Question

Meghan has a jar containing 15 counters. There are only blue counters, green counters and red counters in the jar.

Hector is going to take at random one of the counters from his bag of 12 counters. He will look at the counter and put the counter back into the bag.

Hector is then going to take at random a second counter from his bag. He will look at the counter and put the counter back into the bag.

Meghan is then going to take at random one of the counters from her jar of counters. She will look at the counter and put the counter back into the jar.

The probability that the 3 counters each have a different colour is 7/24

(c) Work out how many blue counters there are in the jar.​

Answer 1

Let the number of blue counters be x. Then the number of green counters and red counters combined is (15-x).

The probability that the first two counters Hector draws are different colors is 1 - (the probability that the first two counters are the same color).

The probability that the first two counters Hector draws are the same color is:

(number of counters of that color in the bag / total number of counters in the bag) * (number of counters of that color in the bag - 1 / total number of counters in the bag - 1)

There are three cases to consider:

1. The first two counters are both blue:

(x/15) * ((x-1)/(15-1))

2. The first two counters are both green:

((15-x)/15) * ((15-x-1)/(15-1))

3. The first two counters are both red:

((15-x)/15) * ((15-x-1)/(15-1))

So the probability that the first two counters are the same color is:

(x/15) * ((x-1)/(14)) + ((15-x)/15) * ((14-x)/(14))

Simplifying:

= (x^2 - x + 210 - 15x^2 + 15x) / (15 * 14)

= (14x^2 + 210) / 210

= (2x^2 + 30) / 30

= (x^2 + 15) / 15

The probability that the three counters each have a different color is 7/24, so:

Probability (three different colors) = Probability (first two are different colors) * Probability (third is a different color)

= (1 - (x^2 + 15) / 15) * (x/15 + (15-x)/15 * x/14)

= [(225 - x^2 - 15) / 225] * [(15x + 210 - x^2) / 210]

= [(210 - x^2) / 225] * [(15x + 210 - x^2) / 210]

= (14 - x^2) / 15 * (3x + 42 - x^2) / 14

= (2 * 7 - x^2/15) * (3x + 42 - x^2) / 14

= (14 - x^2/15) * (3x + 42 - x^2) / 14 = 7/24

Multiplying both sides by 14:

(14 - x^2/15) * (3x + 42 - x^2) = 7/24 * 14

(14 - x^2/15) * (3x + 42 - x^2) = 49/12

Expanding:

42 - 3x^2/15 + 42x/15 - x^2 + x^3/15 = 49/12

Simplifying:

4x^3 - 36x^2 + 168x - 280 = 0

Dividing both sides by 4:

x^3 - 9x^2 + 42x - 70 = 0

By trying integer values for x, we find that x=5 is a solution to the equation.

Therefore, there are 5 blue counters in the jar.