are only drawing 3 and no matter how you do that your odds of getting a "1" is 60%. The odds of the two types of syrup similarly work out to 50% and 50% of 60% is 30%. You've got it right. No need to 2nd guess yourself.

 

[DesertFox]

No, that doesn't work. Another way to got at it is to figure out the probability of NOT GETTING that combination. That turns out to be 70%.

A simple way to do "all combinations" without actually DOING "all" combinations is this:

if you were to draw each of the 5 ice creams types without replacement and indicate whether or not the banana is the one drawn at a particular one, and then write the possible draws down, you get

00001 (banana is drawn last)
00010
00100
01000
10000 (banana is drawn first)

BUT ... you are only drawing 3 and no matter how you do that your odds of getting a "1" is 60%. The odds of the two types of syrup similarly work out to 50% and 50% of 60% is 30%. You've got it right. No need to 2nd guess yourself.

I can choose banana ice-cream during my first, my second or my third choice. But the probability increases after every single choice, because I choose among less types of ice-creams.

In other words, if we take into consideration "the whole picture": for the first choice, the probability is 1/5; the probability for my second choice will be 1/4; the probability for my third choice will be 1/3. The reason is: the portion contains three different types of ice-cream.

That's why I think the overall 3/5 probability of a portion containing banana ice-cream is not valid... it must be different number...

 

[Ray Vickson]

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%. My answer is: 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 can choose banana ice-cream during my first, my second or my third choice. But the probability increases after every single choice, because I choose among less types of ice-creams.

In other words, if we take into consideration "the whole picture": for the first choice, the probability is 1/5; the probability for my second choice will be 1/4; the probability for my third choice will be 1/3. The reason is: the portion contains three different types of ice-cream.

That's why I think the overall 3/5 probability of a portion containing banana ice-cream is not valid... it must be different number...I can choose banana ice-cream during my first, my second or my third choice. But the probability increases after every single choice, because I choose among less types of ice-creams.

In other words, if we take into consideration "the whole picture": for the first choice, the probability is 1/5; the probability for my second choice will be 1/4; the probability for my third choice will be 1/3. The reason is: the portion contains three different types of ice-cream.

That's why I think the overall 3/5 probability of a portion containing banana ice-cream is not valid... it must be different number...

No. The answer is 3/5. Letting B = "banana", we have
$$ \begin{array}{rclr}
P(\text{B first}) &=& \displaystyle \frac{1}{5}& \hspace{4ex} (1) \\
P(\text{B second}) &=& \displaystyle \frac{4}{5} \cdot \frac{1}{4} = \frac{1}{5} & \hspace{4ex} (2) \\
P(\text{B third}) &=& \displaystyle \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{5} & \hspace{4ex} (3)
\end{array}
$$
Equation (1) is true because there is a 1 in 5 chance of picking B first and then it does not matter what happens next. Equation (2) is true because there is a 4 in 5 chance of picking a non-B first; after that we are left with 4 colors, one of which is B, so the probability of picking B next is 1/4. Altogether, the chance of having B second is (4/5)(1/4). Finally, equation (3) is true because there is a 4 in 5 chance of picking a non-B first, then a 3 in 4 chance of picking another non-B next, leaving 3 colors, of which B is one of them.

Adding the different probabilities gives P(B) = P(B first) + P(B second) + P(B third) = (1/5) + (1/5) + (1/5) = 3/5.

Another argument: the number of combinations of 3 things chosen from 5 is the binomial coefficient $C(5,3) = 5!/[3! 2!] = 10.$ That means that there are 10 different 3-flavor cones possible (where the order does not matter). The number of different cones that includes B is the binomial coefficient $C(4,2) = 4!/[2! 2!] = 6$, because after picking B we need to pick two more from the remaining four. Therefore,
$$ P(B) = \frac{6}{10} = \frac{3}{5}.$$

 

[DesertFox]

No. The answer is 3/5. Letting B = "banana", we have
$$ \begin{array}{rclr}
P(\text{B first}) &=& \displaystyle \frac{1}{5}& \hspace{4ex} (1) \\
P(\text{B second}) &=& \displaystyle \frac{4}{5} \cdot \frac{1}{4} = \frac{1}{5} & \hspace{4ex} (2) \\
P(\text{B third}) &=& \displaystyle \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{5} & \hspace{4ex} (3)
\end{array}
$$
Equation (1) is true because there is a 1 in 5 chance of picking B first and then it does not matter what happens next. Equation (2) is true because there is a 4 in 5 chance of picking a non-B first; after that we are left with 4 colors, one of which is B, so the probability of picking B next is 1/4. Altogether, the chance of having B second is (4/5)(1/4). Finally, equation (3) is true because there is a 4 in 5 chance of picking a non-B first, then a 3 in 4 chance of picking another non-B next, leaving 3 colors, of which B is one of them.

Adding the different probabilities gives P(B) = P(B first) + P(B second) + P(B third) = (1/5) + (1/5) + (1/5) = 3/5.

Another argument: the number of combinations of 3 things chosen from 5 is the binomial coefficient $C(5,3) = 5!/[3! 2!] = 10.$ That means that there are 10 different 3-flavor cones possible (where the order does not matter). The number of different cones that includes B is the binomial coefficient $C(4,2) = 4!/[2! 2!] = 6$, because after picking B we need to pick two more from the remaining four. Therefore,
$$ P(B) = \frac{6}{10} = \frac{3}{5}.$$

What about the overall answer of the problem? You say 30%?

 

[Ray Vickson]

What about the overall answer of the problem? You say 30%?

No, I did not say it when I replied, but I do say it now: the correct answer is 3/10.

 

[phinds]

That's why I think the overall 3/5 probability of a portion containing banana ice-cream is not valid... it must be different number...

You REALLY insist on making this harder than it is.

 

[llober]

 

[DesertFox]

View attachment 209354

I can't understand it. :( :( :(
I see your answer is 29.99962%
But I can't grasp the solution itself...
Could you be more explaining, please?

 

[llober]

This Python code is a "Montecarlo" simulation. It does 10.000.000 simulations of the case of getting the icecream and syrup.

The list ice is initialized with 5 different numbers, which represent the 5 different flavours.

The list syr is initialized with 4 different numbers, which represents the 4 different syrups.

The ice.pop(...) instruction, deletes one random item the from the list... so if we begin with the list containing 5 numbers (flavours), and we ice.pop 2 numbers, we end having a list with 3 numbers left (the 3 icecream flavours), because 2 are removed.

The same goes for the syr list.

At the end, if the ice list contains the number (index) of banana, and if the syr list contains the number (index) of caramel, then we account for one more hit, which means that randomly we got to a case in which we have a banana flavour with caramel.

We run this 10.000.000 times ... instead of you doing this tedious work manually, is a Python loop.

The result is 30% (you would have to run an infinite loop to have the exact number).

 

[Ray Vickson]

I can't understand it. :( :( :(
I see your answer is 29.99962%
But I can't grasp the solution itself...
Could you be more explaining, please?

Just for the sake of interest, here is a similar simulation performed using Maple, but drawing only 10,000 ice-cream cones instead of 10,000,000. In the worksheet, B = number containing Banana ice-cream, C = number containing Caramel topping and BC = number containing both.

 

[phinds]

I can't understand it.

You should learn about Monte Carlo simulations. They can be very useful.