# maths question probability hard

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

This discussion is closed.

32 students took part in different events for charity. Some students took part in one of these events, and some students took part in two of these events.

25 students took part in a bake sale.

16 Students took part in a sponsored swim.

10 students went on a bike ride.

What is the probability that a student,chosen at random, took part in two events.

Someone please help. I dont even know how to start.😭😭😭

25 students took part in a bake sale.

16 Students took part in a sponsored swim.

10 students went on a bike ride.

What is the probability that a student,chosen at random, took part in two events.

Someone please help. I dont even know how to start.😭😭😭

0

Report

#2

(Original post by

32 students took part in different events for charity. Some students took part in one of these events, and some students took part in two of these events.

25 students took part in a bake sale.

16 Students took part in a sponsored swim.

10 students went on a bike ride.

What is the probability that a student,chosen at random, took part in two events.

Someone please help. I dont even know how to start.😭😭😭

**Dnfjfdj**)32 students took part in different events for charity. Some students took part in one of these events, and some students took part in two of these events.

25 students took part in a bake sale.

16 Students took part in a sponsored swim.

10 students went on a bike ride.

What is the probability that a student,chosen at random, took part in two events.

Someone please help. I dont even know how to start.😭😭😭

e.g.

0

Report

#4

So to get the probability of doing any 2 specific things, I think you would multiply the fractions for each thing. Not sure where to go from there though?

I tried to delete the image (I forgot not to post actual solutions) but it won't go away now, sorry!

EDIT: ignore this. This is completely the wrong direction. Sorry, I really mucked up this post! Suppose that what comes from multitasking...

I tried to delete the image (I forgot not to post actual solutions) but it won't go away now, sorry!

EDIT: ignore this. This is completely the wrong direction. Sorry, I really mucked up this post! Suppose that what comes from multitasking...

Last edited by lilacdolphin; 1 year ago

0

Report

#5

If you add all the students that participated in each event, you get

25 + 16 + 10 = 51 students

However, you only have 32 students that took part in the events.

From the question, you know that there isn't any student that took part in all three events as it is not mentioned.

Therefore you can calculate the number of students that participated in two events by

51 - 32 = 19 students

So 19 students out of 32 students participated in 2 events.

19 ÷ 32 = 0.59375 -> 0.59 (2s.f)

The probability that a student chosen at random, took part in 2 events is 0.59 to 2 significant figures

25 + 16 + 10 = 51 students

However, you only have 32 students that took part in the events.

From the question, you know that there isn't any student that took part in all three events as it is not mentioned.

Therefore you can calculate the number of students that participated in two events by

51 - 32 = 19 students

So 19 students out of 32 students participated in 2 events.

19 ÷ 32 = 0.59375 -> 0.59 (2s.f)

The probability that a student chosen at random, took part in 2 events is 0.59 to 2 significant figures

0

Report

#6

**Dnfjfdj**)

32 students took part in different events for charity. Some students took part in one of these events, and some students took part in two of these events.

25 students took part in a bake sale.

16 Students took part in a sponsored swim.

10 students went on a bike ride.

What is the probability that a student,chosen at random, took part in two events.

Someone please help. I dont even know how to start.😭😭😭

A: 25/51 = P(A)

B: 16/51 = P(B)

C: 10/51 = P(C)

For two events, the combinations AB, AC and BC are possible, where AB = P(A)*P(B), AC = P(A)*P(C) and BC = P(B)*P(C) is, as the events are dependent on each other for one student. Your calculation is:

P = AB + BC + AC = P(A)*P(B) + P(B)*P(C) + P(A)*P(C)

EDIT: I did the wrong maths too. Did not include the 32 students who take part in two events at the same time. So I think that nazminexd is right in the direction, but not in calculation: it is 32 out of 51, so 32/51, as 32 are in two events, the rest (19) just in one.

Last edited by Kallisto; 1 week ago

0

Report

#7

(Original post by

The total number of students is 51. 25 of those are taken part in bake sale (A), 16 in sponsored swim (B) and 10 in bike ride (C). The probailities are:

A: 25/51 = P(A)

B: 16/51 = P(B)

C: 10/51 = P(C)

For two events, the combinations AB, AC and BC are possible, where AB = P(A)*P(B), AC = P(A)*P(C) and BC = P(B)*P(C) is, as the events are dependent on each other for one student. Your calculation is:

P = AB + BC + AC = P(A)*P(B) + P(B)*P(C) + P(A)*P(C)

EDIT: I did the wrong maths too. Did not include the 32 students who take part in two events at the same time. So I think that nazminexd is right in the direction, but not in calculation: it is 32 out of 51, so 32/51, as 32 are in two events, the rest (19) just in one.

**Kallisto**)The total number of students is 51. 25 of those are taken part in bake sale (A), 16 in sponsored swim (B) and 10 in bike ride (C). The probailities are:

A: 25/51 = P(A)

B: 16/51 = P(B)

C: 10/51 = P(C)

For two events, the combinations AB, AC and BC are possible, where AB = P(A)*P(B), AC = P(A)*P(C) and BC = P(B)*P(C) is, as the events are dependent on each other for one student. Your calculation is:

P = AB + BC + AC = P(A)*P(B) + P(B)*P(C) + P(A)*P(C)

EDIT: I did the wrong maths too. Did not include the 32 students who take part in two events at the same time. So I think that nazminexd is right in the direction, but not in calculation: it is 32 out of 51, so 32/51, as 32 are in two events, the rest (19) just in one.

To start, the total number of students is 32 - the question *says* this explicitly.

Next, you claim that p(A n B) = p(A)p(B), but this is only true when you know A, B are independent events.

Ironically, you now say "I did the wrong maths", but using this method to find P

**would**actually be valid

**if**you could calculate p(A n B) etc. correctly.

Finally, nazminexd is completely correct. 19 are in two events, 13 in just 1.

0

Report

#8

(Original post by

Sorry, but almost everything you do here is wrong.

To start, the total number of students is 32 - the question *says* this explicitly.

Next, you claim that p(A n B) = p(A)p(B), but this is only true when you know A, B are independent events.

Ironically, you now say "I did the wrong maths", but using this method to find P

Finally, nazminexd is completely correct. 19 are in two events, 13 in just 1.

**DFranklin**)Sorry, but almost everything you do here is wrong.

To start, the total number of students is 32 - the question *says* this explicitly.

Next, you claim that p(A n B) = p(A)p(B), but this is only true when you know A, B are independent events.

Ironically, you now say "I did the wrong maths", but using this method to find P

**would**actually be valid**if**you could calculate p(A n B) etc. correctly.Finally, nazminexd is completely correct. 19 are in two events, 13 in just 1.

1

Report

#9

(Original post by

Thank you very much, didn't realise this thread was a year old until after I answered it lol 😂

**nazminexd**)Thank you very much, didn't realise this thread was a year old until after I answered it lol 😂

1

X

Page 1 of 1

Go to first unread

Skip to page:

new posts

Back

to top

to top