Mathematics: Post your doubts here!

[asexamskillme111]

Has been solved. search it

Can you just show me what you got for total no of arrangements and no of arrangements when no G's are next to each other

 

[mukki]

u
u add the probabilities for what is asked so it will be 13/30 + 12/30 = 5/6
or u can split the 12/30 >>>> 13/30 + 9/30 +3/30 = 5/6

can u tell me for which events the listed probabilities are off?

 

[hassam]

weelll for green gage one.....i was doing lykk this that....figured out when twoGS are 2gether....not exactly two...meany two and three both are there.....so got 8!/3!......and then wanted to subtract form when there were 3 Gs 2gether that is 7!/3!.......bt ms says 2* 7!/3!...plxx smbody tell me what i m overcounting Khan_971

 

[Khan_971]

Can you just show me what you got for total no of arrangements and no of arrangements when no G's are next to each other

havent done it . sorry

 

[sea_princess]

can u tell me for which events the listed probabilities are off?

watches "Kops and kids " can either be: 1- a male who watches it >>> probability : 13/30 ( total number of students is 30)
2- a female who watches it >>>> probability 3/ 30
then there are probability of females, but since u already considered the females who watch the show , then here take those who don't watch the show , so the probability becomes 9/30
add them all and u get the required probability in the question

 

[mukki]

watches "Kops and kids " can either be: 1- a male who watches it >>> probability : 13/30 ( total number of students is 30)
2- a female who watches it >>>> probability 3/ 30
then there are probability of females, but since u already considered the females who watch the show , then here take those who don't watch the show , so the probability becomes 9/30
add them all and u get the required probability in the question

oh thanks i tried doing it the way the qs said like first finding Probability of females and then watching kops and then both but cudnt do it :/

 

[sea_princess]

Can you just show me what you got for total no of arrangements and no of arrangements when no G's are next to each other

total number of arrangements = 9 !/ 3!3! =10080
if the three Gs are together = 7! /3! = 840
no Gs are next to each other = 6! / 3! X 7 P3/ 3! = 4200
so the number of arrangements when only 2 Gs are next to each other = 10080 - ( 4200+ 840) =5040

 

[Khan_971]

Simply 'choose' 2 Gs out of 3, and the remaining 7 letters can be completely random!

G G _ _ _ _ _ _ _...............[consider GG to be 'attached' together so that it can alternate in 2! ways]

n = 2! x 3c2 x 7!.................. [but don't forget that there are 3 recurring Es too!]
..............3!
...= 5040...... Q.E.D

Thanks.
hassam asexamskillme111

 

[mukki]

guys if ive got 74 in p1 46 in m1 68 in p3 how much would i need in s1 to get an A* . Grade threshold for A in s1 is between 33 and 38

 

[hassam]

i dint get u ..........i was asking sth different Khan_971

 

[hassam]

guys if ive got 74 in p1 46 in m1 68 in p3 how much would i need in s1 to get an A* . Grade threshold for A in s1 is between 33 and 38

42

 

[mukki]

42

what if i get in 30 wudnt i get an a* overall ?

 

[Wanzi21]

another question.. what is the main condition whether to use binomial or normal dist. in finding probability??

 

[mukki]

np or nq > 20 then u find the probability using normal otherwise binomial.

 

[Aarjit]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_63.pdf
Q6...plzzzz

[Q6] The probability that Sue completes a Sudoku puzzle correctly is 0.75.
(i) Sue attempts n Sudoku puzzles. Find the least value of n for which the probability that she completes all n puzzles correctly is less than 0.06. [3]

S ~ B(n,0.75)

.....P(S = n) < 0.06
or, nCn 0.75^n <0.06......[nCn =1]
or, n. ln(0.75) < ln (0.06)
or, n > 9.77958.................[the < switches to > since we are dividing by a negative number, i.e. ln (0.75) = -2.88..]
.....n = 10

Sue attempts 14 Sudoku puzzles every month. The number that she completes successfully is denotedby X.
(ii) Find the value of X that has the highest probability. You may assume that this value is one of the two values closest to the mean of X. [3]

Every month, the mean number of times she completes the Sudoku correctly is (total #attempts) x (probability of completing correctly)
.................................................................................................................mean...= 14 x 0.75 = 10.5

now, the two possible values closest to the mean (10.5) are 10.0 and 11.0. [Keep in mind that you need the value which gives the highest probability]

X ~ B(14, 0.75)

P(X = 10) = 14C10. (0.75^10). (0.25^4)
...............= 0.22

P(X = 11) = 14C11. (0.75^11). (0.25^3)
...............= 0.24

Since P(X = 11) > P(X = 10), there is a higher chance of Sue getting 'exactly' 11/14 Sudoku puzzles correct than 10/14.

X = 11
(iii) Find the probability that in exactly 3 of the next 5 months Sue completes more than 11 Sudoku puzzles correctly. [5]

If Sue does 14 Sudoku puzzles every month, the probability that she gets >11 puzzles correct is P (S > 11) [where S ~ B(14, 0.75)]

P(S > 11) = P(S = 12) + P(S = 13) + P(S = 14)
................= [14C12. (0.75^12). (0.25^2)] + [14C13. (0.75^13). (0.25)] + [14C14. (0.75^14)]
................= 0.2811..

This means, in every month, the probability of Sue completing more than 11/14 Sudokus correctly is 0.2811 !

Now she needs to do this in exactly 3 months in 5 months of time;

S ~ B(5, 0.2811)

P(S = 3) = 3C15. (0.2811^3). (0.7981^2)
...............= 0.115................QQ.E.D

 

[shan5674]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w09_qp_62.pdf

1(ii)?

 

[Khan_971]

guys if ive got 74 in p1 46 in m1 68 in p3 how much would i need in s1 to get an A* . Grade threshold for A in s1 is between 33 and 38

These are last year's GT for A*. (2nd page)
http://www.xtremepapers.com/papers/...d AS Level/Mathematics (9709)/9709_s11_gt.pdf
http://www.xtremepapers.com/papers/...d AS Level/Mathematics (9709)/9709_w11_gt.pdf
Atm ur total is 188, u need around 35 to make it happen. But it depends if the session's paper easier or harder than last year's?

 

[sea_princess]

np or nq > 20 then u find the probability using normal otherwise binomial.

weren't they supposed to be greater than 5 , not 20?

 

[mukki]

These are last year's GT for A*. (2nd page)
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_gt.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_gt.pdf
Atm ur total is 188, u need around 35 to make it happen. But it depends if the session's paper easier or harder than last year's?

oh wow thanks. didnt knw there was a second page to it Insha Allah we all get A*

 

[Khan_971]

Im hoping too. I hope 75,75,50 in p1,p3 and p4. im worried about p6 now

oh wow thanks. didnt knw there was a second page to it Insha Allah we all get A*