Probability with two dice problem

Homework Statement

If we throw with two dice a and b, than we make two spaces call them space A and B.
Space A = a+b and space B = a*b. Next the question there are a couple of questions:
1) What is the probability P(A > 7)
2) What is the probability P(B = odd)
3) What is the probability P( A n B)
4) What is the probability P(A u B)

My problem is question 3, I know how to get the answers on question 1 and 2 and question 4 is applying Bayes Rule. Could anybody explain me how to get the answer on question 3?

Homework Equations

I placed the tables you get for A and B offcourse A is the first table and B is the second table. Below there are the answers.

The Attempt at a Solution

Because we got P(A >7) and P(B = odd).

Now when i try to get P(A n B) = P(A) * P(B) = (5/12) * (1/4) = 5/48, but the answer above in the picture states the answer is 1/12. I dont understand how to get there. Even when I count all the numbers >7 and are Odd, i dont come on 1/12. I really woul like to understand what i am doing wrong.

I appreciate all the help, thanks in advance

Ray

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andrewkirk
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Each position in the 6 x 6 tables corresponds a possible pair of numbers from a double-roll. What you need to count are the number of positions that :
• contain a number greater than 7 in the first table; and
• contain an odd number in the second table.
There are three such positions, giving a probability of 3/36=1/12.

Well there i do something wrong which is the key to understanding this problem!

I count: 9, 11, 15, 25

Or do you mean the following positions:

15, 15, 25?

FactChecker
Gold Member

Homework Statement

If we throw with two dies a and b, than we make two spaces call them space A and B.
Space A = a+b and space B = a*b. Next the question there are a couple of questions:
1) What is the probability P(A > 7)
2) What is the probability P(B = odd)
3) What is the probability P( A n B)
4) What is the probability P(A u B)

My problem is question 3, I know how to get the answers on question 1 and 2 and question 4 is applying Bayes Rule.
I don't think that Bayes' Rule is used in 4. You may be over-thinking it.
Could anybody explain me how to get the answer on question 3?

Homework Equations

I placed the tables you get for A and B offcourse A is the first table and B is the second table. Below there are the answers.
View attachment 207849

The Attempt at a Solution

Because we got P(A >7) and P(B = odd).

Now when i try to get P(A n B) = P(A) * P(B) = (5/12) * (1/4) = 5/48,
This is only true if A and B are independent. They are not independent.
but the answer above in the picture states the answer is 1/12. I dont understand how to get there. Even when I count all the numbers >7 and are Odd, i dont come on 1/12.
I see 3. -- (5,3), (3,5), and (5,5). That gives their result.

ok so it are the numbers 15, 15, 25.

The thing i dont understand is that i dont see these numbers in the first table. But it is P(A n B), this confuses me a lot.

At the fourth question question the apply this rule:

P( A u B) = P(A) + P(B) - P(A n B) True is this not Bayes Rule

I don't think that Bayes' Rule is used in 4. You may be over-thinking it.
This is only true if A and B are independent. They are not independent.
I see 3. -- (5,3), (3,5), and (5,5). That gives their result.
Sorry for the silly questions but why are they not independent and how can one see that?

andrewkirk
Homework Helper
Gold Member
Well there i do something wrong which is the key to understanding this problem!

I count: 9, 11, 15, 25

Or do you mean the following positions:

15, 15, 25?
I mean the second one, except that those are not positions, but the values in the relevant positions in the second table. The positions are given by FactChecker in the last line of their post.

SammyS
Staff Emeritus
Homework Helper
Gold Member

Homework Statement

If we throw with two dies a and b, than we make two spaces call them space A and B.
Space A = a+b and space B = a*b. Next the question there are a couple of questions:
1) What is the probability P(A > 7)
2) What is the probability P(B = odd)
3) What is the probability P( A n B)
4) What is the probability P(A u B)

My problem is question 3, I know how to get the answers on question 1 and 2 and question 4 is applying Bayes Rule. Could anybody explain me how to get the answer on question 3?

Homework Equations

I placed the tables you get for A and B offcourse A is the first table and B is the second table. Below there are the answers.
View attachment 207849

The Attempt at a Solution

Because we got P(A >7) and P(B = odd).

