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Mathematics: Post your doubts here!

H

hassam

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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
 
sea_princess

sea_princess

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mukki said:
can u tell me for which events the listed probabilities are off?
Click to expand...
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
 
M

mukki

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sea_princess said:
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
Click to expand...
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

sea_princess

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asexamskillme111 said:
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
Click to expand...
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

Khan_971

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Aarjit said:
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
Click to expand...

Thanks.
hassam asexamskillme111
 
M

mukki

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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
 
W

Wanzi21

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

Aarjit

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smartangel said:
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_63.pdf
Q6...plzzzz
Click to expand...

[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
 
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