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 Post subject: CEM probabilityPosted: Tue Feb 11, 2014 6:35 pm

Joined: Sat Apr 14, 2012 9:19 am
Posts: 32
Just browsing through the elevenplusexams Numerical Reasoning book 2 and came a cross a question that involves knowing how to work out the probability of two different outcomes by multiplying fractions. Page 4 no. 3d. Am I wrong in thinking this is closer to a higher level 6 question than anything covered at level 5?

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 Post subject: Re: CEM probabilityPosted: Tue Feb 11, 2014 11:54 pm

Joined: Mon Aug 12, 2013 9:13 am
Posts: 452
Probability questions appear in 11 plus (dice, cards and balls etc).

I think the one you are mentioning is AND type i.e. where one should multiply the probability of two events to happen simultaneously.

Please post the question as that may help.

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 Post subject: Re: CEM probabilityPosted: Wed Feb 12, 2014 9:34 am

Joined: Sat Apr 14, 2012 9:19 am
Posts: 32
There is a diagram showing a selection of 12 balls. Six are black, 4 white and 2 spotted.
Question: What is the probability of picking one black ball and one white ball on the first two rounds, before both are replaced?
This requires that the children not only know 1/2 (6/2) and 4/11 for the two probabilities, but also that to find the probability of both happening you should multiply 1/2 x 4/11 to reach find 4/22 simplified to 2/11. It seems to me that although the children could be taught to do this, the underlying concept must be very advanced. National Curriculum level? Of course this could just be my dimness.

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 Post subject: Re: CEM probabilityPosted: Wed Feb 12, 2014 9:50 am

Joined: Tue Jul 21, 2009 9:56 pm
Posts: 8228
I agree. But I don't know whether it would come up or not. For a child who hasn't been taught this to think it up from scratch in exam time would be remarkable.

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 Post subject: Re: CEM probabilityPosted: Wed Feb 12, 2014 12:41 pm

Joined: Mon Feb 14, 2011 1:42 pm
Posts: 989
Location: Birmingham
Isn't the probability of the white balls 4/12 (not 11), meaning it is the same as 1/3.

So they need to multiply 1/2 by 1/3 which makes it 1/6??

I might be wrong but if that is the case then I think it can be managed without knowing about multiplying fractions.

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 Post subject: Re: CEM probabilityPosted: Wed Feb 12, 2014 12:58 pm

Joined: Wed Jan 18, 2012 11:41 am
Posts: 4599
Location: Essex
No, the probability of picking the white ball in the example as stated is 4/11 not 4/12 - the balls are picked out in separate rounds before being replaced, i.e. you take one ball (probability of black 1/2), then you take another ball without putting the first one back, so there are only 11 balls available at that point.

Unless the question is mis-worded, it does look a bit advanced for the average 11+ paper, even as a 'making use of what you have learnt to figure out a more complex problem' type of question.

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 Post subject: Re: CEM probabilityPosted: Wed Feb 12, 2014 1:02 pm

Joined: Fri Oct 11, 2013 8:55 am
Posts: 500
The probability of the white ball is 4/11 because one black ball has already been taken out leaving 5 black, 4 white and 2 spotted (total 11) when the white ball is picked.

I think this is a difficult question and am surprised that its aimed at 10 year olds. Its more GCSE level IMO.

All the 11plus probability questions I've seen tend to focus on just a single event. e.g. using the same example I would expect to see a question that just asks "What is the probability of choosing a white ball", or similar.

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 Post subject: Re: CEM probabilityPosted: Wed Feb 12, 2014 1:04 pm

Joined: Fri Oct 11, 2013 8:55 am
Posts: 500

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 Post subject: Re: CEM probabilityPosted: Wed Feb 12, 2014 1:29 pm

Joined: Fri Oct 11, 2013 8:55 am
Posts: 500
thisandthat3 wrote:
There is a diagram showing a selection of 12 balls. Six are black, 4 white and 2 spotted.
Question: What is the probability of picking one black ball and one white ball on the first two rounds, before both are replaced?

I've had a re-think and decided that this question is even MORE difficult than I first thought.

The question as stated does not seem to insist that the black ball is chosen first and the white second. They could be picked in either order so long as the end result is one black and one white on the first 2 rounds.

So the probability would be (black,white) + (white, black)

= (6/12 x 4/11) + (4/12 x 6/11)
= 2/11 + 2/11
= 4/11

Does anyone else agree?

Surely you wouldn't expect a 10 year old to work that out?!

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 Post subject: Re: CEM probabilityPosted: Wed Feb 12, 2014 2:10 pm

Joined: Mon Feb 14, 2011 1:42 pm
Posts: 989
Location: Birmingham
Yes, thanks ToadMum and Proud_Dad. In that case my reply was completely useless ! Drat! I thought I could venture into the maths section and actually know how to answer the question for once.

I would say that is beyond the average 10/11 year old and higher than the average CEM exam too. I did some level 6 questions with DS1 and DS2 but more to show them that they might need to push what they already know to a higher level rather than a completely new approach that these questions demand.

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