Mathematics: Post your doubts here!

[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.

 

[reina81]

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.

but then what happens to the other G?

 

[smartangel]

smartangel. Apparently u need a tutor. Too many doubts :O

thats not a very helpful comment a few hours before the paper you know

 

[reina81]

ON 11 P63 Q2, i, I made a correct tree diagram but i keep getting 31/30 as the probability

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_ms_63.pdf

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_ms_63.pdf

 

[mybch]

Can anyone please explain 4 (iii)
it would be of great help
this is mechanics 2 btw

 

[Khan_971]

thats not a very helpful comment a few hours before the paper you know

Im sorry. Well. good uck. But my, p63 w11 is HARD