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please CAN SOMEONE HELP ME? EXAM QUESTION

Please can anyone help meeee
Reply 1
Original post by usernamenew
please can anyone help meeee


please show me how to do this
my exams tomorrow and im freaking out
Original post by usernamenew
please show me how to do this
my exams tomorrow and im freaking out


So the total area of the rectangle is 0.4*1.2m = 0.48m^2
The height of the rectangle is 0.4m.
I presume the circles are touching the edges of the rectangle, therefore, the radius is 0.4/2 = 0.2m.

Area of a circle = πr^2 = π*(0.2)^2 = π0.04
There are 3 circles, total area = 3*0.04*Ï€ = 0.12Ï€.

Area unused = area of rectangle - area of the circles:
0.48 - 0.12Ï€ ~ 0.1030088816m^2
(Area unused / total area of rectangle)*100 = percentage unused
((0.48-0.12Ï€)/0.48)*100 = 21.46018366%
(edited 6 years ago)
Reply 3
Original post by FryOfTheMann
So the total area of the rectangle is 0.4*1.2m = 0.48m^2
The height of the rectangle is 0.4m.
I presume the circles are touching the edges of the rectangle, therefore, the radius is 0.4/2 = 0.2m.

Area of a circle = πr^2 = π*(0.2)^2 = π0.04
There are 3 circles, total area = 3*0.04*Ï€ = 0.12Ï€.

Area unused = area of rectangle - area of the circles:
0.48 - 0.12Ï€ ~ 0.1030088816m^2
(Area unused / total area of rectangle)*100 = percentage unused
(0.48-0.12Ï€/0.48)*100 = 21.46018366%


Thank you so so much!!!:h:
Original post by usernamenew
Thank you so so much!!!:h:


No problem :smile: Good luck with your exam!
Reply 5
Original post by FryOfTheMann
No problem :smile: Good luck with your exam!


aw thank you, sorry if you dont mind could you help me with another question?
the ratio of red to blue counters is 1:4
2 is taken. the probabilityy she takes red is 6/155
how many red were there initially?
Original post by usernamenew
aw thank you, sorry if you dont mind could you help me with another question?
the ratio of red to blue counters is 1:4
2 is taken. the probabilityy she takes red is 6/155
how many red were there initially?


Sure :biggrin:

Say that x is the number of red counters, so the number of blue counters is 4x.
The total number of counters is therefore 5x.

The probability that the first counter will be red is (x)/5x or 1/5.
Without replacement, the total number of counters is now 5x-1.
The total number of red counters is now x-1.

The probability that the second counter is red is then (x-1)/(5x-1).
The probability of both the first and second counter being red is then (1/5) * [(x-1)/(5x-1)], which is equal to 6/155.

(x-1)/5(5x-1) = 6/155
(x-1) * 155 = 6*5(5x-1)
155x - 155 = 6(25x-5)
155x - 155 = 150x - 30
5x - 155 = -30
5x = 125
x = 25
Number of red counters = 25
Reply 7
Original post by FryOfTheMann
Sure :biggrin:

Say that x is the number of red counters, so the number of blue counters is 4x.
The total number of counters is therefore 5x.

The probability that the first counter will be red is (x)/5x or 1/5.
Without replacement, the total number of counters is now 5x-1.
The total number of red counters is now x-1.

The probability that the second counter is red is then (x-1)/(5x-1).
The probability of both the first and second counter being red is then (1/5) * [(x-1)/(5x-1)], which is equal to 6/155.

(x-1)/5(5x-1) = 6/155
(x-1) * 155 = 6*5(5x-1)
155x - 155 = 6(25x-5)
155x - 155 = 150x - 30
5x - 155 = -30
5x = 125
x = 25
Number of red counters = 25


thank you!
i highlighted the part i didnt understand, im not sure how you get x-1?
(edited 6 years ago)
Original post by usernamenew
thank you!
i highlighted the part i didnt understand, im not sure how you get that?


Say you had 50 counters in the bag.
This time, let's call the total number of counters 5J.
So you have J red counters (10) and 4J blue counters (40).

When you remove a counter from the bag, in this example, the person removing the counter does not put it back in the bag. This means that the bag now has 1 counter less than it had before.
So the total number of counters is now 5J-1. (49 counters)
As it was a red counter removed, the total number of red counters will become J-1. (9 counters).

So your probability changes from being just 1/5 to (J-1)/(5J-1), which would be 9/49

Does this explain it enough? :smile:
Reply 9
Original post by FryOfTheMann
Say you had 50 counters in the bag.
This time, let's call the total number of counters 5J.
So you have J red counters (10) and 4J blue counters (40).