Now when i try to get P(A n B) = P(A) * P(B) = (5/12) * (1/4) = 5/48, but the answer above in the picture states the answer is 1/12. I dont understand how to get there. Even when I count all the numbers >7 and are Odd, i dont come on 1/12. I really woul like to understand what i am doing wrong.

I appreciate all the help, thanks in advance

Ray
#3 as it is given in the solution is:
What is the probability that when you throw a pair of dice, the result is that the sum is greater than seven and the product is odd?

I mean the second one, except that those are not positions, but the values in the relevant positions in the second table. The positions are given by FactChecker in the last line of their post.
Thanks i understand. But i dont understand why 9 is not part of the answer? It is >7 and its odd.

andrewkirk
Homework Helper
Gold Member
Thanks i understand. But i dont understand why 9 is not part of the answer? It is >7 and its odd.
The 9 to which you refer is a sum of the numbers from the two dice, which is obtained from the four result pairs (3,6), (4,5), (5,4), (6,3). It is the products of the two numbers that are required to be odd, not the sums. None of the products from those pairs are odd. The products are 18, 20, 20, 18.

Imagine the following process. Place a rectangular piece of paper over table 1 and cut out a small hole above every number greater than 7, so we can see the number through the hole.

Now do the same with table 2 but only cut out holes above odd numbers.

Now put the two pieces of paper on top of one another, so that they are exactly aligned along the edges, and put that on top of a surface that has a different colour from the paper (eg a blue carpet, if the paper is white). Wherever the holes in the two sheets coincide, you will see a patch of the surface underneath. Count the number of patches. That is the number of elements of the intersection of the two sets.

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FactChecker
Gold Member
Thanks i understand. But i dont understand why 9 is not part of the answer? It is >7 and its odd.
There is no combination that sums to 9 whose product is odd. Only one of the sum a+b=9 can be odd, so the other is even. Therefore their product is even.

Ahhhhh offcourse im only looking to the results not to the combinations. Im looking completely wrong :) thanks guys i think i understand it.

@ Andrewkirk you made a very nice explanation!

Hmmm this subject is not so easy IMO, I had an exam a couple of weeks ago but failed badly its a pitty because i like this subject. So now im trying searchong all over internet for extra exercises and found this one. I think i get this one so ready for the next one.

Again thanks a lot!

Ray Vickson
Homework Helper
Dearly Missed
Thanks i understand. But i dont understand why 9 is not part of the answer? It is >7 and its odd.
Ahhhhh offcourse im only looking to the results not to the combinations. Im looking completely wrong :) thanks guys i think i understand it.

@ Andrewkirk you made a very nice explanation!

Hmmm this subject is not so easy IMO, I had an exam a couple of weeks ago but failed badly its a pitty because i like this subject. So now im trying searchong all over internet for extra exercises and found this one. I think i get this one so ready for the next one.

Again thanks a lot!
Another way of finding $P(A \cap B)$ is to use "conditioning": $P(A \cap B) = \sum_i P(A \cap B | D_1 = i) P(D_1 = i)$. Here, $D_1$ is the outcome on Die #1. It is not hard to understand what this means: we must have that $D_1$ is one of the numbers 1, 2, 3, 4, 5, 6, and for any such outcome the event $A \cap B$ will then have a probability conditioned on the assumed value of $D_1$.

(1) If $D_1 = 1$ the event $A$ cannot occur, because we can reach a sum of 7 but not exceed it: $P(A \cap B|D_1=1)=0$.
(2) If $D_1 =2$ the product $D_1 \times D_2$ is even so $B$ cannot happen: $P(A \cap B|D_1=2)=0$.
(3) If $D_1 = 3$ we must have $D_2 \in \{5,6\}$, and only the combination $(D_1,D_2) = (3,5)$ is in $A \cap B$: $P(A \cap B|D_1=3)=1/6$.
(4) If $D_1 = 4$, the product $D_1 \times D_2$ is even, so $B$ does not occur: $P(A \cap B|D_1=4)=0$.
(5) If $D_1 = 5$, $A$ needs $D_2 \in \{3,4,5,6\}.$ Only $D_2 = 3,5$ give an odd product: $P(A \cap B|D_1=5)=2/6$.
(6) If $D_1 = 6$ the product $D_1 \times D_2$ is even, so $B$ does not occur: $P(A \cap B|D_1=6)=0$.
Of course, we could cut out half the work and writing just by noting that if $D_1$ is even the product $D_1 \times D_2$ must also be even, so $P(A \cap B | D_1 \:\text{even} \} = 0.$

