Probability math confusion sample space (I think)

[late347]

Homework Statement

high school has 417 students total
186 of total are athletes (play sports)
136 of total are musicians (play music)
74 of total are musicians and athletes. (play music and play sports)

a) at which probability does randomly chosen athlete also play music (i.e. be a musician)

b) at which probability does randomly chosen student play sports or play music.

Homework Equations

The Attempt at a Solution

[/B]
I think we are given the condition that the student must be an athlete in the beginning.

There are only 186 athletes in the school it seems.
Therefore those athletes who are also musicians. They are 74 in number

Therefore the probability = 74/186
roughly 40%. I wonder if this is the correct method though.


there are overall 417 students. It should be noted that there exist students there, who are neither athletes nor are the musicians. Maybe they are in the debate club. etc...


the randomly chosen student can be

1. non-participant in sports and music
2. pure athlete
3. pure musician
4. athlete-musician

IF the student is (athlete-musician) or (athlete) or (musician) then the condition for event is favorable and fulfilled indeed.

It looks like probability for student is therefore = (186+136-74)/ 417

=248/417
roughly 59%

the probability for non-participant = 1- (248/417) =41%

 

[Ray Vickson]

Homework Statement

high school has 417 students total
186 of total are athletes (play sports)
136 of total are musicians (play music)
74 of total are musicians and athletes. (play music and play sports)

a) at which probability does randomly chosen athlete also play music (i.e. be a musician)

b) at which probability does randomly chosen student play sports or play music.

Homework Equations

The Attempt at a Solution

[/B]
I think we are given the condition that the student must be an athlete in the beginning.

There are only 186 athletes in the school it seems.
Therefore those athletes who are also musicians. They are 74 in number

Therefore the probability = 74/186
roughly 40%. I wonder if this is the correct method though.


there are overall 417 students. It should be noted that there exist students there, who are neither athletes nor are the musicians. Maybe they are in the debate club. etc...


the randomly chosen student can be

1. non-participant in sports and music
2. pure athlete
3. pure musician
4. athlete-musician

IF the student is (athlete-musician) or (athlete) or (musician) then the condition for event is favorable and fulfilled indeed.

It looks like probability for student is therefore = (186+136-74)/ 417

=248/417
roughly 59%

the probability for non-participant = 1- (248/417) =41%


View attachment 101528

Do you know the formulas for conditional probability?

Below, let A = event "student is an Athlete", M = event "student is a Musician".
P(A) = 186/417
P(M) = 136/417
P(M & A) = 74/417.
Now you need to figure out the conditional probability P(M|A) in (a) and P(A or M) in (b).

You said P(M|A) = 74/186; the formula for conditional probability says that P(M|A) = P(M & A)/P(A) = (74/417)/(186/417) = 74/186, so yes, your answer to (a) is OK.

You said "it looks like" P(M or A) = (186+136-74)/ 417. That is correct, but there is no "looks like" about it; it follows from basic probability laws. In fact, the so-called inclusion/exclusion principle gives P(A or M) = P(A) + P(M) - P(A & M).

So, overall you did things correctly.

 

[late347]

My confusion was more about the last question

Originally I had intended to calculate the counter event for (athlete or musician)

The counter event should be that the student is neither athlete nor musician nor athlete musician.

Whats the way to directly calculate p(non participating student either music or athletics)

So student cannot be athlete-musician nor athlete nor musician.

I tried looking at the intersection diagram and it seems that.

1 - [(186+136-74)÷417] =p(nonparticipant)

Why do I need to have the 74 in there included as well?

Why is it wrong if you say 417-136-186 = non participating students

That one had me confused at first and it evidently lead confusion on my part and initially a wrong answer when I made the second part on paper...

 

[late347]

I think I realized it now.

P(pure athlete) = (186-74)÷417

P(pure musician) = (136-74)÷417

P(athlete mussician)= 74÷417

P(nonparticipant) = 1 - (the sum of those above probabilities)

 

[late347]

If there's 186 athletes from total...does that mean that some of those athletes are in fact athlete musicians?

Thats where the problem tricked my understanding.

Therefore the number of pure athletes is 186-74= 112 ?

 

[SammyS]

Homework Statement

high school has 417 students total
186 of total are athletes (play sports)
136 of total are musicians (play music)
74 of total are musicians and athletes. (play music and play sports)

a) at which probability does randomly chosen athlete also play music (i.e. be a musician)

b) at which probability does randomly chosen student play sports or play music.

Homework Equations

The Attempt at a Solution

[/B]
I think we are given the condition that the student must be an athlete in the beginning.

There are only 186 athletes in the school it seems.
Therefore those athletes who are also musicians. They are 74 in number

Therefore the probability = 74/186
roughly 40%. I wonder if this is the correct method though.


there are overall 417 students. It should be noted that there exist students there, who are neither athletes nor are the musicians. Maybe they are in the debate club. etc...


the randomly chosen student can be

1. non-participant in sports and music
2. pure athlete
3. pure musician
4. athlete-musician

IF the student is (athlete-musician) or (athlete) or (musician) then the condition for event is favorable and fulfilled indeed.

It looks like probability for student is therefore = (186+136-74)/ 417

=248/417
roughly 59%

the probability for non-participant = 1- (248/417) =41%


View attachment 101528

That all looks perfectly fine to me.

 

[late347]

That all looks perfectly fine to me.

How would one calculate p(student is only an athlete)

Only an athlete means exactly that (pure athlete)

 

[SammyS]

How would one calculate p(student is only an athlete)

Only an athlete means exactly that (pure athlete)

I would use reasoning similar to what you used in the OP.

 

[late347]

I would use reasoning similar to what you used in the OP.

One could say,

• all athletes are 186 in number

• but some of those athletes are in actual fact athlete-musicians... hiding in sheep's clothing as it were.

• it seems that the number of (pure) athletes is therefore actually 186-74 = 112
• that's where I made an error myself because the book statement clearly said the number of athletes is 186. And I mixed up the terminology from the book, with my own terminology, in which athlete-musician is a separate category.

then when all the three categories for probabilities (athlete, or musician, or athlete-musician) are summed together, then the correct result should be what was required. The sum of those separate three probabilities = p(athlete or musician)

The remaining fourth probability means, p(non-participant).

p(athlete or musician) + p(non-participant) = 1,0