# Difficult understanding this probability problem - help!

#### javiersafer

##### New Member
Folks, first time posting, so first and foremost, thanks for the candid souls who may take mercy at me and help me out! I am taking an online in Business statistics and I have run into the following problem. Believe I have to use Baye's theorem and conditional probabilities but I am struggling to set it up and identify which variable is which. Anyone can suggest how to look at it from the correct perspective? Thanks

A candy store categorizes its stock as either chocolates or fruit candies. Of customers who make a purchase, they find that 65% buy chocolate and 45% buy fruit candies. They further observe that 80% of their customers that buy both chocolate and fruit candies are children, 70% who buy just fruit candies are children, and 50% who buy just chocolate are children. What percent of the stores total customers are children? Round to 3 significant figures

#### javiersafer

##### New Member
To provide some background to what I have tried doing. See the below.

#1 Events to be considered are:
a) Customer can make purchase - either chocolate or fruit candles
b) Customer may be children or adults. Either may buy chocolate or fruit candles
c) Dont think I need to consider whether a customer, regardless of age, makes a purchase or not

#2 - If the above events are accurately identified, then I think (maybe wrong) that then customers either buy chocolcate or fruit candles, so mutually exclusive events and as such the sum of the two probabilities stated at the beginning of the problem should be 100%. It is not, 65% + 45% > 100 %

# When the problem states "They further observe that 80% of their customers that buy both chocolate and fruit candies are children," it does not necessarily need to be interpreted as a customer buying two things at the same time. A child may buy chocolate one day and another day fruit candle. Not sure if this logic is right

#### Karabiner

##### TS Contributor
#1 Events to be considered are:
a) Customer can make purchase - either chocolate or fruit candles
It was not excluded that they can purchase both. The "either-or" distinction refers
to the classification of stock, not to the customers. Therefore, you have three kinds
of customers: chocolate-only, candy-only, chocolate+candy.

With kind regards

Karabiner

#### javiersafer

##### New Member
Thanks Karabiner!

So if I understand it right. My partition consists of
A1= buy both
A2= buy only chocolate
A3=buy fruit candle

Let B the event that the customer is a child, then I will have the following probabilities

P(B/A1)=80%
P(B/A2)=50%
P(B/A3)=70%

So to get P(B) I will need to apply to the rule of total probability but the probabilities that I have for P(A2) and P(A3) are then greater than 1, and again, the probability of a customer buying both products is not given, and I cannot derive it because it would be negative if I assume that P(A1) = 1 - P(A2=045%) - P(A3=65%)

I am really lost. I do not know how to look at this problem to be completely honest. Am I not seeing something that is just there in front of me?

Thanks, much appreciated!

#### Karabiner

##### TS Contributor
A1= buy both
A2= buy only chocolate
A3=buy fruit candle
the probabilities that I have for P(A2) and P(A3) are then greater than 1,
You do not yet have the probabilities for A2 and A3.
It was stated that 65% buy chocolate - but it was NOT stated that 65% are chocolate-only customers.
That means: 65% are either a chocolate-only customer (A2) or a chocolate+fruit -customer (A1).

The same with fruit: those 45% who buy fruits are either fruit-only customers or fruit+chocolate-customers.

Now you have to find out: if 45% buy fruit (either A3 or A1), and 65% buy chocolate (either A2 or A1), how large are the proportions for A1, A2, A3?

With kind regards

Karabiner

#### javiersafer

##### New Member
Hi! if i follow the logic you laid out there. I believe the probability of doing a double purchase will be 10%. So the probs I arrive at are as follows :
10%
55%
35%

If I use this prob vector and multiply it by the conditional prob vector (80%, etc), the result I arrive at is 60%, which based on the system I am using tells me that it is not the correct answer, which all weird unless I did not identify the three probs correctly

#### javiersafer

##### New Member
Hi, so I set a system of 3 equations.

P1 + P2 + P3 =1; the sum of the probability of the three disjoint events should be 1
Then, the problem statement gives me two probabilities but those probabilities are a combination of
Either buy chocolate or both
Either buy fruit or both

So the two other equations are as follows:

0.65=P1 +P3
0.45+P2 + P3

Solving for the system, I arrive at
p1, p2, p3 = 0.55,0.35,01, which in a way, in my mind, makes sense because the 10% is the part that would be shared between the two events described at the beginning of the problem.

Graphically, I also derived it by drawing two vent diagrams overlapping each other, resulting in the overlapped part being 10%

#### javiersafer

##### New Member
to clarify, my equation system consists of:

p1+p2+p3=1
0.65= p1+p3
0.45=p2+p3