Simple problem for probability of 3 indipendent events

Njords

New member
Joined
Aug 29, 2016
Messages
21
I'm struggling to find what method needed to solve this, I don't think it should be hard for someone that is used to this type of question. For me though, I am finding it hard.


2018-08-12.jpg
 
I'm struggling to find what method needed to solve this, I don't think it should be hard for someone that is used to this type of question. For me though, I am finding it hard.


View attachment 9906

This is not primarily a probability problem, and there are no independent events here! "Independent" has a very specific meaning.

What this is, is a Venn diagram problem. Are you familiar with those? There are several ways to solve them, not all of which require actually making a Venn diagram, but that does help to make sense of the data.
 
I have a basic idea of Venn diagram in how to know I'm looking at one.. lol. Maybe not enough to make conclusions on the data.

for the first one a) I figure if it says the probability a student did not walk is 1-(Probability they did) ie. 1-0.4=0.6

b) Used all three, I'm guessing is the product of each of the 3 types of transport.. 0.4 x 0.6 x 0.35

c) Using answer I get from b) I am going with the Intersection of say walking or taking the bus as (0.4 x 0.6) - answer from b)

Since it's not so clear I'm not too confident I'm approaching it correctly or answering with correct figures.
 
I have a basic idea of Venn diagram in how to know I'm looking at one.. lol. Maybe not enough to make conclusions on the data.

for the first one a) I figure if it says the probability a student did not walk is 1-(Probability they did) ie. 1-0.4=0.6

b) Used all three, I'm guessing is the product of each of the 3 types of transport.. 0.4 x 0.6 x 0.35

c) Using answer I get from b) I am going with the Intersection of say walking or taking the bus as (0.4 x 0.6) - answer from b)

Since it's not so clear I'm not too confident I'm approaching it correctly or answering with correct figures.

Now that you have told us your thinking, it's possible to help more directly.

You are guessing. Don't. And you're acting as if the probabilities were independent, which is not true. Never assume that without reason.

I would fill in the Venn diagram, putting an x in the middle (all three). This will be the answer to part (b), but we're just trying to gather information, and to not answer the questions yet.

You will be able to write an expression for the size of the union of the sets, which we are told is 100%. Solve that equation, find x, and then answer the questions

When I did it, I didn't use decimals, just percents. But it doesn't matter either way.
 
Now that you have told us your thinking, it's possible to help more directly.

You are guessing. Don't. And you're acting as if the probabilities were independent, which is not true. Never assume that without reason.

I would fill in the Venn diagram, putting an x in the middle (all three). This will be the answer to part (b), but we're just trying to gather information, and to not answer the questions yet.

You will be able to write an expression for the size of the union of the sets, which we are told is 100%. Solve that equation, find x, and then answer the questions

When I did it, I didn't use decimals, just percents. But it doesn't matter either way.

First of all, I am appreciative you've taken time to help me with this. Lad!

... so I've tried to follow what you've said, this is what I've come up with... either I got it or I've understood nothing. lol

question 2.jpg
 
In essence, as Dr. Peterson has indicated, this is a problem in counting. A Venn Diagram is an aid in counting.

These probabilities were computed as what percentage of the surveyed students fell into which categories. If you were told the number of students surveyed, you could turn all these percentages into integers. In fact, it might help you understand the problem better to assume that the survey consisted of 100 students and turn all those percentages into integers.

So 40 students walked. 60 took the bus. 35 cycled. But there were only 100 students in the survey and

40 + 60 + 35 > 100.

(I can hear you objecting that we dont know the number of students in the survey. True. But multiply those percentages by ANY sample size and you will still get that the sum of the students in each category is greater than the sample size because

40% + 60% + 35% = 135% > 100% so we can work with any sample size that results in no fractional people.)

Now if we add up the number of people in each category and find that totals more than the total number of people, we must have counted some people more than once. How would that happen?

Assuming we counted correctly, it can only be because some people were put in more than one category. Then when we counted the people in those categories, we counted them more than one time. But the problem tells us about people who were in at least two categories. 20 walked and took the bus. 10 walked and cycled. 12 cycled and took the bus. In other words, 42 did not use a single mode of transportation.

Now assume for the moment that no one used three modes of transportation. Then we counted twice those who walked and cycled, once with the walkers and once with the cyclists. We counted twice those who walked and took the bus, once with the walkers and once with the bus takers. And we counted twice those who cycled and took the bus, once with the cyclists and once with the bus takers.

To correct for those we counted twice, we must subtract 42 from the sum of our categories.

135 - 42 = 93, which is not 100.

What happened? Well some students must have used all three modes of transportation. So we initially counted those souls three times when we added up the broad categories of walkers, cyclists, and bus riders. But of course we counted them three times again when we added up the categories of people who took AT LEAST two types of transportation. So when we added up our general categories, we counted the three-mode folks three times, but then when we subtracted the at-least-two-modes people, we subtracted the three-mode people three times. The net is as though we never counted them at all.

So 100 - 93 = 7 people used all three modes.

42 - 7 = 35 used exactly two modes.

35 + 7 = 42 used more than one mode.

100 - 42 = 58 used exactly one mode.

Counting is a bit harder than they taught you in kindergarten.
 
Last edited:
In essence, as Dr. Peterson has indicated, this is a problem in counting. A Venn Diagram is an aid in counting.

These probabilities were computed as what percentage of the surveyed students fell into which categories. If you were told the number of students surveyed, you could turn all these percentages into integers. In fact, it might help you understand the problem better to assume that the survey consisted of 100 students and turn all those percentages into integers.

So 40 students walked. 60 took the bus. 35 cycled. But there were only 100 students in the survey and

40 + 60 + 35 > 100.

(I can hear you objecting that we dont know the number of students in the survey. True. But multiply those percentages by ANY sample size and you will still get that the sum of the students in each category is greater than the sample size because

40% + 60% + 35% = 135% > 100% so we can work with any sample size that results in no fractional people.)

Now if we add up the number of people in each category and find that totals more than the total number of people, we must have counted some people more than once. How would that happen?

Assuming we counted correctly, it can only be because some people were put in more than one category. Then when we counted the people in those categories, we counted them more than one time. But the problem tells us about people who were in at least two categories. 20 walked and took the bus. 10 walked and cycled. 12 cycled and took the bus. In other words, 42 did not use a single mode of transportation.

Now assume for the moment that no one used three modes of transportation. Then we counted twice those who walked and cycled, once with the walkers and once with the cyclists. We counted twice those who walked and took the bus, once with the walkers and once with the bus takers. And we counted twice those who cycled and took the bus, once with the cyclists and once with the bus takers.

To correct for those we counted twice, we must subtract 42 from the sum of our categories.

135 - 42 = 93, which is not 100.

What happened? Well some students must have used all three modes of transportation. So we initially counted those souls three times when we added up the broad categories of walkers, cyclists, and bus riders. But of course we counted them three times again when we added up the categories of people who took AT LEAST two types of transportation. So when we added up our general categories, we counted the three-mode folks three times, but then when we subtracted the at-least-two-modes people, we subtracted the three-mode people three times. The net is as though we never counted them at all.

So 100 - 93 = 7 people used all three modes.

42 - 7 = 35 used exactly two modes.

35 + 7 = 42 used more than one mode.

100 - 42 = 58 used more than one mode.

Counting is a bit harder than they taught you in kindergarten.

Yeah, no kidding!

Thank you JeffM & Dr. Peterson
 
Top