1a casino wants to introduce a new game in this game a...

1)A casino wants to introduce a new game. In this game a player rolls two 4-sided dice and their winnings are determined by the following rules: Otherwise, the player wins nothing. If 3 ≤ sum ≤ 5 the player wins $25. If sum ≥ 6 the player wins $100. a)Determine the probability distribution for this gamble. Probability Distribution X P (X) b)How much should the casino charge for this game if they want to make a profit in the long run? Show the calculations that support your decision c)What is the standard error, σX, for this gamble Complete the probability distribution for X below. Distribution of X X P (X = k) P (X ≤ k) 0 1 0.4219 2 3 0.0469 4 0.0039 Compute: (i) P (X ≤ 3) and (ii) P (X < 3) Compute: P (X > 1) Compute: P (X is odd) Compute the expected value and variance of X.

Transcript

00:01 So in this question we have two four -sided dice and we collect the sum.

00:07 So x is the sum of results of two fair four -sided dice.

00:22 And we have that if x is between three and five inclusive, then we win 25.

00:32 If x is greater than or equal to six, then we win 100.

00:39 So let's get the probability distribution.

00:43 So what we can do is we can look at our sample space, which is the set of results of rolling two fair four -sided dice.

00:53 So we can have 1 -1 -1 -2, 1 -3, 1 -4.

01:00 I'll just list out the results.

01:03 2 -1 -2 -2 -2 -3 -4 3 -1 -3 -2 -3 -3 -3 -3 -4 and finally 4 -1 -4 -2 -4 -4 and each of these is equally probable they're 16 entries so the probability of some element of our sample space is just 1 -16th because they're all equally likely so now x is the sum here.

01:43 So let's write down x of this sample space is defined by x element of x of omega is a plus b for each omega equals a b in omega.

02:11 So we can write that down.

02:15 So for one and one we get two, three, four, five, and then three, four, five, six, and then four, five, six, seven, five, six, seven, eight.

02:33 And what we can do is we can count each of these things, and that's going to give us our probabilities.

02:39 So x can take values two, three, four, five, six, seven, or eight.

02:44 And the probability is just one -sixteenth times the number of times those numbers appear in this set x of omega.

03:00 So for two, there's only one, so it's one -sixteenth.

03:04 For three, we have one -two, so that's one -eighth.

03:10 For four, we have one, two, three, so that's three -sixteenth.

03:15 Five appears one, two, three, four times.

03:19 So that becomes a quarter.

03:24 6 appears 1 to 3 times.