Math dice problem

Question

A four-sided dice with numbers 1-4 is tossed many times and each time the number is written on a table: a) number 1 (frequency: 18); b) number 2 (frequency: x); c) number 3 (frequency: y); d) number 4 (frequency: 22). I need to find: a) if the dice is tossed 100 times, write the equation with x and y that describes sum of all tosses b) If the average is 2,71, write another equation with x and y c) Find x and y

Answer 1

Hi Gabrielius,

The trick here is that parts (a) and (b) are connected - we can use the information in the equation of one part and apply it as information into the equation of the other part.

Question (a): "Write the equation with x and y that describes the sum of all tosses."

In case it isn't clear, the "sum of all tosses" means the sum of all dice face values from the tosses.

We can create an equation: (dice value1)x(number of times d.v.1 was tossed) + (dicevalue2)x(# d.v.2 tosses) ...

This addresses the sum of the values and the number of tosses, which we know to be 100. Therefore, we can create this equation (question (a) answer):

1(18) + 2(x) + 3(y) + 4(22) = sum of all tosses

From the number of tosses information, we also know: 18 + x + y + 22 = 100

Question (b) write the average equation of x and y

For this part, you need to know how to calculate the average (a/k/a "arithmetic mean"). This is critical to know for the SAT! The equation is: Average = (sum of all values) / (number of values)

We have this information here: the sum of all values is the sum of all tosses, and the number of values is the number of tosses we did, i.e., 100. Therefore, we have (question (b) answer):

2.71 = [1(18) + 2(x) + 3(y) + 4(22)] / 100

Question (c) Find the values of x and y

How do we solve for two different variables? As mentioned above, that is the crux of this question. We need to figure out a way to make one of the variables the same as the other - or in other words - make it equal to the other. To do that, we need:

(i) Two different equations that

(ii) include the two variables with

(iii) no other unknowns in the equation

The equation "1(18) + 2(x) + 3(y) + 4(22) = sum of all tosses" doesn't work, because we don't know what the sum of all tosses is.

The equation "18 + x + y + 22 = 100" does work because the only unknowns are x and y. We need one other equation.

The equation "2.71 = [1(18) + 2(x) + 3(y) + 4(22)] / 100" also works because the only unknowns are x and y.

From here, in one equation, we solve for one variable, and for the other equation, we plug the value of that one variable into the variable of the second equation. See below how this works (question (c) answer):

18 + x + y + 22 = 100 (doing basic algebra):

x = 100 - 18 - 22 - y

x = 60 - y (we could have also solved for y instead - it doesn't matter)

2.71 = [1(18) + 2(x) + 3(y) + 4(22)] / 100 (we plug in "60-y" into x, because they are equal, as shown in the first equation)

2.71 = [1(18) + 2(60-y) + 3(y) + 4(22)] / 100 (multiply both sides by 100)

271 = 1(18) + 2(60-y) + 3(y) + 4(22) (multiply the right side - don't forget to distribute the 2!)

271 = 18 + 120 - 2y + 3y + 88 (solve for y)

271 = -2y + 3y + 226 (adding the y's together, subtracting both sides by 226, flipping the equation for ease of reading)

y = 45

We are not done yet! We still need to solve for x. We can plug the value of y back into the average equation to solve for x, or we can plug the value back into the simpler equation of number of tosses:

18 + x + y + 22 = 100

x = 60 - y (from before)

x = 60 - (45)

x = 15

If we plug the value of x into the average equation, we get the same answer - if you don't, you did something wrong.

Hope this helps! Happy mathing,

John A. F., Esq.

Answer 2

To find the average or the mean of a frequency distribution, let's construct a table. (I'm sorry this platform is limited and does not allow me to insert a table, but I will walk you through it.)

In column 1, we will have the value of the random variable, in this problem X = 1, 2, 3, or 4, the face values of the four sided die.

In the next column we have the frequency, how many times x was rolled. We do not know the frequencies of x=2 or x=3, so we are calling those frequencies x and y.

In the third column we will have the product, f·x (frequency times the face value of the die).

The mean of the frequency distribution is the sum of the f·x column divided by the sum of the frequencies (the second column).

average = (18·1 + 2x + 3y + 22·4) / (18 + x + y + 22)

average = (18 + 2x + 2y + 88) / (18 + x + y + 22)

We know the die was rolled 100 times, so 18 + x + y + 22 = 100

We will use this information to express y in terms of x.

y = 100 - 18 - 22 - x y = 60 - x

average = (18 + 2x + 3y + 88) / 100 = (106 + 2x + 3y) / 100

And we are given the average is 2.71

2.71= (106 + 2x + 3y) / 100

substitute the equivalent expression (60 - x ) for y

2.71 = [106 + 2x +3(60-x)] / 100

271 = 106 + 2x + 180 - 3x

271 = 286 - x

x = 15

if x = 15 and y = 60 - x then y = 45

I encourage you to go back to your table, substitute these x and y values into the table, and check the calculations the mean of the distribution.