If you roll a sided die times what is the best prediction possible for the number of times you

y aces or twos, you lose the game immediately. You also lose if you draw picture cards(J,Q,K) more than twice. In this question, you’ll study the probability of winning this game.(a) What is the probability of drawing no aces or twos after thirteen draws?(b) Given you have drawn thirteen times, none of which is aces or twos, what is the probability that you draw at most two picture cards?(c) What is the probability to win this game? 12. Suppose you are tossing an unbiased coin for100times.(a) What is the probability of getting50heads and50tails?(b) LetXbe the random variable counting the number of heads you observe in this exper-iment. What is the expected value ofX? What is the variance ofX? What is thestandard deviation ofX? 13. The following are probability distributions for two random variablesX,Y. kPr(X=k) 0,0.4 1,0.3 2,0.3 kPr(Y=k) 0,0.5 1,0.3 2,0.2 (a) Construct the probability distribution table for the random variableXY.(b) Find E[X],E[Y] and E[XY]. Is is true that E[XY] =E[X]E[Y]?(c) Find the variances σ2X,σ2Y,σ2XY of X,Y and XY. Is it true that σ2XY=σ2Xσ2Y? 14. The aliens who are fond of gambling came back to play another game with you. In this game, you first toss a coin5times. If you observe3or fewer tails, you roll a die3times. If youobserve4or more tails, you roll a die20times. What is the probability that you end up with at most two6’s in your dice rolls? 15. (Challenge question, worth2points) You have two bags, each of which contains10marbles.Each time you remove a marble from a random bag. What is the probability that after one of the bags is emptied, there are still exactly3marbles in the other bag?

it sound like a math problem. The problem : What are the chances of a 4 sided die landing on 1 twice and on 2 twice out of 4 rolls. The solution I came up with originally was (2/4) x (2/4) x (2/4) x (2/4) . Which I realized was wrong as this allowed the die to land on 1 four times in a row. So then I came up with this soultion (which i still think is wrong) (2/4) x (2/4) x (1/4) x (1/4) . So the reasoning behind this is : The first roll obviously has a 50% chance to roll on either 1 or 2. Second roll is the same. BUT, lets say both of them land on 1, and now it HAS to land on 2 the remaining two times. So my problem is with the current solution that I have is what if the die lands on 1 on the first roll, then on two for the second one. then the third roll would still have a 2/4 AKA a 50% chance of landing on either one. I'm sure the last roll is 1/4 but I just dont know if the order matters on the rolls. This has been driving me crazy the last hour. Please help if you can thanks