there are n counters in a bag. 4 of the counters are red and the rest are blue. Ross takes a counter from the bag at random and does not replace it. he then takes another counter at random from the bag. the probability Ross takes two blue counters is 1/3. find the value of n​

n = 10.

Step-by-step explanation:

In the bag, we have n counters.

4 of the counters are red.

the rest are blue, then we have:

(n - 4) blue counters.

Now, the probability that Ross takes a blue counter from the bag is equal to the quotient between the number of blue counters (n - 4) and the total number of counters, n

Then the probability is:

p1 = (n - 4)/n

Now he draws another, and it must be blue again, then we can calculate the probability in the same way as above, but he already take a blue counter, so the number of blue counters is (n - 5) and the total number of counters is (n - 1)

The probability of this event is:

p2 = (n - 5)/(n - 1)

The joint probability (the probability that Ross takes two blue counters) is equal to the product of the individual probabilities, and we know that this is equal to 1/3, then we have the equation:

1/3 = ( (n - 4)/n)*((n - 5)/(n - 1))

Now let's solve this for n.

n*(n - 1)/3 = (n - 4)*(n - 5)

(n^2 - n)/3 = n^2 - 4*n - 5*n + 20

n^2 - n = 3*(n^2 - 9*n + 20)

n^2 - n = 3*n^2 - 27*n + 60

0 = (3*n^2 - n^2) - 27*n + n + 60

0 = 2*n^2 - 26*n + 60

The two solutions of this equation can be found with Bhaskara's equation:

Then the two solutions are:

n = (26 - 14)/4 = 3

This is not an option, because we know for sure that we have 4 red counters, then this option can be discarded.

The other solution is:

n = (26 + 14)/4 = 40/4 = 10

Then we have n = 10, 10 counters in total.

3 is greater then both of those numbers