x Turn on thread page Beta
 You are Here: Home >< Maths

# OCR MEI S1 - help watch

1. I don't understand this - question 2 ii)
Attachment 529317
2. So in part i, the answer is gonna be (5C2 x 5C1)/*10C3) hence the answer to part 1 is 5/12.

In part ii, it turns into a bionomial situation where X ~ B(4, 5/12)

hence the questions just asks to find P(X=>3) which is the same as " 1 - P(X<3)"

Hence we do:

- P(X=0) => 4C0 x (7/12)^4 = 0.1158
- P(X=1) => 4C1 x (5/12)^1 x (7/12)^3 = 0.3308
- P(x=2) => 4C2 x (5/12)^2 x (7/12)^2 = 0.3545

Hence: 1 - (0.1158 + 0.3308 + 0.3545) = 0.1989...= 0.199

NOTE: half way reading writing this post, you can do a much quicker way of just finding P(X=3) and P(X=4) and just adding them together. works either way; whatever floats your boat.
3. (Original post by TheTechnoGuy)
So in part i, the answer is gonna be (5C2 x 5C1)/*10C3) hence the answer to part 1 is 5/12.

In part ii, it turns into a bionomial situation where X ~ B(4, 5/12)

hence the questions just asks to find P(X=>3) which is the same as " 1 - P(X<3)"

Hence we do:

- P(X=0) => 4C0 x (7/12)^4 = 0.1158
- P(X=1) => 4C1 x (5/12)^1 x (7/12)^3 = 0.3308
- P(x=2) => 4C2 x (5/12)^2 x (7/12)^2 = 0.3545

Hence: 1 - (0.1158 + 0.3308 + 0.3545) = 0.1989...= 0.199

NOTE: half way reading writing this post, you can do a much quicker way of just finding P(X=3) and P(X=4) and just adding them together. works either way; whatever floats your boat.
Oh i get it now! But, how are you supposed to recognise that it is a binomial situation?

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: May 9, 2016
Today on TSR

### Four things top students are doing

Over the Easter break

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams