We roll a fair die four times. (a) Describe the sample space...

We roll a fair die four times. (a) Describe the sample space $Omega$ and the probability measure $P$ that model this experiment. To describe $P$, give the value of $P{omega}$ for each outcome $omega in Omega$. (b) Let $A$ be the event that there are at least two fives among the four rolls. Let $B$ be the event that there is at most one five among the four rolls. Find the probabilities $P(A)$ and $P(B)$ by finding the ratio of the number of favorable outcomes to the total (i.e., $P(A) = frac{#A}{#Omega}$ and $P(B) = frac{#B}{#Omega}$). (c) What is the set $A cup B$? What equality should $P(A)$ and $P(B)$ satisfy? Check that your answers to part (b) satisfy this equality.

Transcript

00:01 So in this question we roll a fair die four times.

00:03 So roll a fair die four times.

00:10 Then what is the sample space? well, omega is the set of all events.

00:18 And what is an event? an event is a set x1, x2, x3, x4, where xi is the roll on.

00:33 The ith die.

00:37 So that means that omega is just the set of all quadruples, x1, x2, x3, x4, such that xi is in the set 1, 2, 3, 4, 5, 6 for all i in the set 1, 2, 3 and 4.

00:56 So it's just the set of all quadruples where the xes are drawn from the possible scores you can have on a die.

01:04 Now what's the probability measure for this? well, all outcomes are equally likely, and we have that the size of omega is the number of quadruples from this set that you can have.

01:23 So each one of them can be one of six values.

01:26 So that's six of the power of four, and six of the power of four is 1 ,26.

01:32 So that means that the probability of any outcome omega is 1 over 1296 for all omega in our, in our sample space.

01:44 Part b, let a be the event that there are at least two fives, and b be the event that there are at most one, five.

02:02 Then the probability of a is the number of outcomes, which have result a divided by the total number of outcomes, so the size of omega.

02:14 So how many outcomes are there? so that's the number of outcomes which have two fives plus the number of outcomes with three fives plus the number of outcomes with four fives, divided by 1 ,296, because that's the size of omega.

02:30 So how many outcomes are there with two fives? well, we need to pick our two xs, which are going to be five.

02:36 That's going to be four choose two, times the number of ways that we can pick those two fives, which is just one, times the number of ways that we can pick the remaining two numbers, which is five squared.

02:49 Because we have five choices for the remaining two numbers.

02:56 So then we're going to have the number of ways of choosing three -five, so that's four choose three times one times five, plus four choose four times one times one.

03:12 These ones are telling us the number of ways of choosing the numbers of choosing the numbers for the fives, which is always one, and this is the number of ways of choosing the non -five numbers, which is going to be the remaining numbers that aren't five, of which there are five, raise the power of the number of numbers which aren't five.

03:54 So, and then we divide by 1296.

03:56 So what do we get? 4 choose 2 times 5 squared, plus 4 choose 3 times 5...

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