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Mathematics: Post your doubts here!

emkay

emkay

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okay, so do you guys think its possible for one to appear for complete A'level mathematics including P1,P3,M1 and S1 in mayjune with only 6months in hand while he has to appear for A2 phy and computing too?
 
momo_khan

momo_khan

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V

Vinita Manek

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when u use nd to approx bd u need a cc of +- 0.5.....dat is becuse in binomial distribution u r gvin the probabilty of success and n....here u multiply both n u find ur own mean....also v approximate variance by npq....since frm bd v r usin nd v use +- 0.5....
 
DumbIdiot

DumbIdiot

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Could someone please explain what is done in Q3 (ii), I don't understand what is done in the ms
 

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A

athaan

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DumbIdiot said:
Could someone please explain what is done in Q3 (ii), I don't understand what is done in the ms
Click to expand...
this is how I did my work, hope it clears your problem!

You're asked to find how many 'n' is needed, for P(X>=1) > 0.95
Success is obtained when a person rated 'poor', so p = 0.13
Therefore, it wouldn't be a success if the person rate else than 'poor', so q = 0.87

0.95 < P(X>=1)
---P(X>=1) = 1 - P(X=0)----
0.95 < 1 - P(X=0)
0.05 > P(X=0)

P(X=0) = nC0*0.87^n
P(X=0) = 0.87^n
----since nC0 = 1, we can eliminate it from the equation----

therefore,
0.05 > 0.87^n

use logarithm to find out 'n'. I found it 21.5.
if n = 21, then it wouldn't be 0.95 < P(X>=1)
if n =22, the statement 0.95 < P(X>=1) will be true.

So, the answer is 22.
 
Silver Wing

Silver Wing

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athaan said:
Can someone help me with question number 6(iii) in ON2010 P63 ASAP? -- http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_63.pdf

I tried looking for an explanation but couldn't find one. I'm sorry if this question has been asked and answered. I don't understand why the students is 5 -- or 5/12.
Click to expand...
The reason for why its 5 for students is the question:
What is the diffenerce if there was 1 student instead of 5?
The 5 came from (5 combination 1) the number of ways by which the student sitting in front of Mrs. Lin is Chosen from the five present

Hope it helps, ;)
 
Kirabo Takirambudde

Kirabo Takirambudde

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I miss read th first sentance nd bursted out laughing :ROFLMAO:.... anyhow
Albert Einstein said:
Hi please help me out with this
Click to expand...
Very tricky but also very simple once you see past the bullshit they have put in the question:
The probability of having a girl is 0.5 and a boy is also 0.5, thus here are the 4 possible ways of have 1 girl.... btw G--> girls and B--> boy
(G,B) or (B,G) or (B,B,G) or (B,B,B,G)
= 0.5^2 + 0.5^3 + 0.5^3 + 0.5^4
=11/16

Good luck 2morrow dude
 
Albert Einstein

Albert Einstein

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Kirabo Takirambudde said:
I miss read th first sentance nd bursted out laughing :ROFLMAO:.... anyhow

Very tricky but also very simple once you see past the bullshit they have put in the question:
The probability of having a girl is 0.5 and a boy is also 0.5, thus here are the 4 possible ways of have 1 girl.... btw G--> girls and B--> boy
(G,B) or (B,G) or (B,B,G) or (B,B,B,G)
= 0.5^2 + 0.5^3 + 0.5^3 + 0.5^4
=11/16

Good luck 2morrow dude
Click to expand...
Thnks
 
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