
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?
Answer.
(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 $\displaystyle \frac{5!}{2!(52)!}=10$. Only one combination is a 'match' therefore
P(match 2 numbers in Urn A)=1/10.
P(not match two numbers in A)=11/109/10= P(win GBP 0) (return is 01=1)
P(match urn A AND Green ball)=1/10*4/5 (independent events)=4/50. (Return is 11=0)
P(match urn A AND red ball)=1/10*1/5=1/50 (Return is z+11=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.

All seems good to me (Smile)

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