# Tutor profile: Charis G.

## Questions

### Subject: Discrete Math

Roll five 6-sided dice. (1) What is the probability that at least one of them is a 5 or 6? (2) What is the probability that exactly two of them are a 3? (3) If the first three dice are 6,6,2, what is the conditional probability that all five dice sum to 20?

For this solution we will utilize counting methods, but the problem can also be solved using the binomial distribution. Before starting the problem we can remember that, in a sample space with all equally probably outcomes, the probability of an event is the number of ways we can get that event divided by the total number of outcomes. So while the numerator will differ across these three questions, the denominator will always be the total number of outcomes when rolling 5 dice. Let's calculate that now by multiplying the number of outcomes of each die: $$6 \cdot 6 \cdot 6 \cdot 6 \cdot 6 \cdot 6 = 7776$$ total outcomes. (1) What is the probability that at least one of the 5 rolls is a 5 or a 6? The key words here are "at least one" and "or". Thus, the number of ways we can get this event will span all of the outcomes including one 5 only, three 5s and one 6, two 6s only, and so on. Instead of having to account for the many different types of outcomes (and their potential combinations) that could result in the event of at least one 5 or one 6, we can just solve for its complement: the number of outcomes that do NOT include any 5s or 6s. This is far easier. Assuming we can have no 5s or 6s, the possible number of outcomes would be $$4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 = 1024$$. If there are 7776 total outcomes and 1024 of them have no 5s or 6s, we know that $$7776-1024=6752 $$ outcomes have at least one 5 or 6. Thus the probability of rolling at least one 5 or 6 is $$\frac{6752}{7776}$$. (2) What is the probability that exactly two of them are a 3? First recognize that the two threes could come in a variety of orders. We must determine the number of ways that exactly two dice could land on 3s. $$\left( \begin{array}{c} 5 \\ 2 \end{array} \right) = 10$$ Now for each of those 10 ways we must determine the possible outcomes based on the other three dice. That is easily computed as $$5 \cdot 5 \cdot 5=216$$. (Remember that none of the other three dice can be 3s - the problem states that there must be EXACTLY two 3s.) So our total number of outcomes resulting in the event that two 3s are rolled comes to 640, and the probability of this happening is $$\frac{1250}{7776}$$. (3) Since the first three dice have already been given to us in order, the only part of the roll that involves any probability is the last two dice. This means that, despite what I said at the beginning, our denominator will NOT be the total number of outcomes when rolling five dice. It will be the total number of outcomes when rolling two dice, which is simply $$6 \codt 6 = 36$$. Since the first three dice sum to 14, in order for all five dice to sum to 20 the last two dice must sum to 6. This is relatively easy to count - possible combinations that would give us a sum of 6 are (1,5), (5,1), (2,4), (3,4), (3,3), and (3,3). (Notice that the order here is significant. If we imagine one dice to be green and the other to be red, we will notice that there really are two possible outcomes for each combination.) Thus the probability of all five dice summing to 20 is $$\frac{6}{36}$$.

### Subject: Computer Science (General)

Step through the following psuedocode and give its output. a = 2 b = 3 c = 0 k = 2 if (a % 2 == 0) if (b % 4 == 0) c = 5 else c = 7 end if else if (a % 5 == 0) c = 2 else c = 10 end if while (k < c) print("A") k++ end while

a = 2 b = 3 c = 0 k = 2 if (a % 2 == 0) a=2 and 2%2=0, so this evaluates to T if (b % 4 == 0) b=3 and 3%4=3, so this evaluates to F and the block will not execute c = 5 (will not exectue) else the condition was F so this block WILL execute c = 7 c = 7 end if else if (a % 5 == 0) our original if block was T, so this does not execute c = 2 else our original if block was T, so this does not execute c = 10 end if while (k < c) c = 7, so iterating through this block from 2 to 7 will be 5 times print("A") A will print 5 times k++ end while Our output is AAAAA.

### Subject: Statistics

Let X be a random variable whose image $$X(S)$$ is contained in the set $${1, 2, . . . , n}$$. Show that $$E(X)=\sum_{k=1}^{n} p(X\geq k)$$.

OK, so we have a random variable $$X$$ with a sample space of $$\{1,2,...n\}$$, and we're trying to prove that its expected value is equal to $$\sum_{k=1}^{n} p(X\geq k)$$. In general, we know that $$E(X)=\sum_{k=1}^{n} k\cdot p(X=k)$$. In order to simplify this a bit, we can just note that since here the probability of each outcome is equal, we have $$p=\frac{1}{n} $$ for every $$k$$. This simplifies our expected value to $$E(X)=\sum_{k=1}^{n} \frac{k}{n}$$. Now let's take a look at what we're trying to prove. It looks relatively similar, but we're using a different probability: $$P(X\geq k)$$. Since it might be a bit difficult to immediately understand what that means, let's look at an example. Take the case where $$n=3$$: $$E(X)=\sum_{k=1}^{n} p(X\geq k) = \frac{3}{3} + \frac{2}{3} + \frac{1}{3}$$ Now we can see that this probability can also be written in a much simpler way: $$E(X)=\sum_{k=1}^{n} \frac{n+1-k}{n}$$ Now that we have simplified what we know and what we're trying to prove, let's compare them. Since $$n$$ is constant, we'll also pull the $$\frac{1}{n}$$ to the outside of each sum. We're left with $$\frac{1}{n}\sum_{k=1}^{n} k$$ and $$\frac{1}{n}\sum_{k=1}^{n} n+1-k$$ These two sums can also be written as $$\{1,2,3,...n-1,n\}$$ and $$\{n,n-1...3,2,1\}$$, which are, of course the same. Thus we have $$E(X)=\sum_{k=1}^{n} k\cdot p(X=k)=\sum_{k=1}^{n} p(X\geq k)$$.

## Contact tutor

needs and Charis will reply soon.