4. Consider a game where you roll a 20-sided die, where each side is equally lik

Question

4. Consider a game where you roll a 20-sided die, where each side is equally likely. The sides are numbered 1 through 20, and you "win if you roll a 20. (a) If you can roll 5 times, what is the probability that you win at least once? (b) Assume that one "game allows you to roll the die 7 times, and that the proba bility of a successful game (rolling 20 at least once) is 0.30. What is the expected number of games that you would need to play in order to have a successful game? (c) Using the same scenario as for (b), consider that a successful game means that you win $10. i. What are your expected winnings if you play 15 games? ii. How much would your expected winnings increase by if you could ro the die one more time per game?

Explanation / Answer

a) P(>=1 games) = 1-P(winning games=0)

where p is the probbability of success =1/20 =0.05

and q is the probability of failure =19/20 = 0.95

This is the question of the binoial theorm

So.,P(X=x) = nCx * p^x * q^(n-x)

1-P(X=0)= 1-5C0 * 0.05^0 * 0.95^20 =1-0.95^20 = 0.6415141

So., if you roll the dice 5 times the prob of winning atleast once is 0.6415141

b) number of successful games = 1/0.3 = 3.33

So in order to win min 1 game you need to play 4 games

c) (i) Now when one played 15 games the expected win = 15*0.3*10 = 45

(ii) If you would roll a die one more time then the chances of a win increases by = 0.3 (since rolling a die increases the chance of winning by 0.3)

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

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