Brain Teasers are Back!

a. 1 in 36
b. 1 in 6
d. 2 in 6
c. 1 in 9

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Hi Dozey,

You got off to a brilliant start with a) and b). Well done.

However, you might like to take another look at c) and d).

I notice that your d) could be reduced to 1 in 3 (but neither 1 in 3 nor 2 in 6 accords with my result)

As for your c) at 1 in 9, well again, that doesn’t accord with my result. Your result suggests that you can see four ways of generating 10. For a while, this did make me wonder if my solution is correct. But i’m pretty sure it is.

C) 1 in 12
D) 11/36
E) 10/36

Usual disclaimer, probability and I aren’t a natural mix.

No need for the disclaimer Eoink. Well done.

Dozey’s shot at 1 in 3 for answer (d) was pretty close to the 11/36

The 1 in 12 for answer © is correct. there are only three ways of generating 10. These are 4;6 or 5;5 or 6;4. For a moment or two after seeing Dozey’s answer, I began to wonder if 5;5 could possibly be generated in more ways than one. But only for a moment. Nevertheless, it did make me think. Thank you Dozey !! :grinning:

Eoink,

Just to nit-pick … I reduced [E] to 5 in 18, having already got 5/36 + 5/36 …

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I did think were are two 5:5s.

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I should have put 20/72 to leave you wondering how on earth I’d solved it. :grinning:

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I had to re-write a few pairs just to convince myself there was only one 5:5 pair. Certainly made me think !

Sticking with the gambling …

I have a bag containing six red balls and four black balls. I take four balls out at random, without replacement.

What’s the probability that I will remove at least two red balls ?

I first used the Tree Diagram approach to sort out the 6 red ball/4 black ball problem above. Mrs D could understand it. But it wasn’t pretty.

I then re-did it using combinations. That was pretty, but Mrs D couldn’t understand it.

Here is my probability tree.

Not pretty, I know. But at least Mrs D could understand it.

It needs a bit of (tedious) arithmetic to work out the probability of “at least two red balls”. ie the outcomes that I have circled. I actually worked out the probability of “less than two red balls” ie the outcomes that are not circled and subtracted the result from 1. (it was a bit less tedious).

Of course, with a probability tree, you can fairly easily work out a wide variety of other probabilities such as 4 x red balls or 4 x black balls.

For the chocaholics amongst us !

Barry was given a box containing 125 bars of chocolate. On the wrapper of each bar there was a token, and Barry could exchange 5 tokens at the local shop for an identical bar of chocolate. How many extra bars did he get ?

31 (if the identical bars have the tokens).

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Back to gambling again !

A bag contains a well-mixed large number of discs, some red, some yellow (but no other colours). I need to take some discs out of the bag without looking. How many do I need to remove to be certain that I have four discs of the same colour ?

7

Indeed, the identical bars do have the tokens. I should have said that !

Nicely done.

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Note to myself … Must learn to type faster ! (or set harder brain teasers :sunglasses:)

Well done Eoink.

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“Identical” did imply it, but I just wanted to be explicit in the assumption.

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Start from the * and going horizontally or vertically (not diagonally) spell out the names of seven, plane, geometrical figures.