Green, blue pens in box; 3 more blue than green; more than 12 pens in box. Simon takes 2 pens from box.

pjmcdonnell

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Mar 13, 2019
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Hello, I'm new to the site so sorry if I've done anything wrong😅 I created an account just for this problem because it was bugging me so much. Yes I know it probably has a very simple solution, but it was the last question on one of my GCSE mock exams so I'd like to make sure.

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There are only green pens and blue pens in a box.

There are three more blue pens than green pens in the box.
There are more than 12 pens in the box.

Simon is going to take at random two pens from the box.

The probability that Simon will take two pens of the same colour is 27/55.

Work out the number of green pens in the box.



HELP
 

tkhunny

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Rule #1 - If you don't know what it is, give it a name so we can talk about it.

G = # of Green Pens in the Box.

What's next?

Note: This may strike you as an unusual clue: "There are more than 12 pens in the box." This says to me that your solution is likely to lead you to a quadratic equation of some sort. You may get two possible solutions. One will be 12 or less than 12 and the other will be greater than 12. Pick the right one.
 
Last edited:

pka

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Hello, I'm new to the site so sorry if I've done anything wrong😅 I created an account just for this problem because it was bugging me so much. Yes I know it probably has a very simple solution, but it was the last question on one of my GCSE mock exams so I'd like to make sure.
Do you know the binomial notation \(\displaystyle \binom{G+3}{2}=\dfrac{(G+3)(G+2)}{2\cdot 1}\;?\)

If you do then tell us what this is all about: \(\displaystyle \dfrac{\dbinom{G+3}{2}+\dbinom{G}{2}}{\dbinom{2G+3}{2}}=\dfrac{27}{55}\)
 

freyatown

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I'm also confused by this question. Please can you explain it further?
 

pka

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\(\displaystyle \dfrac{\dbinom{G+3}{2}+\dbinom{G}{2}}{\dbinom{2G+3}{2}}=\dfrac{27}{55}\)
I'm also confused by this question. Please can you explain it further?
The problem is about picking two pens of the same colour. Two blues or two greens. But it is given that "There are three more blue pens than green pens in the box" so \(\displaystyle B=G+3\) Thus there are \(\displaystyle \dbinom{G+3}{2}+\dbinom{G}{2}\) ways to pick two pens of the same colour, either two blue or two green. That also means we have a total of \(\displaystyle 2G+3\) pens in the box.
Lets look at the probability: \(\displaystyle \dfrac{\dbinom{G+3}{2}+\dbinom{G}{2}}{\dbinom{2G+3}{2}}=\dfrac{27}{55}\)

\(\displaystyle \dfrac{\dbinom{G+3}{2}+\dbinom{G}{2}}{\dbinom{2G+3}{2}}=\dfrac{\dfrac{(G+3)(G+2)}{2}+\dfrac{G(G-1)}{2}}{\dfrac{(2G+3)(2G+2)}{2}}=\dfrac{27}{55}\)
Simple algebra should let one solve for \(\displaystyle G\) But remember that \(\displaystyle G+B>12\).

 
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