So,altogether we have
$$P(A \cap B) = \frac{1}{6} \left( 0 +0+\frac{1}{6}+0+\frac{2}{6}+0 \right) = \frac{3}{36} = \frac{1}{12}.$$

Anyway, if we have A=8

haruspex
Homework Helper
Gold Member
Space A = a+b and space B = a*b. Next the question there are a couple of questions:
1) What is the probability P(A > 7)
2) What is the probability P(B = odd)
3) What is the probability P( A n B)
4) What is the probability P(A u B)
Am I the only one who regards the problem statement as severely flawed?
A and B are introduced as "spaces", e.g. A being the event space consisting of the possible values of a+b and their associated probabilities.
In q1 and q2, it seems that A and B are now random variables, being defined respectively as the sum and product of the dice results.
But q3 and q4 appear to redefine A and B as events. That those should be the events "a+b >7" and "a*b is odd" is pure guesswork.

Maybe something was lost in translation, but it does suggest a confusion of concepts on someone's part.

Am I the only one who regards the problem statement as severely flawed?
A and B are introduced as "spaces", e.g. A being the event space consisting of the possible values of a+b and their associated probabilities.
In q1 and q2, it seems that A and B are now random variables, being defined respectively as the sum and product of the dice results.
But q3 and q4 appear to redefine A and B as events. That those should be the events "a+b >7" and "a*b is odd" is pure guesswork.

Maybe something was lost in translation, but it does suggest a confusion of concepts on someone's part.
Im not native speaking english as you have seen yes. But how could i make the translation wrong. Question 3 is written in dutch:

Hoe groot is de kans dat A is groter dan 7 en B is oneven?

In english this is as far as I know:

What is the probability that A is great than 7 and B is odd.

By writing the tables it is not guess work anymore? Or do i miss something here?

Another way of finding $P(A \cap B)$ is to use "conditioning": $P(A \cap B) = \sum_i P(A \cap B | D_1 = i) P(D_1 = i)$. Here, $D_1$ is the outcome on Die #1. It is not hard to understand what this means: we must have that $D_1$ is one of the numbers 1, 2, 3, 4, 5, 6, and for any such outcome the event $A \cap B$ will then have a probability conditioned on the assumed value of $D_1$.

(1) If $D_1 = 1$ the event $A$ cannot occur, because we can reach a sum of 7 but not exceed it: $P(A \cap B|D_1=1)=0$.
(2) If $D_1 =2$ the product $D_1 \times D_2$ is even so $B$ cannot happen: $P(A \cap B|D_1=2)=0$.
(3) If $D_1 = 3$ we must have $D_2 \in \{5,6\}$, and only the combination $(D_1,D_2) = (3,5)$ is in $A \cap B$: $P(A \cap B|D_1=3)=1/6$.
(4) If $D_1 = 4$, the product $D_1 \times D_2$ is even, so $B$ does not occur: $P(A \cap B|D_1=4)=0$.
(5) If $D_1 = 5$, $A$ needs $D_2 \in \{3,4,5,6\}.$ Only $D_2 = 3,5$ give an odd product: $P(A \cap B|D_1=5)=2/6$.
(6) If $D_1 = 6$ the product $D_1 \times D_2$ is even, so $B$ does not occur: $P(A \cap B|D_1=6)=0$.
Of course, we could cut out half the work and writing just by noting that if $D_1$ is even the product $D_1 \times D_2$ must also be even, so $P(A \cap B | D_1 \:\text{even} \} = 0.$

So,altogether we have
$$P(A \cap B) = \frac{1}{6} \left( 0 +0+\frac{1}{6}+0+\frac{2}{6}+0 \right) = \frac{3}{36} = \frac{1}{12}.$$

