[Freddy-Francis]

(Original post by usycool1)
Thanks for these

Is these type of questions that are more likely to come up in the exam?

[bookworm786]

Can anyone please help me with this? It's 4 marks. I know the answer, i just don't know how to do it.

There are 10 students in a class. 6 of them are boys and 4 are girls.

The students are picked out at random from the class to form a team.

Work out the probability that the team consists of 1 girl and 2 boys.

[Hotdog]

(Original post by bookworm786)
Can anyone please help me with this? It's 4 marks. I know the answer, i just don't know how to do it.

There are 10 students in a class. 6 of them are boys and 4 are girls.

The students are picked out at random from the class to form a team.

Work out the probability that the team consists of 1 girl and 2 boys.

is the anwer 54/125??

[mia_hilton]

Hi guys,would anyone be able to look at this problem on cones and spheres i'm really stuck..please?Cones and Spheres.ppt

[bookworm786]

(Original post by Hotdog)
is the anwer 54/125??

No. Shall i tell you it or do you want to give it another go?

[usycool1]

(Original post by mia_hilton)
Hi guys,would anyone be able to look at this problem on cones and spheres i'm really stuck..please?Cones and Spheres.ppt

There are a few mistakes with the answer you've provided

They have divided both sides by 4. So:

(4/3)/4 = 1/3

Then because they have 27 on the RHS, they multiple 27 by 1/3 to get 27/3



I hope this makes sense

And I've attached my working too, if anyone wants to see it

[usycool1]

(Original post by bookworm786)
No. Shall i tell you it or do you want to give it another go?

What is the answer? I think I've got it but I want to make sure

[bookworm786]

(Original post by usycool1)
What is the answer? I think I've got it but I want to make sure

360/720.

If you, or anyone else got this right can you please tell me how you did it? Thanks

[bookworm786]

Oh and i have anew one lol.
Using your result for part a (which i can't give sorry), or otherwise, solve the simultaneous equations.

x=x(x-3)
x squared + y squared=9

[gdunne42]

(Original post by bookworm786)
360/720.

If you, or anyone else got this right can you please tell me how you did it? Thanks

Think of it as taking balls from a bag and not replacing them. Once you've made the first selection the probabilities change and then change again once you made the second pick.

There are three arrangements of picking 2 boys (B) and 1 girl (G)
Pick a boy (6/10) then another boy (5/9) then a girl (4/8) - BBG
the others are BGB or GBB

You need to work out the probabilities of those 3 outcomes and add them together. Though you should know that all 3 will be the same and so a quicker way is to work out one and then times by 3.

For BBG
6/10 x 5/9 x 4/8 = 120/720

[mia_hilton]

(Original post by usycool1)
There are a few mistakes with the answer you've provided

They have divided both sides by 4. So:

(4/3)/4 = 1/3

Then because they have 27 on the RHS, they multiple 27 by 1/3 to get 27/3



I hope this makes sense

And I've attached my working too, if anyone wants to see it

Thanks - i get it now! You're more accurate than the MathsWatch CD , thanks again
P.s love your avatar

[usycool1]

(Original post by bookworm786)
360/720.

If you, or anyone else got this right can you please tell me how you did it? Thanks

Yeah, there are three possibilities: BBG, BGB and GBB. I'd draw a tree diagram.

gdunne42 has explained it quite well If you don't get it, I'll post my working

[usycool1]

(Original post by bookworm786)
Oh and i have anew one lol.
Using your result for part a (which i can't give sorry), or otherwise, solve the simultaneous equations.

x=x(x-3)
x squared + y squared=9

I'm guessing part a is a graph. In that case, find where the lines/curve intersect to get your answer.

Otherwise, you can solve it algebraically. Substitute in y (I presume you mean y=x(x-3)) into the and then solve from there

[usycool1]

(Original post by mia_hilton)
Thanks - i get it now! You're more accurate than the MathsWatch CD , thanks again
P.s love your avatar

No prob, and thanks

[Bigbangtheory010]

(Original post by bookworm786)
Oh and i have anew one lol.
Using your result for part a (which i can't give sorry), or otherwise, solve the simultaneous equations.

x=x(x-3)
x squared + y squared=9

I have done this whole practice paper, you don't really have to solve it simultaneously as you do a graph before and the roots of the graph(points that are on 0) are the two possibilities of x so you then put it into both equations and x is the one that best fits into both.

[bookworm786]

(Original post by Bigbangtheory010)
I have done this whole practice paper, you don't really have to solve it simultaneously as you do a graph before and the roots of the graph(points that are on 0) are the two possibilities of x so you then put it into both equations and x is the one that best fits into both.

Thanks

[bookworm786]

(Original post by usycool1)
I'm guessing part a is a graph. In that case, find where the lines/curve intersect to get your answer.

Otherwise, you can solve it algebraically. Substitute in y (I presume you mean y=x(x-3)) into the and then solve from there

Thanks

[LoyaaaL]

yeah the grade boundaries should be really low, and I agree with everyone saying that the new spec is hard. I was doing the old spec papers and I got A*s in most of them, but in the new spec im getting B's which is frustrating. I'm expecting atleast 140/200 for A* for the upcoming exam.

[bookworm786]

(Original post by gdunne42)
Think of it as taking balls from a bag and not replacing them. Once you've made the first selection the probabilities change and then change again once you made the second pick.

There are three arrangements of picking 2 boys (B) and 1 girl (G)
Pick a boy (6/10) then another boy (5/9) then a girl (4/8) - BBG
the others are BGB or GBB

You need to work out the probabilities of those 3 outcomes and add them together. Though you should know that all 3 will be the same and so a quicker way is to work out one and then times by 3.

For BBG
6/10 x 5/9 x 4/8 = 120/720

Thanks, but i'm still a bit confused. Like how did you get 5/9 for picking another boy, and then 4/8 for a girl? Thanks in advance

[gdunne42]

(Original post by bookworm786)
Thanks, but i'm still a bit confused. Like how did you get 5/9 for picking another boy, and then 4/8 for a girl? Thanks in advance

6 boys and 4 girls
probability of picking a boy 6/10

next pick there only 5 boys and 4 girls left to pick from
probability of a boy is 5/9 probability of a girl is 4/9

etc