2011 HSC Q8(b) Probability (1 Viewer)

[goobi]

So the following is the question that I'm stuck with:

A bag contains seven balls numbered from 1 to 7. A ball is chosen at random and its number is noted. The ball is then returned to the bag. This is done a total of seven times.

(iii) What is the probability that exactly one of the balls is not selected?

And this is the sample solution provided by the BOS:



However, I don't quite understand the third line, "The ball can selected twice could be selected in 7C2 ways." Could anyone please explain this to me?

Thanks in advance for any help.

 

[barbernator]

hmm

 

[Nooblet94]

goobi pls....


Seriously though, I really need to brush up on my probability/perms & combs.

 

[deswa1]

Ok this is how I would do this question.

Consider 7 spaces as follows:
_ _ _ _ _ _ _

and we have 7 balls numbered 1 through 7. We need to make sure that ONE of these numbers does not appear in any of our seven spaces so this obviously implies that we have to use ONE number twice and twice only. Ok, so we can fix which number won't appear in seven different ways. We can pick which number will appear twice now in 6 different ways. Now we have 7 spaces in which to put 7 numbers, two spaces of which have the same number. We can choose two spaces out of seven in 7C2 ways so we can place the 'double' number in those two spaces. Now we have five spaces left for 5 numbers which can be chosen (arranged) in 5! ways so total ways are 7x6x7C2x5!. Then just divide by total arrangements

 

[asianese]

I'm so screwed if there is a harder prob/perm/comb in this years exam T__T.

Also for the '7x6' bit, is it just 7 balls to select from, and then 6 balls to select from?

 

[deswa1]

I'm so screwed if there is a harder prob/perm/comb in this years exam T__T.

Also for the '7x6' bit, is it just 7 balls to select from, and then 6 balls to select from?

Yeah same, I SO hope we just get a standard one early on in the paper rather than a Q16.

Basically we need to choose one ball which won't appear and another ball that will appear twice. These balls can be chosen in 7x6 different ways

 

[asianese]

Yep...Either standard early one, or a MC. If it's MC it's only 1 mark lost - as opposed to a string of 3-5 (or more) marks gone lol

 

[someth1ng]

I'm opposite to you guys, I wouldn't mind a whole section (even Q16) on probability alone LOL

 

[goobi]

Ok this is how I would do this question.

Consider 7 spaces as follows:
_ _ _ _ _ _ _

and we have 7 balls numbered 1 through 7. We need to make sure that ONE of these numbers does not appear in any of our seven spaces so this obviously implies that we have to use ONE number twice and twice only. Ok, so we can fix which number won't appear in seven different ways. We can pick which number will appear twice now in 6 different ways. Now we have 7 spaces in which to put 7 numbers, two spaces of which have the same number. We can choose two spaces out of seven in 7C2 ways so we can place the 'double' number in those two spaces. Now we have five spaces left for 5 numbers which can be chosen (arranged) in 5! ways so total ways are 7x6x7C2x5!. Then just divide by total arrangements

I got it now thanks heaps