Probability help?

[BlsdMama]

I feel a little slow today that I'm coming up with wrong answers. Help?

 

Number line of zero to ten, real numbers represented.

If a number is selected at random, then what is the probability that:

 

a. It is four or more?

c. It is between 5 and 7.

[BlsdMama]

As zero is a real number, probability should be a number out of eleven, yeah?

 

So, I get:

a. 7 of 11

c. 1 / 11

 

Tell me why I'm wrong.

[BlsdMama]

According to Foerester, all options are out of ten possibilities, but zero and ten ought to be " real" possibilities?

 

Moreover, as I count, on a., you could get 4, 5, 6, 7, 8, 9, or 10.... seven total possibles, so 7 out of 11.

 

For c., the only real number between 5 and 7 marked on the number line is 6, so ONE possibility is 1/11.

 

They have correct answers as:

 

a. 3/5

c. 1/5

[Kathy in Richmond]

The key here is that he wants you to consider real numbers, not integers. So you need to think of the number line from 0 to 10 as a continuum with length 10.

 

For the first question, the probability of choosing a real number greater than or equal to four is the length of the line segment from 4 through 10 divided by the total length, or 6/10 = 3/5.

 

For the second part, the probability of choosing a real number between 5 & 7 is equal to the length of that segment (2) divided by the total length (10), or 2/10 = 1/5.

 

Hope that helps!

[regentrude]

I feel a little slow today that I'm coming up with wrong answers. Help?

 

Number line of zero to ten, real numbers represented.

If a number is selected at random, then what is the probability that:

 

a. It is four or more?

c. It is between 5 and 7.

 

I find the easiest approach to be geometrical.

Real numbers form a continuum. They are infinitely close to one another and "cover" the entire number line.

Think of the number line as a paper strip of length 10.

a) the real numbers that are larger than 4 cover a strip of length 6.

That is 6/10 of the total, or 3/5.

 

c) The real numbers between 5 and 7 form a strip of length 2.

That is 2/10 of the total, or 1/5

 

ETA: Just saw your second post: you are thinking about Integers, but the problem asks about real numbers.

[Jann in TX]

You used integers--( lets make it easy and say 'whole numbers'-- not the proper term but you get the idea)-- easy mistake because the number line only labeled the integers as numbering all of the 'real numbers' is impossible!

 

Real numbers include all of the little pieces between the integers (4.001, 4.6, 4.8888... so LOTS of those little buggers).  

 

Now realizing what a 'real number' is vs an 'integer' the paper strip/segment model from the previous posters should make sense... as there are an equal (and technically infinite) number  of 'real numbers' between each integer-- substitute that with a 'segment' and then take the ratio of segments posible out of total segments.

 

 

[BlsdMama]

I do understand how you got from a to b, but zero is a real number, yes? And while you can't see it, the " shaded" area includes both zero and ten as on the line it starts with zero and ends at ten. If ten and zero are not possibilities, then there are only nine possibilities and if both are included, then there are eleven. So, if zero is a real number and the shaded area begins at zero, then there are eleven possibilities you could draw: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. I'm thinking of it as though you were to put all the possibilities into a hat and then draw them out. Is that simply wrong thinking? And how does it vary as opposed to looking on a timeline? Because if zero is a real number, and from what I understand whole numbers that are " real" DO include zero, then drawing it from the hat must be included. And when I say numbers BETWEEN 5 and 7, doesn't that exclude 5 and 7 thereby only leaving one possible draw - 6?

 

I promise I really do see what you're saying. For one, I suspect I'm overthinking, but to explain it to the boy I used that stupid hat example.

 

I am learning with Foerester to check what I THINK is right with what IS right before teaching.

[Kathy in Richmond]

Yes, 0 and 10 are real numbers and are included in this example. But 1/3, 2.143, pi, etc are also real numbers. Integers, whole numbers, rationals (fractions),and irrationals (like pi) are all real numbers, too. So put your pencil tip on ANY point on the number line between 0 and 10 and you've located a real number. In all, there are an infinite number of choices! You can't write them all down or even count them. You can't use the numbers in the hat analogy for this kind of problem.

 

So to compare how many real numbers lie between 5 & 7 versus how many real numbers lie between 0 and 10, we turn to measuring the lengths of the line segments between those numbers.

 

If the problem instead asked for the probability of choosing a number between 1/3 and 1/2, we'd calculate it as

 

(1/2 -1/3) / (10 -0)  or  1/60

 

If the problem asked for the probability of choosing "5", the answer would be

 

(5 -5) / (10 -0)  or 0   ....when there are an infinite number of choices, the probability of drawing one specific choice is zero :-)

[Arcadia]

So, if zero is a real number and the shaded area begins at zero, then there are eleven possibilities you could draw: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. I'm thinking of it as though you were to put all the possibilities into a hat and then draw them out. Is that simply wrong thinking? And how does it vary as opposed to looking on a timeline?

Draw a number line. Because you are looking at real numbers, you would be drawing a line from one number to another, not marking out dots/crosses for integers.

 

The confusion is because you are counting integers instead.

 

For 4 or more, it would be a dot at 4 with a line going from 4 to 10 and a dot at 10.

[MrSmith]

I apologize in advance.  My comment is totally off topic, but this:

 

....when there are an infinite number of choices, the probability of drawing one specific choice is zero :-)

 

is precisely why no one actually exists in the Universe, and anyone you might actually meet is but a figment of your imagination :laugh:

[Julie of KY]

I find it easiest to "see" the answer to this kind of question if you draw a number line and then shade it in. As previous posters said you are dealing with the entire number line (for all real numbers), not just the integers.

 

For the first problem, if you shade everything above four (from 4.00001... to 10) then it is easy to see that the probability is 3/5. For the second problem you shade the entire area between 5 and 7 which shades 1/5 of the area.