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
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 21×n−1n/2−1=2(n−1)n/2−1=4(n−1)n−2
However, we know that this probability is 3/13. This tells us that 4(n−1)n−2=133 as required.
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
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 21×n−1n/2−1=2(n−1)n/2−1=4(n−1)n−2
However, we know that this probability is 3/13. This tells us that 4(n−1)n−2=133 as required.