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A* maths question? :)

lolhenry

2

http://i40.tinypic.com/25ps1uv.png

The c part :smile:

, how do you do it?

Reply 1

CullenLoverX

9

oooh, I was doing this earlier and had no idea how to do it...

Reply 2

M_E_X

14

lolhenry

http://i40.tinypic.com/25ps1uv.png

The c part :smile:

, how do you do it?

The answer is 14, right? :smile:

edit for method:
ok: look at P(W,W), the probability of her winning then winning. This is 1/4 , from the probability tree.

Then look at P(L,L), which is 1/6.

So we can see that she is 1.5 times more likely to WW than to LL. (to win both than to lose both).
As such - she will have WW 1.5 times more than LL

So we do 21 divided by 1.5
= 14.

Does that kinda make sense? :s-smilie:

Reply 3

PeterMcPete

1

(1/3)x(3/4)xnumber of fridays altogether=21

so number of fridays lost = (2/3)x(1/4)x((21)/(1/3)(3/4))

I think

Reply 4

User12399

5

dont worry i got it lol

Reply 5

chris2pyle

5

you do 21 divided by 3 (3/12) is the probability she wins both games and then times 7 by 2(2/12) which is the probability that she wins neither game so the answer is 14

Reply 6

lolhenry

OP

2

Ahahahaha, I looked at the mark scheme, but i have no idea how you wuold do it in the first place. Yes it is 14 :smile:

Reply 7

neomilan

17

b) (win x not win) + (not win x win)

Reply 8

M_E_X

14

CullenLoverX

oooh, I was doing this earlier and had no idea how to do it...

I'm quoting you so that you see my solution, hehe :smile:

James

Reply 9

M_E_X

14

lolhenry

Ahahahaha, I looked at the mark scheme, but i have no idea how you wuold do it in the first place. Yes it is 14 :smile:

hehe (Y)

I was furiously scribbling on my paper to try and be the first :P:.

Reply 10

chris2pyle

5

User12399

how do u do part b?

Is it 3/4 * 1/3?

youwork outhe probability of winning first and then losing and then add it to losing first then winning.

Reply 11

neomilan

17

b)

(win

\times

not win) + (not win

\times

win)

(\frac{3}{4} \times \frac{2}{3}) + (\frac{1}{4} \times \frac{1}{3})

=

\frac{7}{12}

Reply 12

thethinker

1

21 times is the number of times she won at both. work out the probability of winning both games and that fraction is equal to 21 days. 3/4*1/3= 1/4 . if 1/4 = 21, 1= 84. The P( loses) is 1/4* 2/3 = 1/6. Therfore she loses both times 1/6 * 84 = 14.

I think this is it but I haven't seen a question like this before.

Reply 13

dbmag9

15

a) P(win)+P(notwin)=1

b) either win-notwin or notwin-win; work out each and sum

c) the probability of winning both is 3/4 * 1/3 = 3/12 = 1/4. If, with probability 1/4, she wins 21 nights, how many nights must she have played? Based on this work out how many nights she won neither.

Reply 14

CullenLoverX

9

M_E_X

I'm quoting you so that you see my solution, hehe :smile:

James

Thank you, I think I get it now =)

Reply 15

chris2pyle

5

is this really an A* star question lol. I have maths tomorrow and I'm sure this is pretty easy, even for GCSE

Reply 16

pete.mcfc

1

lolhenry

Ahahahaha, I looked at the mark scheme, but i have no idea how you wuold do it in the first place. Yes it is 14 :smile:

please can you post a link to the mark scheme

Reply 17

OnDaRunFromPoor

2

chris2pyle

is this really an A* star question lol. I have maths tomorrow and I'm sure this is pretty easy, even for GCSE

Same Here.

I am absoloutly craping myself.

Edexcel I persume?

Heh.. do you know the UMS or is it 100 like AQA? Only I haven't had any luck getting hold of edexcel past papers, being the bellends they are...

Reply 18

Wildstylerandymoss

Its 21 x 4= 84

84 divide by 6= 14

Reply 19

chris2pyle

5

OnDaRunFromPoor

Same Here.

I am absoloutly craping myself.

Edexcel I persume?

Heh.. do you know the UMS or is it 100 like AQA? Only I haven't had any luck getting hold of edexcel past papers, being the bellends they are...

yeah edexcel, and im not sure about ums, i think u need around 80ish for an a* tho

A* maths question? :)

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