Maths - Probability Questions

stevebirch2002

stevebirch2002

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Help. I have been tutoring a couple of kids to pass their GCSE. I love maths and enjoy the tutoring but I have never been taught Probability. Understand most but two questions have me stumped.

1. A bag contains 6 Green Balls, 4 Red Balls and 2 Blue Balls

Four balls are picked at random. From these, two balls are randomly selected. What is teh probability that these two balls are Blue?


2. Trevor and his two brothers and five friends are seated at random in a row of eight seats at the cinema. What is the probability that Trevor has one brother on his immediate left and one on his immediate right?

There is an answer to the first question at the back of the book that doesnt agree with my answer!

Second question is from a book without any answers.... Please help, I teaching Probabilities tonight!

Steve
 
colvic987 said:
I'm not a maths person, but this might help you if not plenty on line, if you can understand it...

http://www.mathsrevision.net/gcse/
Click to expand...

I think I understand how to do it but my answer doesnt agree with the one in the book on Question 1!

My brain hurts after 2 hours of Possibility Trees and fraction calculations!

S
 
I think the first one is 1 in 66

You have a 2 in 12 chance of drawing a bue on the first go and a 1 in 11 change on the second.

Not a clue on the second question unles it's

Trevor has a 1 in 6 change of not sitting on the end

Bro 1 then has a 2 in 7 chance of sitting next to him

Bro 2 has a 1 in 6 change of sitting on the other side


1/6 * 2/7 * 1/6 = 2/252 = 1 / 126 chance overall
 
stevebirch2002 said:
Help. I have been tutoring a couple of kids to pass their GCSE. I love maths and enjoy the tutoring but I have never been taught Probability. Understand most but two questions have me stumped.

1. A bag contains 6 Green Balls, 4 Red Balls and 2 Blue Balls

Four balls are picked at random. From these, two balls are randomly selected. What is teh probability that these two balls are Blue?


2. Trevor and his two brothers and five friends are seated at random in a row of eight seats at the cinema. What is the probability that Trevor has one brother on his immediate left and one on his immediate right?

There is an answer to the first question at the back of the book that doesnt agree with my answer!

Second question is from a book without any answers.... Please help, I teaching Probabilities tonight!

Steve
Click to expand...

First question looks like an elaborate method of choosing two balls from the set of twelve. I make that 2 in 12 for the first ball and 1 in 11 for the second one. Overall probability is then 1 in 66 or 0.01515151. Is this what the book says?
 
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Second question looks like the two end seats don't work so the probability of not being in one of those is 6 in 8 then 1 in 7 for the first brother being next him and 1 in 6 for the second brother being on the other side. Overall probability by multiplying these together gives 1 in 56 or 0.017857.
 
Exactly the answer I got. 1/66

Book says 1/99

???

Question H27c in Essential Exam Practice Key Stage 4 Mathematics Higher Brookworth Books

Agree with the Trevor, Brothers and 5 friends at 1/56 This book doesnt have answers so I was after confirmation. CGP GCSE Mathematics Workbook - Higher


S
 
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These are two events that have dependency.... the key to understanding statistics... the probability of selecting two balls in the second round depends entirely on the likelihood of there being 2 blue balls in there to select to start with.

The chance of picking 2 blue balls in 4 attempts is 2/33

so if there are 2 blue balls in there, the probability of selecting 2 at the 2nd round is 1/6

1/6 x 2/33 is 1/99
 
"You have a 2 in 12 chance of drawing a blue on the first go and a 1 in 11 change on the second."

If your first ball is not a blue on you have a 2 in 11 chance with the second, not 1 in 11. So it's 1/6 for the first, 1/11 or 2/11 for the second, 0/10 or 1/10 or 2/10 for the third and 0/9 or 1/9 or 2/9 for the fourth.

I reckon that's a 0.5141414r chance of getting one blue out of the twelve with four picks. to get two blues i would guess the probability is half that = 0.2570707r ie 25.7%

So now you have four balls. the probability of selecting a blue first is 1/4 x 25.7% = 6.425%. The second is either 0/3 x 0.257070 (if the blue ball has already been chosen) or 1/3 x 0.257070 the average of which is 1/6 x 0.257070 = 0.042845 = 4.2845%. Combined probability of selecting two blue balls = 10.71%
Probably a load of rubbish but it passed 10 mins !
 
Morgana said:
The chance of picking 2 blue balls in 4 attempts is 2/33
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Thats the bit I can't get. There are 495 possible combinations of 4 balls from a set of 12, of these there are six ways of getting 2 blue balls so I get 6/495 or 2/165 instead of 2/33. All work has ceased here now so you'll have to tell me what I'm doing wrong!
 
Morgana said:
These are two events that have dependency.... the key to understanding statistics... the probability of selecting two balls in the second round depends entirely on the likelihood of there being 2 blue balls in there to select to start with.

The chance of picking 2 blue balls in 4 attempts is 2/33

so if there are 2 blue balls in there, the probability of selecting 2 at the 2nd round is 1/6

1/6 x 2/33 is 1/99
Click to expand...

I agree with the second stage being 1/6 = (2/4 x 1/3)

But I still cant get teh first stage to be 2/33 I get 1/11 Is it me just being thick?

S
 
I concur with 1/11 for problem 1.

Problem 2 I get to be 1/28 using DJE's working at #6 but doubling it as the brothers can be in two sequences.
 
rallyveteran said:
I concur with 1/11 for problem 1.

Problem 2 I get to be 1/28 using DJE's working at #6 but doubling it as the brothers can be in two sequences.
Click to expand...
On reflection I agree as the chance is 2/7 for the first brother not 1/7 as I had used earlier. This is why I never liked stats.
 
Sorry... nipped out for tea....

I've gone through my calcs again quickly, and I can't get back to my original answer... will have to try again later!

The problem however is solved using conditional probability, not independent probability which is where all your confusion lays...
 
Can anyone explain how taking 4 balls out then taking 2 of those is different to just taking 2 from the original 12?

I always struggled with probability at school, perhaps because we had to use roman numerals that long ago.
 
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