a. 1 in 36
b. 1 in 6
d. 2 in 6
c. 1 in 9
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 !!
Eoink,
Just to nit-pick … I reduced [E] to 5 in 18, having already got 5/36 + 5/36 …
I did think were are two 5:5s.
I should have put 20/72 to leave you wondering how on earth I’d solved it.
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).
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.
Note to myself … Must learn to type faster ! (or set harder brain teasers )
Well done Eoink.
“Identical” did imply it, but I just wanted to be explicit in the assumption.