Question
A bag contains n counters
One counter is blue and the rest are red.
Two counter are taken at random from the bag.
Express in terms of n, the probability a counter of each colour is taken. Give your answer as a fraction in its simplest form.
The probability that a counter of each colour is taken is 0.125. How many red counters in the bag?
One counter is blue and the rest are red.
Two counter are taken at random from the bag.
Express in terms of n, the probability a counter of each colour is taken. Give your answer as a fraction in its simplest form.
The probability that a counter of each colour is taken is 0.125. How many red counters in the bag?
Answers
GPT 3.5
Let's say there are r red counters in the bag.
The probability of selecting a blue counter on the first draw is 1/n since there is only 1 blue counter out of n counters in total. The probability of selecting a red counter on the first draw is (n-1)/n since there are (n-1) red counters out of n counters in total.
After the first draw, there are (r-1) red counters remaining in the bag out of (n-1) counters in total. So the probability of selecting a red counter on the second draw is (r-1)/(n-1). The probability of selecting a blue counter on the second draw is 1/(n-1) since there is only 1 blue counter remaining out of (n-1) counters in total.
According to the given information, the probability of selecting a counter of each color is 0.125. So:
(1/n) * (r-1)/(n-1) = 0.125
Multiplying both sides by n and (n-1):
(r-1) = 0.125 * n * (n-1)
Expanding and simplifying:
r - 1 = 0.125n^2 - 0.125n
0.125n^2 - 0.125n - r + 1 = 0
Multiply both sides by 8 to eliminate the decimal:
n^2 - n - 8r + 8 = 0
We need to find values of r and n that satisfy this equation and also make r/(n-1) = 1/7. This means we need to find values of r and n such that after simplifying the fraction r/(n-1), its numerator (r) is divisible by 7.
Checking through values of r and n, we find that when r = 4 and n = 9, the numerator of the fraction r/(n-1) is divisible by 7: 4/(9-1) = 4/8 = 1/2 = 0.125.
Therefore, there are 4 red counters in the bag.
The probability of selecting a blue counter on the first draw is 1/n since there is only 1 blue counter out of n counters in total. The probability of selecting a red counter on the first draw is (n-1)/n since there are (n-1) red counters out of n counters in total.
After the first draw, there are (r-1) red counters remaining in the bag out of (n-1) counters in total. So the probability of selecting a red counter on the second draw is (r-1)/(n-1). The probability of selecting a blue counter on the second draw is 1/(n-1) since there is only 1 blue counter remaining out of (n-1) counters in total.
According to the given information, the probability of selecting a counter of each color is 0.125. So:
(1/n) * (r-1)/(n-1) = 0.125
Multiplying both sides by n and (n-1):
(r-1) = 0.125 * n * (n-1)
Expanding and simplifying:
r - 1 = 0.125n^2 - 0.125n
0.125n^2 - 0.125n - r + 1 = 0
Multiply both sides by 8 to eliminate the decimal:
n^2 - n - 8r + 8 = 0
We need to find values of r and n that satisfy this equation and also make r/(n-1) = 1/7. This means we need to find values of r and n such that after simplifying the fraction r/(n-1), its numerator (r) is divisible by 7.
Checking through values of r and n, we find that when r = 4 and n = 9, the numerator of the fraction r/(n-1) is divisible by 7: 4/(9-1) = 4/8 = 1/2 = 0.125.
Therefore, there are 4 red counters in the bag.