Alice and Bob play the following game with a fair dice.
Alice rolls the dice six times, and wins Rs. 100 if she gets a six in at least one of the six rolls; she wins nothing in all other cases.
Bob rolls the dice twelve times, and wins Rs. 100 if he gets a six in at least two of the twelve rolls; he wins nothing in all other cases.
i. What is the expected number of sixes rolled by Alice?
ii. What is the expected number of sixes rolled by Bob?
iii. Which of the two players, if any, has a higher chance of winning something?
To answer this question, find the probability that Alice wins Rs. 100, and the probability that Bob wins Rs. 100.
(Hint: You may find it useful to find the probability of not winning Rs. 100 in either case.)
How can I solve this question?
Expected number of sixes by alice can be given as E[X]=n*p where n is trials and p is success probability so E[X] becomes 6 * 1/6=1 right?
Similarly for Bob it will be E[X]=12 * 1/6=2
For the final answer we can find the probability of Alice winning by 1-P(x=0) i.e she lands on no 6 and similarly for bob we can calculate probability of winning as 1-P(x=0)-P(x=1) since he need 2 six to win.
On comparing both values it shows Alice has a higher probability of winning. Is my approach correct?