# Homework Help: Probability Problem- im still stuck

1. Oct 15, 2006

### ruby_duby

Probability Problem- im still stuck !!

Hi i need help on the following question:

In a teashop 70% of customers order tea with milk, 20% tea with lemon and 10% with neither. of those taking tea with milk 3/5 take sugar, of those taking tea with lemon 1/4 take sugar, and of those taking tea with neither milk or lemon 11/20 take sugar. a customer is then chosem at random.

i need to find the probability that the customers take suggar or milk or both. i know that the answer is o.525 but i dont know how.

2. Oct 15, 2006

### JasonRox

Try writing them in term of sets.

Have you done this before?

Let A represent the customers that order tea with milk.
Let B represent the customers that order tea with lemon.
Let C represent... and so on.

Then with the question you should write the answer in terms of sets. Then use the rules that follow to find the solution.

Not sure if you've done this before though.

3. Oct 15, 2006

### quasar987

This is what I get...

Define the events

E: he takes suggar
F: he takes milk
G:he takes lemon
H:he takes neither milk nor lemon

Then what you're looking for is

$$P(E\cup F\cup EF)=P(E)+P(F)+P(EF)-P(EF)-P(EFE)-P(FEF)+P(EFEF) = P(E)+P(F)-P(EF)$$

You know that P(F)=0.7, P(G)=0.2, P(H)=0.1, P(E|F)=3/5, P(E|G)=1/4, P(E|H)=11/20.

Since {F,G,H} forms a partition of the fundamental set, P(E) = P(E|F)P(F)+P(E|G)P(G)+P(E|H)P(H) = 3/5(0.7)+1/4(0.2)+11/20(0.1)=21/40=0.525

P(EF)=P(F)P(E|F)=3/5(0.7)=21/50=0.42

So,

$$P(E\cup F\cup EF)=0.525+0.7-0.42=0.805$$

Note that 0.525 is only the probability that he takes suggar.