Maths - Probability Questions

choppy said:
Fooled by randomness worth a read
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Oh yes.

You're at a magic show and the magician shuffles a pack of cards (excluding jokers) and asks a member of the audience to name a card.

What's the probability of the named card being at the top of the pack?
 
I am a maths teacher, I'm also extremely tired and brain fogged, so here goes.
stevebirch2002 said:
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?
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It is 1/11 for the first 4 having two blues, and 1/6 finally. The picking 4 and then picking 2 is a red herring, if you calculate it in two stages you get the same result if you hadn't, 1/66. You end up with 2 balls from 12 whatever the process and no bias is introduced in either stage.

Your book has a typo, extremely common in maths text books. Imagine some data entry person typing in reams of numbers. Imagine an author checking his proofs. Yeah really.


stevebirch2002 said:
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?
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Trevor has to sit in a seat not at the edge, so 6/8 before we even begin, then 2/7 chance on his right and then 1/6 for his left (or vice versa), multiplying gives 12/336 or 1/28.

You don't need to double as the 2 in 7 was for either brother.

Sorry it's too late though.
 
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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.



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
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I think it's a lot more probable that Trev would choose to sit next to his friends than his brothers....
depends on the film!
 
DJE said:
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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First, spot on, good question, logic over formulae. Second is fiendish. There are eight seats. There are only seven sitters. There is one empty seat out of the eight. Chance NOT in edge seat = 6/8. Chance brother first side = 2/7. Chance other brother other side = 1/6. 6/8x2/7x1/6 =12/306.
 
I am struggling to imagine how the answer to the first one can possibly be 99, because a quick look at any of the old betting odds permutation tables will confirm that any 4 from 12 is 495 bets. A simple calculation will show that if you were to number the 12 balls and call the blue ones 1 and 2 they would appear together 45 times in those 495 groups of four. 495 divided by 45 equals 11. The chances of picking the two blue balls out of four are 1 in 6 and 6 times 11 = 66, Surprise, surprise the same betting permutation table will reveal that a perm of any 2 from 12 requires 66 bets.

Useful permutation table here http://www.systembetting.co.uk/calculator/sysbperm.htm
 
stevebirch2002 said:
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
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Using the method on that website


"Single Events
Example
There are 6 beads in a bag, 3 are red, 2 are yellow and 1 is blue. What is the probability of picking a yellow?

The probability is the number of yellows in the bag divided by the total number of balls, i.e. 2/6 = 1/3."



(1) Would be 6 divided by 2 = 3:1
(2) Would be 8 divided by 3 = 8/3 :1
 
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