1. ## Probability 2

A shopping mall cafe offers the following choices for its" two course meal plus beverage" package :
Choice of the three main course - Beef,Chicken or fish
Choice of two sweets - Apple-pai or Cheesecake
Followed by choice of - Tea or Coffee

Q1. List the possible equally like outcome (choices) that each costomer does have the complete package.

Q2.Determine the probability that the customer chosen at random has
(a) Chicken, Apple-pai and coffee
(b) Beef, Cheesecake and Tea
(c) Chicken and Coffee
(d) Beef but not Cheesecake
(e) Cheesecake but not Beef
(f) Fish and Apple-Pie

Sorry i'm totally stuck here...

need help
nikk

2. Originally Posted by nikk
A shopping mall cafe offers the following choices for its" two course meal plus beverage" package :
Choice of the three main course - Beef,Chicken or fish
Choice of two sweets - Apple-pai or Cheesecake
Followed by choice of - Tea or Coffee

Q1. List the possible equally like outcome (choices) that each costo,er does have the complete package.

Q2.Determine the probability that the customer chosen at random has
(a) Chicken, Apple-pai and coffee
(b) Beef, Cheesecake and Tea
(c) Chicken and Coffee
(d) Beef but not Cheesecake
(e) Cheesecake but not Beef
(f) Fish and Apple-Pie

Sorry i'm totally stuck here...

need help
nikk
Part 1:
Each of the main courses has a $1/3$ probability of being chosen. Each of the sweets has a $1/2$ chance of being chosen. And each of the drinks has a $1/2$ chance of being chosen too. So all possible combinations of meals have $\frac{1}{(2^2)(3)}$ chance of being chosen.

Part 2:
(a) Chance of getting chicken: 1/3, chance of getting apple pie: 1/2, chance of getting coffee: 1/2. So the total chance is the product of the individual probabilities.

Repeat with the rest

3. Originally Posted by Last_Singularity
Part 1:
Each of the main courses has a $1/3$ probability of being chosen. Each of the sweets has a $1/2$ chance of being chosen. And each of the drinks has a $1/2$ chance of being chosen too. So all possible combinations of meals have $\frac{1}{(2^2)(3)}$ chance of being chosen.

Part 2:
(a) Chance of getting chicken: 1/3, chance of getting apple pie: 1/2, chance of getting coffee: 1/2. So the total chance is the product of the individual probabilities.

Repeat with the rest
so , according to Last_Singularity Part 1, i will come out with tree diagram as below. is it correct what i have get??

4. so, for
Q1. List the possible equally like outcome (choices) that each costomer does have the complete package.

the outcome
S = {(Beef,Apple Pai,Tea) , (Beef,Apple Pai, coffee) ,
(Beef,Cheesecake,Tea), (Beef,Cheesecake, coffee),
(Chicken,Apple Pai,Tea) , (Chicken,Apple Pai, coffee),
(Chicken,Cheesecake,Tea), (Chicken,Cheesecake, coffee)
(Fish,Apple Pai,Tea) , (Fish,Apple Pai, coffee),
(Fish,Cheesecake,Tea), (Fish,Cheesecake, coffee)}

Q2.Determine the probability that the customer chosen at random has

(a) Chicken, Apple-pai and coffee
P(Chicken) = 1/3
P(Apple-pai) = 1/2
P(coffee) = 1/2
So, P (Chicken, Apple-pai and coffee) = 1/3 x 1/2 x 1/2 = 1/12

(b) Beef, Cheesecake and Tea
P(Beef) = 1/3
P(Cheesecake) = 1/2
P(Tea) = 1/2
So, P (Beef, Cheesecake and Tea) = 1/3 x 1/2 x 1/2 = 1/12

(c) Chicken and Coffee
P(Chicken) = 1/3
P(coffee) = 1/2
So, P (Chicken and Coffee) = 1/3 x 1/2 =1/6

(d) Beef but not Cheesecake
P(Beef) = 1/3
P (not Cheesecake) = P (Apple-pai) = 1/2

So, P ( Beef but not Cheesecake ) = P ( Beef and Apple-pai )
= 1/3 x 1/2
=1/6
(e) Cheesecake but not Beef
P(Cheesecake) = 1/2
P (not Beef) = P (Chicken or fish) = 1/3 + 1/3 = 2/3

So P( Cheesecake but not Beef) = P(Cheesecake) and P (Chicken or fish)
= 1/2 x 2/3
= 2/6
= 1/3

(f) Fish and Apple-Pie
P ( Fish) = 1/3
P ( Apple-Pie) = 1/2
So, P (Fish and Apple-Pie) = 1/3 x 1/2 = 1/6

but i'm not sure my ans.. so answer given on the books for this question. but i totally confuse on Q2(d) and Q2(e).
any one can help me to reconfirm all my ans

tq for your time and help