Anyway, if we have A=8
Thanks for the mathematical explanation i really appreciate it but here i got totaly lost unfortunately :) Im new to this subject and realy would like to undertand it better but this looks to complicated im sorry

haruspex
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Gold Member
Hoe groot is de kans dat A is groter dan 7 en B is oneven?
In english this is as far as I know:
What is the probability that A is great than 7 and B is odd
Ah, but that is not what you wrote in post #1. P(AnB) means the probability of both event A and event B, but A and B are not events here. The events are A>7 and B odd, so a correct form would have been P((A>7) n (B is odd)).
And to be consistent with that, A and B should have been defined as the random variables a+b and a*b.
By writing the tables it is not guess work anymore
Sure, but the tables were your interpretation of the problem statement. My issue was with the problem statement itself.

You might think I am just being picky, but the way you expressed the problem in English does hint at some confusion regarding the basic concepts of random variables, events and event spaces. You will the find the subject rather easier if you are clear on what these things are and how they relate.

No problem at all I fully agree that the question needs to be written correct. That is indeed very important. But please note that the posted tables are not my interpretation but a copy of the solutions manual from my book. I do see the difference in you post regaring the events. But the answers and tables come from the book.

Yes interpreting the question correct is very important to find the correct answer so again im lost completely lost. For your answer i see that there is a difference between

P(AnB) and P(A) n P(B)

The nice explanations from the people before showed the correct answer and made sense to me when looking to the tables. But did they interpret is as:

P(A n B)

or

P(A) n P(B)
?

haruspex
Homework Helper
Gold Member

P(AnB) and P(A) n P(B)
No, that's not the point I am making, and again points to some confusion.

You need to distinguish random variables, events and event spaces, and only use operators that are meaningful on each type of object.

The outcomes of dice rolls can be considered random variables. Algebraic combinations of the results, such the sum and product, are also random variables.
An atomic event is a particular outcome, i.e. a random variable taking a certain value on some trial.
An event more generally is a set of outcomes considered equivalent for some purpose. Thus, if we only care whether the sum is > 7 and X is the random variable for the sum then this event is (X>7).
We can combine events using ∩, ∪ etc. to create other events. (There are limitations on this when we get into infinite spaces.)
An event space is the set of all possible outcomes for some random variable.
A probability function is a mapping from the set of possible events (i.e. all subsets of the event space) to [0,1].

In short, an event specifies a condition that the outcome of a random variable might satisfy in a trial; the probability of the event is the probability that the outcome will satisfy the condition.

In your initial post you wrote that A and B are spaces, but then you specified A=a+b, B=a*b. That means they are not spaces, they are random variables.
1) What is the probability P(A > 7)
2) What is the probability P(B = odd)
Again, that means A and B are random variables. If we ignore the original statement that they are spaces, all is fine up to here.

But then you wrote
3) What is the probability P( A n B)
4) What is the probability P(A u B)
Applying operators ∩, ∪ to random variables does not mean anything. Those operators apply to events. If I now try to interpret A and B as events, I don't know what those events are. Q3 and Q4 are posed as though independent of Q1 and Q2.
(In your post #18, you wrote P(A) n P(B). This also means nothing. P(A) and P(B) are numbers.)

There are a couple of ways this could have been written that make sense. E.g. with the more common usage of X, Y, Z.. for random variables and A, B, C... for events:
X is the r.v. a+b
Y is the r.v. a*b
A is the event X>7
B is the event Y is odd
q1: P(A)
q2: P(B)
q3: P(A∩B)
q4: P(A∪B)

FactChecker
Gold Member
P(A n B) is correct.

P(A) n P(B) does not make sense, since both P(A) and P(B) are real numbers. They can be added, multiplied, subtracted, divided, etc. But they can not be 'anded' in this context.
CORRECTION EDIT: As @haruspex says, P( A ∩ B ) is not correct. It should be something like P( (A(X)>7) and (B(X) is odd)), where A and B are redefined to be functions of the random result of two die X=(a,b).

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haruspex
Homework Helper
Gold Member
P(A n B) is correct.
Only if A and B are events. They have been defined as random variables, a+b and a*b.
If you disagree with my posts, let's sort it out on a private thread rather than confuse ElectricRay.

FactChecker