When you remove a counter from the bag, in this example, the person removing the counter does not put it back in the bag. This means that the bag now has 1 counter less than it had before.
So the total number of counters is now 5J-1. (49 counters)
As it was a red counter removed, the total number of red counters will become J-1. (9 counters).

So your probability changes from being just 1/5 to (J-1)/(5J-1), which would be 9/49

Does this explain it enough? :smile:


yep thank you so much! im so sorry im going to ask one more question. can i ask whats the difference between these 2, i copied this from the markscheme
but why does the 2nd question need 32 to be subtracted, whereas in the first 10 isnt subtracted from the total?
Original post by usernamenew
yep thank you so much! im so sorry im going to ask one more question. can i ask whats the difference between these 2, i copied this from the markscheme
but why does the 2nd question need 32 to be subtracted, whereas in the first 10 isnt subtracted from the total?


Could you tell me what paper this is from? If you tell me the paper details, I can have a look at the question itself and explain how to do the parts as I am a bit confused at what you're asking :smile:
Original post by FryOfTheMann
Could you tell me what paper this is from? If you tell me the paper details, I can have a look at the question itself and explain how to do the parts as I am a bit confused at what you're asking :smile:


im not sure sorry i cant find the actual paper, i found these questions ages ago but i rewrote these ones so it was neater, but the 2nd question was
there were 120 coins
T= coins from 20th century
B=british coins
find the probability a coin is chosen from the 20th century

i cant find the other one sorry! but how would you know when to subtract the number outside the circles
Original post by usernamenew
im not sure sorry i cant find the actual paper, i found these questions ages ago but i rewrote these ones so it was neater, but the 2nd question was
there were 120 coins
T= coins from 20th century
B=british coins
find the probability a coin is chosen from the 20th century

i cant find the other one sorry! but how would you know when to subtract the number outside the circles


Sorry for the late reply!

I think I found your question:
http://i.imgur.com/vzva5Ti.jpg
http://i.imgur.com/Ezemxu3.jpg

Side note:
P(B|A) is the possibilty that B happens given A has/will happen.
so P(T|B) is the probability that a coin is from the 20th century given that it is from Britain.
Original post by FryOfTheMann
Sorry for the late reply!

I think I found your question:
http://i.imgur.com/vzva5Ti.jpg
http://i.imgur.com/Ezemxu3.jpg

Side note:
P(B|A) is the possibilty that B happens given A has/will happen.
so P(T|B) is the probability that a coin is from the 20th century given that it is from Britain.



Thank you so much, i was worrying about how do you know when to subtract the number outside the circle and when not to? Because in this example you added on the 32 and made it equal 120 whereas with the other question i sent, the method was different. Im sorry im makikg no sense
Original post by usernamenew
Thank you so much, i was worrying about how do you know when to subtract the number outside the circle and when not to? Because in this example you added on the 32 and made it equal 120 whereas with the other question i sent, the method was different. Im sorry im makikg no sense


Ah, I understand what you mean now.
So the whole x(x-15) + x + (x-2) + 32 = 120 was:
x(x-15) is P(T) - the intersection of T and B (T n B)
x is the intersection of T and B (T n B)
and x-2 is P(B) - the intersection of T and B (T n B)

32 is the number of coins that are not from the 20th century AND are not British.

You know the total number of coins is 120 so all of the numbers in the circle and outside of the circle must add up to a total of 120. If the circles and the 32 outside of the circles had been fractions, the probabilities would add up to 1.

By equating the x equation with 32 to 120, you work back to find x. Obviously x>0 as you cannot have a negative number of coins.

I hope you understand now :smile:
Original post by FryOfTheMann
Ah, I understand what you mean now.
So the whole x(x-15) + x + (x-2) + 32 = 120 was:
x(x-15) is P(T) - the intersection of T and B (T n B)
x is the intersection of T and B (T n B)
and x-2 is P(B) - the intersection of T and B (T n B)

32 is the number of coins that are not from the 20th century AND are not British.

You know the total number of coins is 120 so all of the numbers in the circle and outside of the circle must add up to a total of 120. If the circles and the 32 outside of the circles had been fractions, the probabilities would add up to 1.

By equating the x equation with 32 to 120, you work back to find x. Obviously x>0 as you cannot have a negative number of coins.

I hope you understand now :smile:


Thank you, i get it now! I just realised the attachment that i sent of the 2 diagrams werent even sent? Sorry about that, i thought it got uploaded.
Original post by usernamenew
Thank you, i get it now! I just realised the attachment that i sent of the 2 diagrams werent even sent? Sorry about that, i thought it got uploaded.


Haha, no problem. I was just a little confused without the rest of the question. I hope your exam went well!

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