# maths question probability hard

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#1
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.

0
1 year ago
#2
(Original post by 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.

Could you draw a venn diagram with 3 circles? e.g.
0
1 year ago
#3
Moved to the maths forum 0
1 year ago
#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...
Last edited by lilacdolphin; 1 year ago
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1 week ago
#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
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1 week ago
#6
(Original post by 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.

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.
Last edited by Kallisto; 1 week ago
0
1 week ago
#7
(Original post by 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.
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.
0
1 week ago
#8
(Original post by 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.
Thank you very much, didn't realise this thread was a year old until after I answered it lol 😂
1
1 week ago
#9
(Original post by nazminexd)
Thank you very much, didn't realise this thread was a year old until after I answered it lol 😂
True, I did not do this too. And I guess it is already to late for your answer. It is time for a report.
1
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