1. ## lottery game question

Question. Consider the following lottery game involving two urns:
- Urn A: five balls, labelled 1 to 5
- Urn B: five balles, four green and one red.

Tickets cost GBP1 and players select to numbers between 1 and 5 (repetition not permitted). The lottery draw consists of two balls taken without replacement from Urn A, and one ball taken from Urn B. For a given ticket, there are three possible outcomes:

- numbers on ticket do not match draw from Urn A ->win GBP 0
- numbers on ticket match draw from Urn A and a green ball is drawn from Urn B ->win GBP1
- numbers on ticket match draw from Urn A and a red ball is drawn from Urn B -> win GBP (z+1).

(a) evaluate probability of each of these three outcomes.

(b) define random variable R as the return from playing the game with a single ticket; so R is our winnings minus the GBP 1 cost of the ticket. Write down the probability mass function of R and evaluate the expected return E(R). What value must z take for the expected return to be zero?

(it is a simple question but I almost always make mistakes in questions involving evaluating probabilities... could you pls check my answer).

(a) probability of each outcome

Urn A: total number of ways to pick 2 numbers out of 5 is 5(Choose)2 $\displaystyle \frac{5!}{2!(5-2)!}=10$. Only one combination is a 'match' therefore

P(match 2 numbers in Urn A)=1/10.

P(not match two numbers in A)=1-1/10-9/10= P(win GBP 0) (return is 0-1=-1)

P(match urn A AND Green ball)=1/10*4/5 (independent events)=4/50. (Return is 1-1=0)

P(match urn A AND red ball)=1/10*1/5=1/50 (Return is z+1-1=z)

(b) Return mass function shows probability of each possible random value ie

P(R=-1)=9/10
P(R=0)=4/50
P(R=z)=1/50

Through applying these probabilities to each return value and finding the sum of all products, we get E(R)=0 when z=45.

2. All seems good to me

3. Oh really? I am pleasantly surprised ))) Thanks!