Mathematics: Post your doubts here!

[Wanzi21]

how to do this? last question is always the hardest of all!!!

 

[Khan_971]

but please check june 11 63 question 5)iii) they said normal distribution but in the ms they said with or without cc, can u see this and make it clearer to me??

i dont know

 

[Ashleyxoxo93]

for approximation questions, when do you +0.5 and when do you -0.5? im always confused with this. e.g using a suitable approximation, find the probability that more than 155 people wore a watch on their left wrist. what if they asked for less than 155? or at least? or at most?

 

[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_97171 plx anser thisss

 

[omar hazem]

i dont know

ok i hope it won't come confusing problem in our exam tomorrow :/

 

[omar hazem]

View attachment 10595

how to do this? last question is always the hardest of all!!!

find the probability of ronnie winning the game from first draw so P(GY) and add to P(GGGY) ronnie winning it after his 2nd draw and then add it to P(GGGGGY) ronnie winning it after his 3rd and last draw as 3 yellow balls are only found so 3 trials are max

 

[mukki]

can nybdy pls solve q6 b part 1 n 2
and q7 part 3
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf

q7 (iii) event X is the answer from (i) and event Y is the probability of getting a ball numbered 2 from bag A so P = 1/4
now you need to find the probability of (X and Y) i.e p(exactly two balls selected have the same number and ball selected from bag A is 2) so what this means is that you will have to find the probability that the ball selected from bag A is 2 and the ball selected from bag C is also 2. which leads to =1/4 *1/7 =1/28
P(X and Y) is P(x)*P = 47/140 *1/4. since these two probabilities dont match events are not independent

 

[Khan_971]

View attachment 10595

how to do this? last question is always the hardest of all!!!

Probability of getting a 3= 1/6
P of getting att least one 3= 1-P(0 3s)

P(0)= (5/6)^9
therefore AnsWER: 0.806

part 11) against, if prob of getting at least one 3 >.9. then Prob of not getting any is
1-p(0) >.9
-p(o)>-.1
p(0)<.1
(5/6)^n <.1
nlog(5/6) > log .1
n>12.63. Therefore least n=13

 

[Khan_971]

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_97171 plx anser thisss

i dont know (

 

[mukki]

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

1(ii)? please?

check previous pages it has been answere already

 

[Ashleyxoxo93]

what do u mean by cc?

 

[mukki]

what do u mean by cc?

continuity correction

 

[Ashleyxoxo93]

what do u mean by cc?

 

[mukki]

what do u mean by cc?

cute and cuddly

 

[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 Aarjit???question is Find the number of different ways in which the 9 letters of the word GREENGAGE can be
arranged if exactly two of the Gs are next to each other.
just tell my mistake......

 

[Ashleyxoxo93]

what do u mean by cc?

 

[shan5674]

check previous pages it has been answere already

im sorry i cant find it :/

 

[starlight123]

thnks

q7 (iii) event X is the answer from (i) and event Y is the probability of getting a ball numbered 2 from bag A so P = 1/4
now you need to find the probability of (X and Y) i.e p(exactly two balls selected have the same number and ball selected from bag A is 2) so what this means is that you will have to find the probability that the ball selected from bag A is 2 and the ball selected from bag C is also 2. which leads to =1/4 *1/7 =1/28
P(X and Y) is P(x)*P = 47/140 *1/4. since these two probabilities dont match events are not independent

thnk alot <3

 

[Wanzi21]

help me on 4 ii)

 

[Aarjit]

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 Aarjit???question is Find the number of different ways in which the 9 letters of the word GREENGAGE can be
arranged if exactly two of the Gs are next to each other.
just tell my mistake......

Well, this was how I'd solved it before:

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


P.S. If you're still uncertain, let me know. I'll work in it one more time.