help with 8th grade math question please

[ColinCodyMom]

I love math but probability is not my strong point!! Here are a couple questions from an 8th grade homework sheet that I could really use some help with for my DS.

The question is:
Suppose two 1-6 number cubes are rolled. Find the probability of each outcome.
Probability of rolling a 3 or 5
Probability of rolling both even numbers
Probability of rolling an odd product
Probability of a sum more than 10
Probability of rolling both the same number
Probability o rolling numbers whose product is a square

Thank you for any help and advice!!

 

[Mickey'snewestfan]

First you need to figure out how many potential combinations there are. For example you can have:

a 1 on the first die and a 1 on the second (we'll call that 1 + 1 = 2)
a 1 on the first die and 2 on the second (1 + 2 = 3)

Write down that list of addition problems. 1 + 2 is different from 2 + 1.

Count how many problems are on the list. For every probability that will be the denominator.

Then take a criteria, for example "the probability of rolling a sum of more than 10". Well, if you look at your list you'll probably see 3 problems that meet that criteria -- 5 + 6, 6 + 5, and 6 + 6. So the probability of rolling a sum more than 10 is 3/the total number of combinations. That happens to be a fraction that you can simplify, so go ahead and do this.

Similarly, if you had a criteria that said "the probability of rolling a sum of 2" you'd see that there's only one combination that equals 2, so the answer is 1/the total number of combinations. If ther criteria said "The probability of rolling an even number, you'd write down all the combos that are even: 1+1, 1+3, 1+5, 2+2, 2+4, 2+6, 3+1 . . . . count them and generate a fraction with that number on the top, and the total number of combinations on the bottom.

 

[DMass]

Oh no..a pop quiz...I didn't study...

 

[ColinCodyMom]

First you need to figure out how many potential combinations there are. For example you can have:

a 1 on the first die and a 1 on the second (we'll call that 1 + 1 = 2)
a 1 on the first die and 2 on the second (1 + 2 = 3)

Write down that list of addition problems. 1 + 2 is different from 2 + 1.

Count how many problems are on the list. For every probability that will be the denominator.

Then take a criteria, for example "the probability of rolling a sum of more than 10". Well, if you look at your list you'll probably see 3 problems that meet that criteria -- 5 + 6, 6 + 5, and 6 + 6. So the probability of rolling a sum more than 10 is 3/the total number of combinations. That happens to be a fraction that you can simplify, so go ahead and do this.

Similarly, if you had a criteria that said "the probability of rolling a sum of 2" you'd see that there's only one combination that equals 2, so the answer is 1/the total number of combinations. If ther criteria said "The probability of rolling an even number, you'd write down all the combos that are even: 1+1, 1+3, 1+5, 2+2, 2+4, 2+6, 3+1 . . . . count them and generate a fraction with that number on the top, and the total number of combinations on the bottom.

Thanks for the detailed explanations! I will try that!

 

[ColinCodyMom]

Sooooo confused still!!

Would the denominator (total number of combinations) for these problems be 12 or 36????

 

[brerrabbit]

Outcomes are all based on 36 possible combinations because 6 independent outcomes on one dice times the six independent out comes on the other dice so everything is something over 36.

So how many ways to get 2? Only one, a 1 on 1 and a 1 on the other. So the possibility of getting a 2? 1 in 36 or 2.77% chance of outcome. Same numbers hold for a 12 (6 +6).

Why is a 7 craps at the tables in Vegas? Highest probability of out come at 6/36 or 16.66% chance of occurance. 2 and 5, 3 and 4, 6 and 1. Double it becasue of the two die and you get 6 out of 36 probable outcomes.

 

[brerrabbit]

Denominator should always be 36 because you have two independent actions, rolling two dice and each has 6 outcomes, so denominator is 36. From there count the number of outcomes that get what you looking for and the answer is that over 36.


To confuse you even more, if you flip a coin 50 times and it comes up heads every time, what is the probability that the next flip will yield a heads?

 

[Tumblwd501]

To confuse you even more, if you flip a coin 50 times and it comes up heads every time, what is the probability that the next flip will yield a heads?

50% because the previous coin flips are independent events that don't affect the probability of getting heads on the current flip.

 

[brerrabbit]

50% because the previous coin flips are independent events that don't affect the probability of getting heads on the current flip.

Correct.

 

[SAHDad]

50% because the previous coin flips are independent events that don't affect the probability of getting heads on the current flip.

Technically, it's not 50% - but it has nothing to do with the previous results. There is the slim chance that it will land on edge (1 in 6000, or thereabouts), the act of flipping it creates about a 51/49 split (being slightly more likely to land on the same face it was launched - start with heads up, and you are slightly more likely to end up heads up), there are imperfections and uneven weighting of the coin surface, etc.

But 50% is an accepted (and generally correct) answer.

(If you really want to game the thing, spin it, don't flip it - it can land heavy-side down over 75% of the time!)

 

[scrapquitler]

I love math but probability is not my strong point!! Here are a couple questions from an 8th grade homework sheet that I could really use some help with for my DS.

The question is:
Suppose two 1-6 number cubes are rolled. Find the probability of each outcome.
Probability of rolling a 3 or 5
Probability of rolling both even numbers
Probability of rolling an odd product
Probability of a sum more than 10
Probability of rolling both the same number
Probability o rolling numbers whose product is a square

Thank you for any help and advice!!

My husband is really good at this stuff, me, not so much. The only ones I can tell you for sure:
The probability of rolling both even numbers is 1 in 3 (because either they will both be even, both odd, or one of each).
The probability of rolling an odd product is 1
in 2 (either it will be even or it will be odd, only two possibilities)

 

[PrincesCJM]

Those are some good questions! I should put them on the probability test I'm giving tomorrow.