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.
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?
1 answer