pheasantchick wrote:1) 4440
2) 2/9 chance
3) 2/9 chance
Nb. Hated probablility at school, so those answers are probably wrong.)
4440 is correct. The missing digits must end in 0 or 5, be divisible by four and add up, with the 4 and 4 you are given, to a number divisible by three. The first two criteria limit it to 20, 40, 60 or 80, and the third criterion limits it to 40 (4+4+4+0 is 12, which is divisible by 3).
The proof of why "sum of the digits divisible by three" is equivalent to "number is divisible by three" is very nice, by the way, although I suspect it's not straightforward even at GCSE (it relies on the fact that 10^x is congruent to 1 modulo 3 for all non-negative integers x).
You're correct in thinking your probability answers are wrong. Those are the chances of drawing a white or a red bean in one trial. One problem with those probability questions is the issue of replacement (if I pull out a ball other than the colour I'm waiting for, do I replace it?) Let's assume you do replace it, as otherwise the answers are self-evidently "1" --- after nine trials, the bag is empty and you have all the beans in your hand. Another problem is that they don't state whether you're after the chance of drawing a white (or red) bean exactly once, or whether in ten (or twenty) trials you draw at least one bean of the right colour.
So let's assume the question is "I've got a bag with nine beans in it, two of them white, and I make ten attempts in which I draw out a bean and then replace it. What is the chance that at least one of those trials draws a white bean?" It's obviously more than 2/9 --- you're making ten attempts. And it's obviously less than one --- you might never draw a white bean. It isn't 2/9 x 2/9 x 2/9... 10 times (ie (2/9)^10), because that's the chance of drawing a white bean every time. And it isn't 2/9 + 2/9 ... ten times, because you can't add independent trials like that (it's almost, but not quite, a version of
the gambler's fallacy).
In fact, allowing for all the options (do I draw the white bean once, twice, three times, every time?) the answer is 1-(7/9)^10, or about 92%. That's worked out by saying that the chance of _not_ drawing a white bean in any one trial is 7/9, so the chance of not drawing a white bean in all ten trials is (7/9)^10, or about 8%. So the chance of that not happening, or the chance that at least one trial yields a white bean, is one minus that, or 92%.
For twenty trials, it's 99.3%.
This is asymptotic to one (ie, as you make more trials, the chances get closer and closer to one, but never reach it). Those of a literary bent will immediately think of the first scene of Rosencrantz and Guildenstern are Dead. So if you drew a bean out of a bag containing nine beans, two of them white, fifty times, you'd have a 99.999651% chance of getting a white bean at least once. In other words, if a million people each performed this fifty-drawing experiment, you would expect about three of them to not get a white bean in any of those fifty drawings. Which is not zero (the odds of it happening are a lot better than winning the lottery, for example) but is quite a small number.
Unless there's something missing for the question, or it's a multiple choice question where all the answers bar one are obviously wrong, this is quite clearly not 11+ material.