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There are n beads in a bag - Probability and Quadratics

This is a question from the Edexcel 2002 GCSE Calculator Paper and it has got me quite confused.

There are n beads in a bag.
6 beads are black and all the rest are white.
Heather picks one bead at random from the bag and does not replace it.
She picks a second bead at random from the bag.
The probability that she will pick 2 white beads is 0.5.

Show that n²-25n+84=0

Any idea how to solve it??

I've worked out that 6/n are black and (n-6)/n are white.
Therefore with a tree diagram

1st Pick

6/n - Black
(n-6)/n - White

2nd Pick

5/(n-1)Black (after a black has been picked)
(n-6)/(n-1)White (after a black has been picked)

6/(n-1)Black (after a white has been picked)
(n-6)/(n-1)White (after a white has been picked)

And I believe that:

(n-6)²/n*(n-1)=0.5

But after that, I'm lost.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: There are n beads in a bag - Probability and Quadratics

Nearly all correct so far, but the second probability for white should be (n-7)/(n-1), because one white has already been removed. So you have (n-6)(n-7)/n(n-1) = 0.5.
From there you just manipulate your equation to get it into standard quadratic form.

(n-6)(n-7)/n(n-1)=0.5

(n-6)(n-7) = 0.5n(n-1)

n²-13n+42 = 0.5n²-0.5n

0.5n² - 12.5n + 42 = 0

n² - 25n + 84 = 0

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Why did the vector cross the road?
It wanted to be normal.