Question

There is a game called Bishops bank where on the table had numbers 1 through 25 painted on it. Players place a dime on a number they choose and a fair gaming whell is spun. If the wheel selects your number, you get to select a card from a bundle of 100 card. The card will reveal how much cash you win($20.00, $10.00, $5.00, $1.00, or $0.50) with corresponding probabilities (0.01, 0.02, 0.10, 0.20, 0.67). If your number fails to come up you lose your bet. Assume all wheel spins and bets are independent.

(a) Thirty bets are placed. What is the probability that exactly three bets will win on the next spin of the wheel?

(b) On the next spin of the wheel, thirty bets are placed and you are the only winner(you made one bet). What will be the average intake/pay-out for this concession given these circumstances?

Answer 1

(a) Thirty bets are placed. What is the probability that exactly three bets will win on the next spin of the wheel?

probability of win =p= 1/25

Now we want to find the probability of exactly 3 bet wins when 30 bets are placed

So we need to use the binomial theorm

P(X=x) = nCx * p^x *(1-p)^(1-x)

here for us

P(X=3) = 30C3 * (1/25)^3 * (1-1/25)^(1-3)

=0.2819444

So., we say that 0.2819444 is the probability that exactly 3 bets out of 30 bets will win in the next spin of the wheel.

(b) Now when you are the only winner the payout is nothing but the sum product of the payout with their respective probabilities

So Payout= 20*0.01 + 10*0.02+5*0.1+1*0.2+0.5*0.67= 1.435$

Total intake = 30*0.1 = 3$

So.,  the average intake/pay-out for this concession given these circumstances

=3/1.435 =2.090592

Hope the above answer has helped you in understanding the problem. Please upvote the ans if it has really helped you. Good Luck!!