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?
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.
Let's solve this step-by-step.
Step 1: The total number of counters in the bag is n.
Step 2: There is only one blue counter and the rest are red. So, there are (n - 1) red counters in the bag.
Step 3: When two counters are taken at random from the bag, the total number of possible outcomes is given by the formula nC2, which represents the number of ways to choose 2 counters from n.
Step 4: The probability of selecting one blue counter and one red counter can be calculated as follows: (1 blue counter / n total counters) * ((n - 1) red counters / (n - 1) total counters) = 1/n * (n - 1)/(n - 1) = 1/n.
Step 5: Similarly, the probability of selecting one red counter and one blue counter is also 1/n.
Step 6: Since the problem states that the probability of selecting one counter of each color is 0.125, we can set up the equation:
(1/n) + (1/n) = 0.125
Step 7: Simplify the equation:
2/n = 0.125
Step 8: Multiply both sides of the equation by n:
2 = 0.125n
Step 9: Divide both sides of the equation by 0.125:
n = 2 / 0.125 = 16
Therefore, there are 16 red counters in the bag.
To solve this problem, we need to use the concept of probability and combinations.
Let's first consider the case where there are n counters in the bag, with one blue counter and the rest red. When picking two counters at random, there are a total of n choose 2 ways to do this, which is given by the combination formula:
nC2 = n! / (2! * (n-2)!)
Now, to calculate the probability of choosing one counter of each color, we need to determine the number of successful outcomes and divide it by the total number of possible outcomes.
The number of successful outcomes is simply 1, since we have only one blue counter in the bag.
The total number of possible outcomes is nC2, which we have already calculated.
So, the probability of choosing one counter of each color can be written as:
1 / (nC2) = 0.125
To find the number of red counters (n) in the bag, we can rearrange the equation:
1 = 0.125 * (nC2)
Now, let's solve this equation for n:
nC2 = n! / (2! * (n-2)!) = n * (n-1) / 2
0.125 * n * (n-1) / 2 = 1
0.0625 * n * (n-1) = 1
n * (n-1) = 16
n^2 - n - 16 = 0
Using the quadratic formula, we can solve for n:
n = (-(-1) ± √((-1)^2 - 4(1)(-16))) / (2(1))
n = (1 ± √(1 + 64)) / 2
n = (1 ± √65) / 2
Since the number of counters cannot be negative, we discard the negative solution:
n = (1 + √65) / 2
So, the number of red counters in the bag is given by (1 + √65) / 2.