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Hi juicy probability question ;)

A bag contains n coloured counters.
Olga picks a counter at random from the bag.
The probability that the counter is green is ½.
Olga does not replace the first counter.
She then picks another counter at random from the bag.
The probability that both counters are green is 3/13.
a) Show that n-2/4(n-1) = 3/13
b) Hence or otherwise find the value of n.

a) I found out that the second chance of picking out a green counter must be 6/13
b) I got n = 14 =D I know that is right but I didn’t understand what to do with a...=S

Thanks!:cool:
Reply 1
If P(first counter is green) is 1/2 then there must be n/2 green counters in the bag initially.

Once the first counter has been removed, there would be n/2 - 1 green counters and n-1 counters in total in the bag. So, the probability of a second green being removed would be (n/2 - 1)/(n-1)

Therefore, the probability of both counters being removed being green would be 12×n/21n1=n/212(n1)=n24(n1)\dfrac{1}{2} \times \dfrac{n/2 - 1}{n-1} = \dfrac{n/2 - 1}{2(n-1)} = \dfrac{n-2}{4(n-1)}

However, we know that this probability is 3/13. This tells us that n24(n1)=313\dfrac{n-2}{4(n-1)} = \dfrac{3}{13} as required.
Reply 2
a) you have n counters. chance is 1/2 so you have a chance of (n/2)/n. then you only have n-1 left. so you have [(n/2)-1]/n-1

multiply these 2 and then cancelling this gives the result in a)

b) 14 is right.
Reply 3
*Faiqah*
A bag contains n coloured counters.
Olga picks a counter at random from the bag.
The probability that the counter is green is ½.
Olga does not replace the first counter.
She then picks another counter at random from the bag.
The probability that both counters are green is 3/13.
a) Show that n-2/4(n-1) = 3/13
b) Hence or otherwise find the value of n.

a) I found out that the second chance of picking out a green counter must be 6/13
b) I got n = 14 =D I know that is right but I didn’t understand what to do with a...=S

Thanks!:cool:

i sent a message :biggrin: ~
Reply 4
lovmaths
i sent a message :biggrin: ~


Thanks! :biggrin:
Reply 5
ttoby
If P(first counter is green) is 1/2 then there must be n/2 green counters in the bag initially.

Once the first counter has been removed, there would be n/2 - 1 green counters and n-1 counters in total in the bag. So, the probability of a second green being removed would be (n/2 - 1)/(n-1)

Therefore, the probability of both counters being removed being green would be 12×n/21n1=n/212(n1)=n24(n1)\dfrac{1}{2} \times \dfrac{n/2 - 1}{n-1} = \dfrac{n/2 - 1}{2(n-1)} = \dfrac{n-2}{4(n-1)}

However, we know that this probability is 3/13. This tells us that n24(n1)=313\dfrac{n-2}{4(n-1)} = \dfrac{3}{13} as required.


Thank you :cool:
Reply 6
danny111
a) you have n counters. chance is 1/2 so you have a chance of (n/2)/n. then you only have n-1 left. so you have [(n/2)-1]/n-1

multiply these 2 and then cancelling this gives the result in a)

b) 14 is right.


Thank you!

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